11th Physics Chapter 1 Exercise Units and Measurements Solutions Maharashtra Board

Units and Measurements Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 1 Units and Measurements Textbook Exercise Questions and Answers.

Class 11 Physics Chapter 1 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 1 Exercise Solutions

1. Choose the correct option.

Question 1.
[L1M1T-2] is the dimensional formula for
(A) Velocity
(B) Acceleration
(C) Force
(D) Work
Answer:
(C) Force

Question 2.
The error in the measurement of the sides of a rectangle is 1%. The error in the measurement of its area is
(A) 1%
(B) \(\frac{1}{2}\)%
(C) 2%
(D) None of the above.
Answer:
(C) 2%

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 3.
Light year is a unit of
(A) Time
(B) Mass
(C) Distance
(D) Luminosity
Answer:
(C) Distance

Question 4.
Dimensions of kinetic energy are the same as that of
(A) Force
(B) Acceleration
(C) Work
(D) Pressure
Answer:
(C) Work

Question 5.
Which of the following is not a fundamental unit?
(A) cm
(B) kg
(C) centigrade
(D) volt
Answer:
(D) volt

2. Answer the following questions.

Question 1.
Star A is farther than star B. Which star will have a large parallax angle?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 1
i). ‘b’ is constant for the two stars
∴ θ ∝ \(\frac{1}{D}\)

ii) As star A is farther i.e., DA > DB
⇒ θA < θB
Hence, star B will have larger parallax angle than star A.

Question 2.
What are the dimensions of the quantity l \(\sqrt{l / g}\), l being the length and g the acceleration due to gravity?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 2
[Note: When power of symbol expressing fundamental quantity appearing in the dimensional formula is not given, ills taken as 1.]

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 3.
Define absolute error, mean absolute error, relative error and percentage error.
Answer:
Absolute error:
a. For a given set of measurements of a quantity, the magnitude of the difference between mean value (Most probable value) and each individual value is called absolute error (∆a) in the measurement of that quantity.
b. absolute error = |mean value – measured value|
∆a1 = |amean – a1|
Similarly, ∆a2 = |amean – a2|
. . . ..
. . . .
. . . .
∆an = |amean – an|

Mean absolute error:
For a given set of measurements of a same quantity the arithmetic mean of all the absolute errors is called mean absolute error in the measurement of that physical quantity.
∆amean = \(\frac{\Delta \mathrm{a}_{1}+\Delta \mathrm{a}_{2}+\ldots \ldots .+\Delta \mathrm{a}_{n}}{\mathrm{n}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{n} \Delta \mathrm{a}_{\mathrm{i}}\)

Relative error:
The ratio of the mean absolute error in the measurement of a physical quantity to its arithmetic mean value is called relative error.
Relative error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\mathrm{mean}}}\)

Percentage error:
The relative error represented by percentage (i.e., multiplied by 100) is called the percentage error.
Percentage error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\mathrm{mean}}}\) × 100%
[Note: Considering conceptual conventions question is modified to define percentage error and not mean percentage error.]

Question 4.
Describe what is meant by significant figures and order of magnitude.
Answer:
Significant figures:

  1. Significant figures in the measured value of a physical quantity is the sum of reliable digits and the first uncertain digit.
    OR
    The number of digits in a measurement about which we are certain, plus one additional digit, the first one about which we are not certain is known as significant figures or significant digits.
  2. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
  3. If one uses the instrument of smaller least count, the number of significant digits increases.

Rules for determining significant figures:

  1. All the non-zero digits are significant, for example if the volume of an object is 178.43 cm3, there are five significant digits which are 1,7,8,4 and 3.
  2. All the zeros between two nonzero digits are significant, eg., m = 165.02 g has 5 significant digits.
  3. If the number is less than 1, the zero/zeroes on the right of the decimal point and to the left of the first nonzero digit are not significant e.g. in 0.001405, significant. Thus the above number has four significant digits.
  4. The zeroes on the right hand side of the last nonzero number are significant (but for this, the number must be written with a decimal point), e.g. 1.500 or 0.01500 both have 4 significant figures each.
    On the contrary, if a measurement yields length L given as L = 125 m = 12500 cm = 125000 mm, it has only three significant digits.

Order of magnitude:
The magnitude of any physical quantity can be expressed as A × 10n where ‘A’ is a number such that 0.5 ≤ A < 5 then, ‘n’ is an integer called the order of magnitude.
Examples:

  1. Speed of light in air = 3 × 108 m/s
    ∴ order of magnitude = 8
  2. Mass of an electron = 9.1 × 10-31 kg
    = 0.91 × 1030 kg
    ∴ order of magnitude = -30

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 5.
If the measured values of two quantities are A ± ∆A and B ± ∆B, ∆A and ∆B being the mean absolute errors. What is the maximum possible error in A ± B? Show that if Z = \(\frac{A}{B}\)
\(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)
Answer:
Maximum possible error in (A ± B) is (∆A + ∆B).
Errors in divisions:
i) A
Suppose, Z = \(\frac{A}{B}\) and measured values of A and B are (A ± ∆A) and (B ± ∆B) then,
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 5
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 6
∴ Maximum relative error of \(\frac{\Delta Z}{Z}=\pm\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}\right)\)

ii) Thus, when two quantities are divided, the maximum relative error in the result is the sum of relative errors in each quantity.

Question 6.
Derive the formula for kinetic energy of a particle having mass m and velocity v using dimensional analysis
Answer:
Kinetic energy of a body depends upon mass (m) and velocity (v) of the body.
Let K.E. ∝ mx vy
∴ K.E. = kmx vy …….. (1)
where,
k = dimensionless constant of proportionality. Taking dimensions on both sides of equation (1),
[L2M1T-2] – [L0M1T0]x [L1M0T-1]y
= [L0MxT0] [LyM0T-y]
= [L0+yMx+0T0-y]
[L2M1T2] = [LyMxT-y] …………. (2)
Equating dimensions of L, M, T on both sides of equation (2),
x = 1 and y = 2 ,
Substituting x, y in equation (1), we have
K.E. = kmv2

3. Solve numerical examples.

Question 1.
The masses of two bodies are measured to be 15.7 ± 0.2 kg and 27.3 ± 0.3 kg. What is the total mass of the two and the error in it?
Answer:
Given: A ± ∆A = 15.7 ± 0.2kg and
B ± ∆B = 27.3 ± 0.3 kg.
To find: Total mass (Z), and total error (∆Z)
Formulae: i. Z = A + B

ii) ±∆Z = ±∆A ± ∆B
Calculation: From formula (i),
Z = 15.7 + 27.3 = 43 kg
From formula (ii),
± ∆Z (± 0.2) + (± 0.3)
=±(0.2 + 0.3)
= ± 0.5 kg
Total mass is 43 kg and total error is ± 0.5 kg.

Question 2.
The distance travelled by an object in time (100 ± 1) s is (5.2 ± 0.1) m. What is the speed and it’s relative error?
Answer:
Given: Distance (D ± ∆D) = (5.2 ± 0.1) m,
time(t ± ∆t) = (100 ± 1)s.
To find: Speed (v), maximum relative error \(\left(\frac{\Delta \mathrm{v}}{\mathrm{v}}\right)\)

Formulae: i. v = \(\frac{\mathrm{D}}{\mathrm{t}}\)
ii. \(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\pm\left(\frac{\Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{t}}{\mathrm{t}}\right)\)

Calculation: From formula (i),
v = \(\frac{5.2}{100}\) = 0.052 m/s
From formula (ii),
\(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\pm\left(\frac{0.1}{5.2}+\frac{1}{100}\right)\)
= \(\pm\left(\frac{1}{52}+\frac{1}{100}\right)=\pm \frac{19}{650}\)
= ± 0.029 rn/s
The speed is 0.052 m/s and its maximum relative error is ± 0.029 m/s.
[Note: Framing of numerical is modified to make it specific and meaningful.]

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 3.
An electron with charge e enters a uniform. magnetic field \(\vec{B}\) with a velocity \(\vec{v}\). The velocity is perpendicular to the magnetic field. The force on the charge e is given by
|\(\vec{F}\)| = Bev Obtain the dimensions of \(\vec{B}\).
Answer:
Given: |\(\vec{F}\)| = B e v
Considering only magnitude, given equation is simplified to,
F = B e v
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 3
∴ B = [L0M1T-2I-1]
[Note: The answer given above is calculated in accordance with textual method considering the given data.]

Question 4.
A large ball 2 m in radius is made up of a rope of square cross section with edge length 4 mm. Neglecting the air gaps in the ball, what is the total length of the rope to the nearest order of magnitude?
Answer:
Volume of ball = Volume enclosed by rope.
\(\frac{4}{3}\) π (radius)3 = Area of cross-section of rope × length of rope.
∴ length of rope l = \(\frac{\frac{4}{3} \pi r^{3}}{A}\)
Given:
r = 2 m and
Area = A = 4 × 4 = 16 mm2
= 16 × 10-6 m2
∴ l = \(\frac{4 \times 3.142 \times 2^{3}}{3 \times 16 \times 10^{-6}}\)
= \(\frac{3.142 \times 2}{3}\) × 10-6 m
≈ 2 × 106 m.
Total length of rope to the nearest order of magnitude = 106 m = 103 km

Question 5.
Nuclear radius R has a dependence on the mass number (A) as R = 1.3 × 10-16 A\(\frac{1}{3}\) m. For a nucleus of mass number A=125, obtain the order of magnitude of R expressed in metre.
Answer:
R= 1.3 × 10-16 × A\(\frac{1}{3}\) m
For A = 125
R= 1.3 × 10-16 × (125)\(\frac{1}{3}\)
= 1.3 × 10-16 × 5
= 6.5 × 10-16
= 0.65 × 10-15 m
∴ Order of magnitude = -15
[Note: Taking the standard value of nuclear radius R = 1.3 × 10-155 m, the order of magnitude comes to be 10-14 m.]

Question 6.
In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measured value of the length.
Answer:
Given: a1 = 3.11 cm, a2 = 3.13 cm,
a3 = 3.14 cm. a4 = 3.14cm
Least count L.C. = 0.01 cm.
To find.
i. Mean length (amean)
ii. Mean absolute error (∆amean)
iii. Percentage error.

Formulae: i. amean = \(\frac{a_{1}+a_{2}+a_{3}+a_{4}}{4}\)
ii. ∆an = |amean – an|
iii. ∆amean = \(\frac{\Delta \mathrm{a}_{1}+\Delta \mathrm{a}_{2}+\Delta \mathrm{a}_{3}+\Delta \mathrm{a}_{4}}{4}\)
iv. Percentage error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\text {mean }}}\) × 100

Calculation: From formula (i),
amean = \(\frac{3.11+3.13+3.14+3.14}{4}\)
= 3.13 cm
From formula (ii),
∆a1 = |3.13 – 3.11| = 0.02 cm
∆a2 = |3.13 – 3.13| = 0
∆a3 = |3.13 – 3.14| = 0.01 cm
∆a4 = |3.13 – 3.14| = 0.01 cm
From formula (iii),
∆amean = \(\frac{0.02+0+0.01+0.01}{4}\) = 0.01 cm
From formula (iii).
% error = \(\frac{0.01}{3.13}\) × 100
= \(\frac{1}{3.13}\)
= 0.3196
……..(using reciprocal table)
= 0.32%

i. Mean length is 3.13 cm.
ii. Mean absolute error is 0.01 cm.
iii. Percentage error is 0.32 %.
[Note: As per given data of numerical, percentage error calculation upon rounding off yields percentage error as 0.32%]

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 7.
Find the percentage error in kinetic energy of a body having mass 60.0 ± 0.3 g moving with a velocity 25.0 ± 0.1 cm/s.
Answer:
Given: m = 60.0 g, v = 25.0 cm/s.
∆m = 0.3 g, ∆v = 0.1 cm/s
To find: Percentage error in E
Formula: Percentage error in E
\(\left(\frac{\Delta \mathrm{m}}{\mathrm{m}}+2 \frac{\Delta \mathrm{v}}{\mathrm{v}}\right)\) × 100%

Calculation: From formula,
Percentage error in E
= \(\left(\frac{0.3}{60.0}+2 \times \frac{0.1}{25.0}\right)\) × 100%
= 1.3%
The percentage error in energy is 1.3%.

Question 8.
In Ohm’s experiments, the values of the unknown resistances were found to be 6.12 Ω , 6.09 Ω, 6.22 Ω, 6.15 Ω. Calculate the mean absolute error, relative error and percentage error in these measurements.
Answer:
Given: a1 = 6.12 Ω, a2 = 6.09 Ω, a3 = 6.22 Ω, a4 = 6.15 Ω,

To find:

i) Absolute error (∆amean)
ii) Relative error
iii) Percentage error

Formulae:

i) amean = \(\frac{a_{1}+a_{2}+a_{3}+a_{4}}{4}\)
ii) ∆an = |amean – an|
iii) ∆amean = \(\frac{\Delta \mathrm{a}_{1}+\Delta \mathrm{a}_{2}+\Delta \mathrm{a}_{3}+\Delta \mathrm{a}_{4}}{4}\)
iv) Percentage error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\text {mean }}}\) × 100

From formula (ii),
∆a1 = |6.145 – 6.12|= 0.025
∆a2 = |6.145 – 6.09| = 0.055
∆a3 = |6.145 – 6.22| = 0.075
∆a4 = |6.l45 – 6.15| = 0.005
From formula (iii),
∆amean = \(\frac{0.025+0.055+0.075+0.005}{4}=\frac{0.160}{4}\)
= 0.04 Ω
From formula (iv),
Relative error = \(\frac{0.04}{6.145}\) = 0.0065 Ω
From formula (v).
Percentage error = 0.0065 \frac{0.04}{6.145} 100 = 0.65%

i. The mean absolute error is 0.04 Ω.
ii. The relative error is 0.0065 Ω.
iii. The percentage error is 0.65%.
[Note: Framing of numerical is modified to reach the answer given to the numerical.]

Question 9.
An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. If v depends upon gravitational acceleration g and distance, prove with dimensional analysis that v = k\(\sqrt{g h}\) where k is a constant.
Answer:
Given = v = k\(\sqrt{g h}\)
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 4
k being constant is assumed to be dimensionless.
Dimensions of L.H.S. = [v] = [L1T-1]
Dimension of R.H.S. = [latex]\sqrt{g h}[/latex]
= [L1T-2]\(\frac{1}{2}\) × [L1]\(\frac{1}{2}\)
= [L2T-2]\(\frac{1}{2}\)
= [L1T-1]
As, [L.H.S.] = [R.H.S.],
=> v = k\(\sqrt{g h}\)is dimensionally correct equation.

Question 10.
v = at + \(\frac{b}{t+c}\) + v0 is a dimensionally valid equation. Obtain the dimensional formula for a, b and c where v is velocity, t is time and v0 is initial velocity.
Answer:
Solution: Given: y = at + \(\frac{b}{t+c}\) + + v0

As only dimensionally identical quantities can be added together or subtracted from each other, each term on R.H.S. has dimensions of L.H.S. i.e., dimensions of velocity.

∴ [LH.S.] = [v] = [L1T-1]
This means, [at] = [v] = [L1T-1]
Given, t = time has dimension [T-1]
∴ [a] = \(\frac{\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right]}{[\mathrm{t}]}=\frac{\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right]}{\left[\mathrm{T}^{1}\right]}\) = [L1T-2] = L1M0T-2]
Similarly, [c] = [t] = [T1] = [L0M0T1]
∴ \(\frac{[\mathrm{b}]}{\left[\mathrm{T}^{1}\right]}\) = [v] = [L1T-1]
∴ [b] = [L1T-1] × [T1] = [L1] = [L1M0T0]

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Given: l = 4.234 m, b = 1.005 m,
t = 2.01 cm = 2.01 × 10-2 m = 0.0201 m

To find:
i) Area of sheet to correct significant figures (A)
ii) Volume of sheet to correct significant figures (V)
Formulae: i. A = 2(lb + bt + tl)
iii) V = l × b × t

Calculation: From formula (i),
A = 2(4.234 × 1.005 + 1.005 × 0.0201 +0.0201 × 4.234)
= 2 |[antilog(log 4.234 + logl .005) + antiiog(log 1.005 + log0.0201) + antilog(log 0.0201 + log 4.234)]}
= 2{[antilog(0.6267 + 0.0021) + antilog(0.0021 + \(\overline{2}\) .3010) + antilog (\(\overline{2}\) .3010 + 0.6267)]}
= 2 {[antilog(0.6288) + antilog (\(\overline{2}[/latex .3031) +antilog([latex]\overline{2}\) .9277)]}
= 2 [4.254 + 0.02009 + 0.08467]
= 2 [4.35876]
= 8.71752m2

In correct significant figure,
A = 8.72 m2 From formula (ii),
V =4.234 × 1.005 × 0.0201
= antilog [log (4.234) + log (1.005) + log (0.0201)]
= antilog [0.6269 – 0.0021 – \(\overline{2}\).3032]
= antilog [0.6288 – \(\overline{2}\).3032]
= antilog [ 2 .9320]
= 8.551 × 10-2
= 0.08551 m3
In correct significant figure (rounding off),
V = 0.086 m3

i.) Area of sheet to correct significant figures is 8.72 m2.
ii) Volume of sheet to correct significant figures is 0.086 m3.
[Note: The given solution is arrived to by considering a rectangular sheet.]

Question 12.
If the length of a cylinder is l = (4.00 ± 0.001) cm, radius r = (0.0250 ± 0.001) cm and mass m = (6.25 ± 0.01) gm. Calculate the percentage error in the determination of density.
Answer:
Given: l = (4.00 ± 0.001) cm,
In order to have same precision, we use, (4.000 ± 0.001),
r = (0.0250 ± 0.00 1) cm,
In order to have same precision, we use, (0.025 ± 0.001)
m = (6.25 ± 0.01) g
To find: percentage error in density

Formulae:
i) Relative error in volume, \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}\)
….(∵ Volume of cylinder, V = πr2l)

ii) Releative error \(\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{\Delta V}{V}\)
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 10

iii) Percentage error= Relative error × 100%

Calculation.
From formulae (i) and (ii),
∴ \(\frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}\)
= \(\frac{0.01}{6.25}+\frac{2(0.001)}{0.025}+\frac{0.001}{4.000}\)
= 0.00 16 + 0.08 + 0.00025
= 0.08 185
From formula (iii).
% error in density = \(\frac{\Delta \rho}{\rho}\) × 100
= 0.08185 × 100
= 8.185%
Percentage error in density is 8.185%.

Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements

Question 13.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of the Jupiter.
Answer:
Given: Angular diameter (a) = 35.72″
= 35.72″ × 4.847 × 10-6 rad
1.73 × 10-4 rad

Distance from Earth (D)
= 824.7 million km
= 824.7 × 106 km
= 824.7 × 109 m.

To find: Diameter of Jupiter (d)
Formula: d = α D
Calculation: From formula,
d = 1.73 × 10-4 × 824.7 × 109
= 1.428 × 108 m
= 1.428 × 105 km
Diameter of Jupiter is 1.428 × 105 km.

Question 14.
If the formula for a physical quantity is X = \(\frac{a^{4} b^{3}}{c^{1 / 3} d^{1 / 2}}\) and if the percentage error in the measurements of a, b, c and d are 2%, 3%, 3% and 4% respectively. Calculate percentage error in X.
Answer:
Given X = \(\frac{a^{4} b^{3}}{c^{1 / 3} d^{1 / 2}}\)
Percentage error in a, b, c, d is respectively 2%, 3%, 3% and 4%.
Now, Percentage error in X
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 7

Question 15.
Write down the number of significant figures in the following: 0.003 m2, 0.1250 gm cm-2, 6.4 × 106 m, 1.6 × 10-19 C, 9.1 × 10-31 kg.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 8

Question 16.
The diameter of a sphere is 2.14 cm. Calculate the volume of the sphere to the correct number of significant figures.
Answer:
Volume of sphere = \(\frac{4}{3}\) πr3
= \(\frac{4}{3}\) × 3.142 × (\(\frac{2.14}{2}\))3 ………….. (∵ r = \(\frac{d}{2}\))
= \(\frac{4}{3}\) × 3.142 × (1.07)3
= 1.333 × 3.142 × (1.07)3
= {antilog [log (1.333) + log(3.142)+3 log(1.07)]}
= {antilog [0.1249 + 0.4972 + 3 (0.0294)])
= {antilog [0.6221 + 0.0882]}
= {antilog [0.7103]}
= 5.133cm3
In multiplication or division, the final result should retain as many significant figures as there are in the original number with the least significant figures.
Volume in correct significant figures
∴ 5.13 cm3

11th Physics Digest Chapter 1 Units and Measurements Intext Questions and Answers

Can you recall (Textbook Page No. 1)

Question 1.
i) What is a unit?
ii) Which units have you used in the laboratory for measuring
a. length
b. mass
c. time
d. temperature?
iii. Which system of units have you used?
Answer:

  1. The standard measure of any quantity is called the unit of that quantity.
  2. Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 11
  3. MKS or SI system is used mostly. At times. even CGS system is used.

Can you tell? (Textbook Page No. 8)

Question 1.
If ten students are asked to measure the length of a piece of cloth upto a mm, using a metre scale, do you think their answers will be identical? Give reasons.
Answer:
Answers of the students are likely to be different. Length of cloth needs to be measured up to a millimetre (mm) length. Hence, to obtain accurate and precise reading one must use measuring instrument having least count smaller than 1 mm.

But least count of metre scale is 1 mm. As a result, even smallest uncertainty in reading would vary reading significantly. Also, skill of students doing measurement may also introduce uncertainty in observation.
Hence, their answers are likely to be different.

Activity (Textbook Page No. 10)

Perform an experiment using a Vernier callipers of least count 0.01cm to measure the external diameter of a hollow cylinder. Take 3 readings at different positions on the cylinder and find (i) the mean diameter (ii) the absolute mean error and (iii) the percentage error in the measurement of diameter.
Answer:
Given: L.C. = 0.01 cm
To measure external diameter of hollow cylinder readings are taken as follows:
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 12
[Note: The above table is made assuming zero error in Vernier calipers. If caliper has positive or negative zero error, the zero error correction needs to be introduced into observed reading.]
Maharashtra Board Class 11 Physics Solutions Chapter 1 Units and Measurements 13

Internet my friend (Textbook Page No. 12)

i. ideoiectures.net/mit801f99_lewin_lec0l/
ii. hyperphysicsphy-astr.gsu.ed u/libase/hfra me. html

[Students can use links given above as a reference and collect information about units and measurements]

Maharashtra State Board Class 11 Physics Textbook Solutions 

Metallurgy Class 10 Questions And Answers Maharashtra Board

Std 10 Science Part 1 Chapter 8 Metallurgy Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 1 Chapter 8 Metallurgy Question Answer Maharashtra Board

Question 1.
a. Alloy of sodium with mercury.
Answer:
Silver amalgam.

b.Molecular formula of common ore of aluminium.
Answer:
Al2O3.nH2O

c. The oxide that forms salt and water by reacting with both acid and base.
Answer:
Aluminium oxide (Al2O3).

d. device used for grinding an ore.
Answer:
The device used for grinding an ore is grinding mill.

e. The nonmetal having electrical conductivity.
Answer:
Graphite having electrical conductivity.

f. The reagent that dissolves noble metals.
Answer:
Aqua regia is the reagent that dissolves noble metals like gold and platinum.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 2.
Make pairs of substances and their properties.

Column IColumn II
SubstanceProperty
(1) Potassium bromide(a) Combustible
(2) Gold(b) Soluble in water
(3) Sulphur(c) No chemical reaction
(4) Neon(d) High ductility
(e) Magnetic ingredient

Answer:
(1) Potassium bromide – Soluble in water
(2) Gold – High ductility
(3) Sulphur – Combustible
(4) Neon – No chemical reaction

Question 3.
Identify the pairs of metals and their ores from the following.

Column I (ores)Column II (metals)
(1) Bauxite(a) Mercury
(2) Cassiterite(b) Aluminium
(3) Cinnabar(c) Tin
(d) Copper

Answer:
(1) Bauxite – Aluminium
(2) Cassiterite – Tin
(3) Cinnabar – Mercury

Question 4.
Explain the terms.
a. Metallurgy
Answer:
Metallurgy: The process used for extraction of metals in their pure form from their ores, then metals are further purified by different methods of purification. All the process is called metallurgy.

b. Ores.
Answer:
Ores: The minerals from which metals are extracted profitably and conveniently are called ores.
Examples: Bauxite (Al2O3.H2O), Cinnabar (HgS).

c. Minerals.
Answer:
Minerals: The naturally occurring compounds of metals along with other impurities are known as minerals.
Examples: Rocks are composed of mixtures of minerals. Talc and granite are minerals.

d. Gangue.
Answer:
Gangue: Ores contain metal compounds with some of the impurities like soil, sand, rocky material, etc. These impurities are called gangue.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 5.
Write scientific reasons.
a. Lemon or tamarind is used for cleaning copper vessels turned greenish.
Answer:

  • Copper undergoes oxidation in air to form black copper oxide. Copper oxide reacts slowly with carbon dioxide in air and gains a green coat. This green substance is copper carbonate.
  • Lemon and tamarind contain acid. The acid dissolves the green coating of basic copper carbonate present on the surface of a tarnished copper utensil and makes it shiny again.

b. Generally the ionic compounds have high melting points.
Answer:

  • The ionic compounds exist in solid state and are hard due to strong electrostatic force of attraction between oppositely charged ions.
  • The intermolecular force of attraction is high in ionic compounds and large energy is required to overcome it. Therefore, ionic compounds have high melting points.

c. Sodium is always kept in kerosene.
(OR)
Why is sodium stored in kerosene?
Answer:

  • Sodium reacts so vigorously with atmospheric oxygen that it catches fire if kept in the open.
  • It does not react with kerosene and sinks in it. Hence, to protect sodium and to prevent accidental fires it is always kept in kerosene.

d. Pine oil is used in the froth floatation process.
Answer:

  • In the concentration of an ore by froth floatation process, the ore is mixed with water and pine oil. When air is bubbled through the mixture a froth is formed.
  • The mineral particles in the ore are preferentially wetted by the oil and float on the top in the froth.
  • The gangue particles are wetted by water and settle down. Thus the mineral can be separated from the gangue and the ore is concentrated.

e. Anodes need to be replaced from time to time during the electrolysis of alumina.
Answer:

  • During electrolysis of alumina, the oxygen liberated at the carbon anode reacts with graphite rods (carbon anode) and forms carbon dioxide.
  • As the anodes get oxidised during electrolysis of alumina, they are continuously eroded. Hence, it is necessary to replace anodes from time to time.

Question 6.
When a copper coin is dipped in silver nitrate solution, a glitter appears on the coin after some time. Why does this happen? Write the chemical equation.
Answer:
When a copper coin is dipped in a silver nitrate solution, more reactive copper displaces silver from silver nitrate solution. The silver so liberated deposits on the copper coin. As a result, a shiny coat of silver is formed on the coin.
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 7.
The electronic configuration of metal ‘A’ is 2, 8, 1 and that of metal ‘B’ is 2, 8, 2. Which of the two metals is more reactive? Identify these metals. Write their reaction with dilute hydrochloric acid. (Practice Activity Sheet – 1)
Answer:
If the number of electrons in the outermost orbit is less, then the metal is more reactive. Metal A contains one electron in the outermost shell, while metal B contains two electrons. Hence, metal A is more reactive than metal B.

Metal A is sodium and metal B is magnesium. Reactions of Na and Mg with dil. HCl are,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 1

Question 8.
Draw a neat labelled diagram.
a. Magnetic separation method.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 2

b. Forth floatation.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 3

c. Electrolytic reduction of alumina.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 4

d. Hydraulic separation method.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 5

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 9.
Write chemical equation for the following events.
a. Aluminium came in contact with air.
Answer:
When aluminium is exposed to air, it develops a thin oxide layer of aluminium.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 6

b. Iron filings are dropped in aqueous solution of copper sulphate.
Answer:
When iron filings are dropped in copper sulphate solution, more reactive iron displaces copper from copper sulphate solution. The iron filings get coated with reddish brown copper metal and the blue colour of copper sulphate fades gradually and ferrous sulphate is formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 7

c. A reaction was brought about between ferric oxide and aluminium.
Answer:
The reaction between ferric oxide and iron produces aluminium oxide and iron. It is a thermite reaction and is highly exothermic.
It produces a large amount of heat, which is released to melt oxygen and aluminium. This reaction is used in welding of machineries. It is also used in warfare to make grenades.

The chemical reaction for the above is as follows:
3Fe3O2 + 4Al → 2Al2O3 + 6Fe

d. Electrolysis of alumina is done.
Answer:
During electrolysis of alumina, aluminium is deposited at the cathode. Molten aluminium being heavier than the electrolyte, is collected at the bottom of the tank. Oxygen gas is liberated at the anode.
Anode reaction: 2O → O2 + 4e (Oxidation)
Cathode reaction: Al3+ + 3e → Al(l) (Reduction)

e. Zinc oxide is dissolved in dilute hydrochloric acid.
Answer:
Zinc oxide is dissolved in dilute hydrochloric acid, zinc chloride and water are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 8

Question 10.
Complete the following statement using every given options.
During the extraction of aluminium
a. Ingredients and gangue in bauxite.
b. Use of leuching during the concentration of ore.
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.
d. Heating the aluminium ore with concentrated caustic soda.
Answer:
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 11.
Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li into three groups, namely, reactive metals, moderately reactive metals and less reactive metals.
Answer:
Reactive metals: Na, Li, Ca
Moderately reactive metals: Zn, Fe, Mg,
Less reactive metals: Cu

Project: (Do it your self)

Collect metal vessels and various metal articles. Write detailed informciton. write the steps in the procedure that can be done in the laboratory for giving glitter to these. Seek guidance from your teacher.

Can you recall? (Text Book Page No. 93)

Question 1.
what are the physical properties of metals and nonmetals?
Answer:
Properties of metals:

  1. Solid state (Exception: Mercury and gallium)
  2. Typical lustre
  3. Malleability and ductility
  4. Hardness (Exception: Lithium, sodium and potassium)
  5. Good conductors of heat and electricity
  6. High melting and boiling points (On the other hand, the melting and boiling points of the metals sodium, potassium, mercury and galium are very low.)
  7. Sonorous and produce sound on striking a hard surface.

Properties or nonmetals:

  1. Gaseous or solid state (Exception: Bromine in liquid state)
  2. Lack of any typical lustre (Exception: Iodine and Diamond)
  3. Brittleness in the solid state (Exception: Diamond is the hardest natural substance)
  4. Bad conductors of heat and electricity (Exception: Graphite) (Diamond is good conductor of heat)
  5. Low melting and boiling points.

Can you recall? (Text Book Page No. 102)

Question 1.
What is the electronic definition of oxidation and reduction?
Answer:
When a metal loses çlectrons the process is called an oxidation while when a nonmetal gains electrons, it is called a reduction,
Na → Na+ + e (oxidation)
Cl + e → Cl (reduction)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Can you recall? (Text Book Page No. 106)

Question 1.
What is meant by corrosion?
Answer:
Corrosion is degradation of a material due toreaction with its environment.

Question 2.
Have you observed the following things?
(1) Old iron bars in the builthngs.
Answer:
When old iron bars in the buildings are exposed to moist air for a long time, they acquire a coating of browm nlaky substance called rust. (Fe2O3.H2O)

(2) Copper vessels not cleaned for a long time.
Answer:
If copper vessels are not cleaned for a long time, they react with moist carbon dioxide in the air, lose their shine and gain a green coat of copper carbonate. (CuCO3)

Question 3.
Silver ornaments or idols exposed to air for a long time.
Answer:
When silver ornaments or idols are kept exposed to air for a long time, silver reacts with sulphur in the air to form a coating of black silver sulphide. (Ag2S)

Question 4.
Old vehicles fit to be thrown away.
Answer:
The metallic parts of the body of old cars are corroded, eaten up and sometimes become perforated. The old cars also lose the original colour due to formation of flakes of rust.

Use your brain power! (Text Book Page No. 98)

Question 1.
In the reaction between chlorine and HBr a transformation or HBr into Br2 takes place. Can this transformation be called oxidation? What Is the oxidant that brings about this oxidation?
Answer:
The conversion of HBr to Br2 is an oxidation process. In the above reaction, Cl2 in the oxidant.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Can you tell? (Text Book Page No. 104)

Question 1.
what are the moderately reactive metals?
Answer:
In the middle of the reactivity series, metals such as iron. zinc, lead, copper are moderately reactive.

Question 2.
In which form to the moderately reactive metals occur in nature?
Answer:
The moderately reactive metals which occur in nature are in the form of their sulphide salts or carbonate salts.

Think about it (Text Book Page No. 106)

Question 1.
Why do silver articles turn blackish while copper vessels turn greenish on keeping in air for long time?
Answer:

  1. Silver articles turn blackish on exposure to air for a long time. This is because of silver sulphide (Ag2S) laver formed on the silver articles by the reaction of silver with hydrogen sulphide.
  2. Carbon dioxide in moist air reacts with copper vessel. Copper loses its lustre due to formation of greenish layer of copper carbonate (CuCO3) on its surface.

Question 2.
Why do pure gold and platinum always glitter?
Answer:
Gold and platinum are noble metals as they do not react with moisture, O2 and CO2 from air also acids and alkalis, therefore, pure gold and platinum always glitter.

Use your brain power! (Text Book Page No. 103)

Question 1.
Write the electrode reaction for electrolysis of molten magnesium chloride and calcium chloride.
Answer:
(1) Magnesium chloride (MgCl2):
MgCl2 → Mg2+ + 2Cl
At the cathode: Mg2+ + 2e → Mg
At the anode: 2Cl → Cl2 + 2e

(2) Calcium chloride (CaCl2):
At the cathode: Ca2+ + 2e → Ca
At the anode: 2Cl → Cl2 + 2e

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Can you tell? (Text Book Page No. 106)

Question 1.
Which measures would you suggest to stop the corrosion of metallic articles or not allow the corrosion to start?
Answer:
Various types of methods are used to protect metals from corrosion. Almost in all the methods, special attention is paid so that iron does not rust. It is possible to lower the rate of the process of rusting of iron. Corrosion of metals can be stopped by detaching the air from metals.

Some methods are as follows :

  1. To fix a layer of some substance on the metal surface so that the contact of the metal with moisture and oxygen in the air is prevented and no reaction would occur between them.
  2. To prevent corrosion of metals by applying a layer of paint, oil, grease or varnish on their surface. For example, corrosion of iron can be prevented by this method.

Question 2.
What is done so to prevent rusting of iron windows and iron doors of your house?
Answer:
To prevent rusting of iron windows and iron doors in the house, they are painted so that they do not rust.

Question 3.
What is done so to prevent rusting or iron windows and iron doors of your house?
Answer:
To prevent rusting of iron windows and iron doors in the house, they are painted so that they do not rust.

Use your brain power! (Text Book Page No. 107)

Question 1.
Can we permanently prevent the rusting of an iron article by applying a layer of paint on its surface?
Answer:
The method of painting is alright for some time. We cannot protect the articles permanently from rusting by painting them.

Question 2.
Why do new iron sheets appear shiny?
Answer:
The new iron sheets appear shiny because a layer of non-corrosionable metal is fixed on the surface of corrosionable metal.

Collect information. (Text Book Page No. 108)

Question 1.
What are the various alloys used in daily life? Where are those used?
Answer:

Various alloysUses
1. BronzeIt is used to prepare: Coins, utensils, medals, statues
2. BrassPipes, condenser tubes, utensils worshipping God.
3. Stainless steelUtensils, tools, dairy equipment, boilers.
4. SteelConstruction of bridges and buildings, cutting tools, blades.
5. Tungsten steelHigh speed cutting tools
6. AmalgamSilver amalgam used by dentists
7. DuraluminBodies of aircraft, buses, kitchenwares
8. Aluminium bronzePigment in ink and paint
9. German silverElectrical heaters, resistors
10. Gun metalGuns, boiler fittings
11. MagneliumBeams of scientific balances, aircraft parts.
12. Gold with copper or nickel or silver or platinumJewellery

Question 2.
What are the properties that the alloy used for minting coins should have?
Answer:
The alloy used for minting coins should have excellent wear resistance and anti-corrosion properties.

Fill in the blanks:

Question 1.
……………has the highest melting point.
Answer:
Tungsten has the highest melting point.

Question 2.
Mercury and…………are two metals in the liquid state at room temperature.
Answer:
Mercury and galium are two metals in the liquid state at room temperature.

Question 3.
…………is the hardest natural substance.
Answer:
Diamond is the hardest natural substance.

Question 4.
The naturally occurring compounds of metals along with other impurities are known as………….
Answer:
The naturally occurring compounds of metals along with other impurities are known as minerals.

Question 5.
The minerals from which metals are extracted profitably and conveniently are called…………..
Answer:
The minerals from which metals are extracted profitably and conveniently are called ores.

Question 6.
An ore contains some of the impurities like soil, sand, etc. These impurities are called…………
Answer:
An ore contains some of the impurities like soil, sand, etc. These impurities are called gangue.

Question 7.
The process of extraction of a metal from its ore is called……….
Answer:
The process of extraction of a metal from its ore is called metallurgy.

Question 8.
Bauxite is a common ore of………..
Answer:
Bauxite is a common ore of aluminium.

Question 9.
…………. process is used for the purification of bauxite.
Answer:
Bayer’s process is used for the purification of bauxite.

Question 10.
During the electrolysis of alumina, ………..is liberated at the anode.
Answer:
During the electrolysis of alumina, oxygen is liberated at the anode.

Question 11.
The reaction of iron oxide with aluminium is known as…………..reaction.
Answer:
The reaction of iron oxide with aluminium is known as thermit reaction.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 12.
The process of coating a thin layer of zinc on iron is known as…………
Answer:
The process of coating a thin layer of zinc on iron is known as galvanising.

Question 13.
The metal that produces a sound on striking a hard surface is said to be………….
Answer:
The metal that produces a sound on striking a hard surface is said to be sonorous.

Question 14.
The process in which carbonate ores are changed into oxides by heating strongly in limited air is known as …………….
Answer:
The process in which carbonate ores are changed into oxides by heating strongly in limited air is known as calcination.

Question 15.
…………compounds are insoluble in solvents like kerosene and petrol.
Answer:
Ionic compounds are insoluble in solvents like kerosene and petrol.

Question 16.
…………… is used to obtain pure metals from impure metals.
Answer:
Electrolys is method is used to obtain pure metals from impure metals.

Question 17.
Corrosion can be prevented-by putting a layer of…………metal on corrosionable metal.
Answer:
Corrosion can be prevented by putting a layer of non-corrosionable metal on corrosionable metal.

Rewrite the following statements by selecting the correct options:

Question 1.
………… is a metal.
(a) Mg
(b) S
(c) P
(d) Br
Answer:
Mg is a metal.

Question 2.
………. is a nonmetal.
(a) Au
(b) Hg
(c) Br
(d) Cu
Answer:
Br is a nonmetal.

Question 3.
………… is a metalloid.
(a) Aluminium
(b) Antimony
(c) Zinc
(d) Mercury
Answer:
Antimony is a metalloid.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 4.
Metalloids have properties of ………..
(a) metals
(b) nonmetals
(c) both metals and nonmetals
(d) neither metals nor nonmetals
Answer:
Metalloids have properties of both metals and nonmetals.

Question 5.
…………. is a good conductor of electricity.
(a) Bromine
(b) Iodine
(c) Graphite
(d) Sulphur
Answer:
Graphite is a good conductor of electricity.

Question 6.
………. is a metal which is in liquid form at ordinary temperature and pressure.
(a) Magnesium
(b) Sodium
(c) Scandium
(d) Mercury
Answer:
Mercury is a metal which is in liquid form at ordinary temperature and pressure.

Question 7.
………. is an amphoteric oxide.
(a) Na2O
(b) MgO
(c) ZnO
(d) SO2
Answer:
ZnO is an amphoteric oxide.

Question 8.
……….. is an acidic oxide.
(a) Na2O
(b) CO2
(c) FeO3
(d) H2O
Answer:
CO2 is an acidic oxide.

Question 9.
………. is a basic oxide.
(a) CO2
(b) K2O
(C) SO2
(d) Al2O3
Answer:
K2O is a basic oxide.

Question 10.
………… is an ore of aluminium.
(a) Cryolite
(b) Bauxite
(c) Haematite
(d) Aluminium carbonate
Answer:
Bauxite is an ore of aluminium.

Question 11.
Bronze is an alloy of ………..
(a) copper and tin
(b) copper and zinc
(c) copper and iron
(d) iron and nickel
Answer:
Bronze is an alloy of copper and tin.

Question 12.
An alloy prepared from iron, nickel and chromium is known as …………
(a) brass
(b) bronze
(c) stainless steel
(d) amalgam
Answer:
An alloy prepared from iron, nickel and chromium is known as stainless steel.

Question 13.
…………. is an allotropic form of a nonmetal which conducts electricity.
(a) Sulphur
(b) Graphite
(c) Chlorine
(d) Iodine
Answer:
Graphite is an allotropic form of a nonmetal which conducts electricity.

Question 14.
………….. has an oxide which is soluble in sodium hydroxide.
(a) Calcium
(b) Magnesium
(c) Iron
(d) Zinc.
Answer:
Zinc has an oxide which is soluble in sodium hydroxide.

Question 15.
………… prevents the rusting of iron.
(a) Copper
(b) Zinc
(c) Aluminium
(d) Silver
Answer:
Zinc prevents the rusting of iron.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 16.
………….. is obtained by the reduction of its oxide by carbon.
(a) Zinc
(b) Aluminium
(c) Sodium
(d) Potassium
Answer:
Zinc is obtained by the reduction of its oxide by carbon.

Question 17.
………….. is used as an anode during the electrolytic reduction of bauxite.
(a) Sulphur
(b) Graphite
(c) Platinum
(d) Aluminium
Answer:
Graphite is used as an anode during the electrolytic reduction of bauxite.

Question 18.
Silver gets corroded due to ………… in air.
(a) oxygen
(b) hydrogen sulphide
(c) carbon dioxide
(d) nitrogen
Answer:
Silver gets corroded due to hydrogen sulphide in air.

Question 19.
…………. is the hardest substance and has the highest melting and boiling points.
(a) Iodine
(b) Sulphur
(c) Diamond
(d) Phosphorus
Answer:
Diamond is the hardest substance and has the highest melting and boiling points.

Question 20.
Jewellery articles are gold plated ………….
(a) to prevent corrosion
(b) to prevent rusting of the base metal
(c) to make articles attractive
(d) all of these
Answer:
(d) all of these

Question 21.
To show that zinc is more reactive than copper, the correct procedure is to ………..
(a) prepare copper sulphate solution and dip a zinc strip in it
(b) prepare zinc sulphate solution and dip a copper strip in it
(c) heat together zinc and copper strips
(d) add dil. nitric acid to both the strips
Answer:
To show that zinc is more reactive than copper, the correct procedure is to prepare copper sulphate solution and dip a zinc strip in it.

Question 22.
Iron is ………
(a) more reactive than zinc
(b) more reactive than aluminuium
(c) less reactive than copper
(d) less reactive than aluminium
Answer:
Iron is less reactive than aluminium.

Question 23.
A solution of Al2(SO4)3 in water is …………
(a) blue
(b) pink
(c) green
(d) colourless
Answer:
A solution of Al2(SO4)3 in water is colourless.

Question 24.
A solution of ………… in water is blue in colour.
(a) CuSO4
(b) FeSO4
(c) ZnSO4
(d) Al2(SO4)3
Answer:
A solution of CuSO4 in water is blue in colour.

Question 25.
A solution of …………. n water is green in colour.
(a) CuSO4
(b) FeSO4
(c) ZnSO4
(d) Al2(SO4)3
Answer:
A solution of FeSO4 in water is green in colour.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 26.
What would be the correct order if Zn, Fe, Al and Cu are arranged in increasing order of reactivity?
(a) Cu, Fe, Zn, Al
(b) Al, Cu, Fe, Zn
(c) Zn, Al, Cu, Fe
(d) Fe, Zn, Al, Cu (Practice Activity Sheet – 2)
Answer:
(a) Cu, Fe, Zn, Al

Question 27.
During the extraction of aluminium ……….
(a) Ingredients and gangue in bauxite
(b) Use of leaching during the concentration of ore
(c) Chemical reaction of transformation of bauxite into alumina by Hall’s process.
(d) Heating the aluminium ore with concentrated caustic soda.
Answer:
During the extraction of aluminium Chemical reaction of transformation of bauxite into alumina by Hall’s process.

Question 28.
A solution of CuSO4 in water is ………… in colour.
(a) pink
(b) blue
(c) colourless
(d) green
Answer:
A solution of CuSO4 in water is blue in colour.

Question 29.
Which of the following process is to be carried out to avoid the formation of greenish layer on brass vessels due to corrosion?
(a) Plating
(b) Anodization
(c) Tinning
(d) Alloying (Practice Activity Sheet – 3)
Answer:
(c) Tinning

State whether the following statements are True or False (If a statement is false, correct it and rewrite it.):

Question 1.
Metals are known as sonar metals.
Answer:
True.

Question 2.
Diamond is the softest natural substance.
Answer:
False. (Diamond is the hardest natural substance.)

Question 3.
Electrolysis method is used to obtain pure metals from impure metals.
Answer:
True.

Question 4.
Iodine and diamond are lustrous substances.
Answer:
True.

Question 5.
Aqua Regia is a mixture of conc. HCl and conc. HNO3 in the ratio of 1:3.
Answer:
False. (Aqua Regia is a mixture of conc. HCl and conc. HNO3 in the ratio of 3:1.)

Question 6.
Corrosion of metals can be stopped by detaching the air from metals.
Answer:
True.

Question 7.
Due to corrosion a greenish layer forms on the surface of copper or brass vessel.
Answer:
True.

Question 8.
Ionic compounds are soluble in kerosene.
Answer:
False. (Ionic compounds are soluble in water and insoluble in kerosene.)

Question 9.
Ionic compounds in the solid state conduct electricity.
Answer:
False. (Ionic compounds in the solid state do not conduct electricity.)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 10.
Mercury, silver and gold are very reactive metals.
Answer:
False. (Mercury, silver and gold are least reactive metals.)

Question 11.
In electroplating a metal is coated with another metal using electrolysis.
Answer:
True.

Question 12.
In anodising method. the copper or aluminium article is used as anode.
Answer:
True.

Question 13.
Silver plated spoon, gold plated ornaments are the examples of alloying.
Answer:
False. (Silver plated spoon, gold plated ornaments are the examples of electroplating)

Question 14.
silver amalgam is mainly used by dentists.
Answer:
True.

Question 15.
Aluminium oxide is an acidic oxide.
Answer:
False. (Aluminium oxide is an amphoteric oxide.)

Question 16.
copper reacts with moist carbon form copper carbonate.
Answer:
True.

Question 17.
Corrosion is degradation of a reaction with its environment.
Answer:
True.

Find the correlation in the given pair and rewrite the answer:

Question 1.
Brass : Copper and Zinc :: Bronze :………….
Answer:
Brass : Copper and Zinc :: Bronze : Copper and tin

Question 2.
Tinning : Tin :: Galvanizing :…………….
Answer:
Tinning : Tin :: Galvanizing : Zinc

Question 3.
Pressure cooker : Anodizing :: Silver plated spoons :…………..
Answer:
Pressure cooker : Anodizing :: Silver plated spoons : Electro-plating

Question 4.
The sulphides ores are strongly heated in air : Roasting :: The carbonates ores are strongly heated in a limited supply of air :………….
Answer:
The sulphides ores are strongly heated in air : Roasting :: The carbonates ores are strongly heated in a limited supply of air : Calcination.

Question 5.
Sulphide ores : Froth floatation method : Cassiterite ore :………..
Answer:
Sulphide ores : Froth floatation method : Cassiterite ore : Magnetic separation method.

Find the odd one out:

Question 1.
Sodium, Potassium, Silver, Sulphur
Answer:
Sulphur. (All except sulphur, others are metals.)

Question 2.
Boron, Chlorine, Bromine, Fluorine
Answer:
Boron. (All except boron, others are nonmetals.)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 3.
Copper, Iron, Mercury, Brass
Answer:
Brass. (All except brass, others are metals.)

Question 4.
Brass, Bronze, Phosphorus, Stainless steel
Answer:
Phosphorus. (All except phosphorus, others are alloys.)

Question 5.
Magnesium chloride, Sodium chloride, Water, Zinc chloride
Answer:
Water. (All except water, others are ionic compounds.)

Question 6.
Tinning, Anodization, Alloying, Froth floatation (March 2019)
Answer:
Froth floatation. (All except froth floatation, others are processes of coating a thin layer of metal on the surface of other metals.)

Match the following:

Question 1.

Column IColumn II
(1) ZnS(a) Cuprous sulphide
(2) HgS(b) Bauxite
(3) Cu2S(c) Zinc blend
(4) Al2O3.H2O(d) Cinnabar
(e) Cryolite

Answer:
(1) ZnS – Zinc blend
(2) HgS – Cinnabar
(3) Cu4S – Cuprous sulphide
(4) Al2O3.H2O – Bauxite.

Question 2.

Column IColumn II
(1) Copper and zinc(a) Stainless steel
(2) Copper and tin(b) Zinc amalgam
(3) Iron, nickel and chromium(c) Bronze
(4) Mercury and zinc(d) Brass
(e) Steel

Answer:
(1) Copper and zinc – Brass
(2) Copper and tin – Bronze
(3) Iron, nickel and chromium – Stainless steel
(4) Mercury and zinc – Zinc amalgam.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 3.

Column IColumn II
(1) Galvanising(a) Pressure cooker
(2) Tinning(b) Silver plated spoons
(3) Electroplating(c) Coating of tin on copper
(4) Anodizing(d) Coating of Zn on iron

Answer:
(1) Galvanising – Coating of Zn on iron
(2) Tinning – Coating of tin on copper
(3) Electroplating – Silver plated spoons
(4) Anodizing – Pressure cooker.

Translate the following statements into chemical equations and then balance them:

Question 1.
steam is passed over aluminium.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 9

Question 2.
Extraction of copper from its sulphide ore.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 10
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 11

Question 3.
Thermit reaction.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 12

Question 4.
Magnesium reacts with hot water.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 13

Question 5.
what happens when aluminium oxide dissolves in aqueous sodium hydroxide?
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 14

Question 6.
Zinc reacts with sulphuric acid.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 15

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 7.
Iron reacts with sulphuric acid.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 16

Name the following:

Question 1.
A metal which forms an amphoteric oxide.
Answer:
Aluminium forms an amphoteric oxide.

Question 2.
An alloy of copper and zinc.
Answer:
An alloy of copper and zinc is termed as brass.

Question 3.
A compound which is added to lower the fusion temperature.
Answer:
Cryolite (AlF3, 3NaF) and fluorspar (CaF2) are added to lower the fusion temperature.

Question 4.
A metal which does not react with cold water but reacts with steam.
Answer:
Aluminium does not react With cold water but reacts with Steam.

Question 5.
A common ore of aluminium.
Answer:
Bauxite (Al2O3.H2O) is a common ore of aluminium.

Question 6.
A metal which is in liquid state at ordinary temperature.
Answer:
Mercury is in liquid state at ordinary temperature.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 7.
Two metals which are malleable.
Answer:
Iron and aluminium are malleable metals.

Question 8.
Two metals which are ductile.
Answer:
Gold and silver are ductile metals.

Question 9.
Two metals which are good conductors of heat.
Answer:
Silver and copper are good conductors of heat.

Question 10.
Two metals which are good conductors of electricity.
Answer:
Copper and aluminium are good conductors of electricity.

Question 11.
Two metals which are used for making cooking vessels.
Answer:
Copper and aluminium are used in making cooking vessels.

Question 12.
Two metals having low melting points.
Answer:
Sodium and potassium have low melting points.

Question 13.
Two highly reactive metals.
Answer:
Sodium and potassium are highly reactive metals.

Question 14.
A nonmetal which is in liquid state at room temperature.
Answer:
Bromine is in liquid state at room temperature.

Question 15.
Two ionic compounds.
Answer:
Sodium chloride (NaCl) and magnesium chloride (MgCl2) are ionic compounds.

Question 16.
The process of heating the sulphide ore to a high temperature in the excess of air.
Answer:
In roasting, sulphide ore is heated to a high temperature in the excess of air.

Question 17.
The process of heating the carbonate ore to a high temperature in limited air.
Answer:
In calcination, carbonated ore is heated to a high temperature in limited air.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 18.
The compound formed by the reaction between aluminium oxide and sodium hydroxide.
Answer:
Sodium aluminate is formed by the reaction between aluminium oxide and sodium hydroxide.

Question 19.
Two metals which are found in the free state in nature.
Answer:
Gold (Au) and silver (Ag) are found in the free state in nature.

Question 20.
A metal which has the highest melting point.
Answer:
Tungsten has the highest melting point.

Question 21.
Two nonmetals which are lustrous.
Answer:
Iodine and diamond are lustrous in nature.

Answer the following questions in one sentence each:

Question 1.
State the property of the metals due to which they can be drawn into wires.
Answer:
The property of the metal due to which they can be drawn into wires is called ductility.

Question 2.
State the property of the metals due to which they can be beaten into thin sheets.
Answer:
The property of the metals due to which they can be beaten into thin sheets is called malleability.

Question 3.
Which is the hardest substance?
Answer:
Diamond which is a form of carbon is the hardest substance.

Question 4.
What material is used to coat electrical wires?
Answer:
PVC (Polyvinyl chloride) is used to coat electrical wires.

Question 5.
State two metals which can be cut easily with a knife.
Answer:
Sodium and potassium are soft metals and can be cut easily with a knife.

Question 6.
Which of the following metals react with cold water?
Sodium, iron, copper, potassium.
Answer:
Sodium and potassium metals react with cold water.

Question 7.
Which of the following metals do not react with cold water or hot water?
Sodium, potassium, aluminium, iron.
Answer:
Aluminium and iron do not react with cold water or hot water.

Question 8.
State two metals which displace hydrogen from dilute acids and two metals which do not do so.
Answer:
Metals which displace hydrogen from dilute acids are: Magnesium and zinc.
Metals which do not displace hydrogen from dilute acids are: Copper and silver.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 9.
Arrange the following metals in the increasing order of their activity:
Copper, Silver, Aluminium, Iron. (Practice Activity Sheet – 1)
Answer:
The arrangement of metals in the increasing order of their activity:
Silver < Copper < Iron < Aluminium

Question 10.
Write the chemical equation for the reaction of hot iron with steam.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 17

Question 11.
Complete the following reactions:
(1) Zn(s) + H2O(g) → _____________
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 18
Answer:
(1) Zn(s) + H2O(g) → ZnO(s) + H2(g)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 19

Question 12.
Complete the following reactions:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 20
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 21

Question 13.
3MnO2 + 4Al → 3Mn + 2Al2O3 + heat.
Identify the substances undergone oxidation and reduction reactions.
Answer:
MnO2 is reduced to Mn.
Al is oxidised to Al2O3

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 14.
State the impurities present in the bauxite ore.
Answer:
The main impurities present in the bauxite ore are silica (SiO2) and iron oxide (Fe2O3).

Question 15.
Write the formula of (i) bauxite (ii) cryolite.
Answer:
The formula of bauxite is Al2OH2O and that of cryolite is (Na3AlF6).

Question 16.
What is galvanization?
Answer:
The process of coating a thin layer of zinc on iron or steel is called galvanization.

Question 17.
Name the reaction in which aluminium is used as a reducing agent.
Answer:
The thermite reaction in which iron oxide is reduced by aluminium. Aluminium is used as a reducing agent in the thermit reaction.

Question 18.
What are the constituents of bronze?
Answer:
Copper and tin are the constituents of bronze.

Question 19.
State the term used to express the purity of gold.
Answer:
The purity of gold is expressed in carat.

Question 20.
What is meant by amalgam?
Answer:
The amalgam is an alloy in which one of the metals is mercury.

Question 21.
What is meant by electroplating?
Answer:
A process in which a less reactive metal is coated on a more reactive metal by electrolysis is called electroplating.

Question 22.
Why are metals called electropositive elements?
Answer:
Metals are reactive. They lose electrons and become positively charged ions. Therefore, metals are called electropositive elements.

Answer the following questions:

Question 1.
Distinguish between the physical properties of metals and nonmetals with respect to the following points:
(1) Physical state (2) Lustre (3) Ductility and malleability (4) Conduction of heat and electricity (5) Hardness (6) Melting and boiling points.
Answer:
(1) Physical state: Under ordinary conditions, metals are generally solids. Exceptions: mercury and gallium are liquids. Under ordinary conditions, nonmetals may be solids or gases. Exception: bromine is in liquid state.

(2) Lustre: Metals usually have a high lustre (called metallic lustre). They can be polished to give a highly reflective surface. With the exceptions of gold and copper, metals usually have silvery grey colour. Nonmetals lack lustre, exceptions: graphite and iodine. Some nonmetals are colourless and others possess a variety of colours.

(3) Ductility and malleability: Metals are ductile and malleable. Nonmetals are not ductile and mallfeable.

(4) Conduction of heat and electricity: Metals are good conductors of heat and electricity. Nonmetals are bad conductors of heat and electricity. Exception: Graphite is a good conductor of electricity.

(5) Hardness: Metals are usually hard, but not brittle, exceptions: sodium, potassium, lead, zinc. Nonmetals are brittle in the solid state, exception: diamond.

(6) Melting and boiling points: The melting and boiling points of metals are high, exceptions: sodium, potassium, mercury, gallium. The melting and boiling points of nonmetals are low, exceptions: carbon, silicon.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 2.
Write any three physical properties of nonmetals.
Answer:

  1. Nonmetals may be solid or gaseous.
  2. Nonmetals lack lustre. They are not ductile and malleable.
  3. The melting and boiling points of nonmetals are low.
  4. Nonmetals are bad conductors of heat and electricity.

Question 3.
Metals are good conductors of heat. Explain why.
Answer:
(1) The electrons in the outermost shells of atoms of a metal are free to move throughout the metal.
(2) When a metal is heated, these electrons start moving with higher velocity and conduct heat. Hence, metals are good conductors of heat.

Question 4.
Metals are good conductors of electricity. Explain why.
Answer:
(1) The electrons in the outermost shells of atoms of a metal are free to move throughout the metal.
(2) When a potential difference is applied between the ends of a metal wire, the net movement of the electrons in a particular direction, from a point at lower potential to a point at higher potential, constitutes an electric current. Hence, metals are good conductors of electricity.

Question 5.
A metal can be drawn into a wire. Explain why.
Answer:

  1. The property due to which a substance can be drawn into a thin wire without cracking or breaking is called ductility.
  2. Metals are ductile. Thus, a metal can be drawn into a wire.

Question 6.
A metal can be hammered into a thin sheet. Explain why.
Answer:

  1. The property due to which a substance can be hammered (or rolled) into a thin sheet without cracking is called malleability.
  2. Metals are malleable. Thus, a metal can be hammered to form a thin sheet.

Question 7.
How do metals react with oxygen?
Answer:
Metals combine with oxygen on heating in air and metal oxides are formed.
Metal + Oxygen → Metal oxide
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 22

Question 8.
How does a metal react with water?
Answer:
Sodium and potassium react vigorously with water to evolve hydrogen. Calcium reacts with water slowly and less vigorously to evolve hydrogen and the metal floats on water. Magnesium reacts with hot water to evolve hydrogen. Aluminium, iron and zinc do not react with cold or hot water but they react with steam to evolve their oxides and hydrogen.
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) + heat energy
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) + heat energy
2Ca(s) + 2H2O(l) → 2Ca(OH)2(aq) + H2(g)
Mg(s) + 2H2O(hot) → Mg(OH)2(aq) + H2(g)
2Al(s) + 3H2O steam → Al2O3(s) + 3H2(g)
3Fe(s) + 4H2O steam → Fe3O4 + 4H2(g)
Zn(s) + H2O steam → ZnO(s) + H2(g)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 9.
(9) How does a metal react with an acid?
Answer:
Reaction of metals with acids: Metals react with dilute hydrochloric acid or dilute sulphuric acid to form metal chloride or metal sulphate and hydrogen gas. The rate of evolution of H2 is maximum in case of magnesium. The reactivity decreases in the order
Mg > Al > Zn > Fe.
Mg(s) + 2HCl(aq) → MgCl2(s) + H2(g)
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
Fe(s) + H2SO4(aq) → FeSO4 + H2(g)
Zn(s) + H2SO4(aq) → ZnSO4 + H2(g)
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

Question 10.
How does a metal react with nitric acid?
Answer:
Metals react with nitric acid to form nitrate salts. Depending on the concentration of nitric acid, various oxides of nitrogen (NO, NO2) are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 23

Question 11.
Arrange the following metals in the decreasing order of chemical reactivity:
Cu, Mg, Fe, Ca, Zn, Na.
Answer:
The reactivity of metal decreases in the following order:
Na > Mg > Ca > Zn > Fe > Cu.

Question 12.
What is meant by aqua regia?
Answer:
Aqua regia is a highly corrosive and fuming liquid. It is a freshly prepared mixture of conc. HCl and conc. HNO3 in the ratio of 3:1. Most of the substances dissolve in it. Aqua regia is a reagent which dissolves gold and platinum.

Question 13.
How does a metal react with salts of other metals?
Answer:
The reaction of metals with solutions of salts of other metals is the displacement reaction. If a metal A displaces other metal B from the solution of its salt, it means that the metal A is more reactive than the metal B.
Metal A + Salt solution of metal B → Salt solution of metal A + metal B
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 24
In this reaction Fe has displaced Cu from CuSO4. It means Fe is more reactive than Cu.

Question 14.
Explain the reactivity series of the metals.
Answer:
The arrangement of metals in decreasing order of their reactivity in the form of a series is called the reactivity or activity series of the metals.
The most reactive metal is placed at the top of the list and least reactive metal is placed at the bottom of the list.

On the basis of reactivity, we can classify metals into the following categories:

  1. High reactivity metals
  2. Moderately reactive metals
  3. Less reactive metals.

1. Extraction of High reactivity metals: The metals which are placed at the top of the reactivity series are very reactive. They are never found in nature as free elements, e.g., sodium, potassium, calcium and aluminium. These metals are obtained by electrolytic reduction.

2. Extraction of Moderately reactive metals: The metals in the middle of reactivity series such as iron, zinc, lead, copper are moderately reactive. These elements are present as sulphides or carbonates in nature. Generally metals are obtained from their oxide as compared to their sulphides and carbonates.

3. Extraction of Less reactive metals: The metals which are placed at the bottom of the reactivity series are least reactive. They occur in free state, e.g. gold, silver and copper. Copper and silver are also found in the combined state as sulphide and oxide ores. These metals are obtained from their ores by just heating the ores in air.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 15.
Atomic number of metal “A” is 11, while atomic number of metal “B” is 20. Which of them will be more reactive? Write the chemical reaction of dilute HCl with metal “A”. (Practice Activity Sheet – 2)
Answer:
Metal ‘A’ is more reactive than metal ‘B’.
Atomic number of metal ‘A’ is 11, hence it is Na.
2Na + 2HCl → 2NaCl(aq) + H2(g)

Question 16.
How does a metal react with a nonmetal?
Answer:
By oxidation of a metal, cations are formed, on the other hand by reduction of a nonmetal, anions are formed. The ionic compound is formed due to the metal losing electrons while the nonmetal accepts the electrons. The ionic compound of sodium chloride is formed as sodium loses one electron while chlorine accepts one electron.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 25
Similarly Mg and K form ionic compounds MgCl2 and KCl.

Question 17.
How do nonmetals react with oxygen?
Answer:
Nonmetals combine with oxygen to form acidic oxides. In some cases, neutral oxides are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 26

Question 18.
How do nonmetals react with water?
Answer:
Nonmetals do not react with water, (exception : halogen). Chlorine dissolves in water giving hypochlorous acid.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 27

Question 19.
How do nonmetals react with dilute acids?
Answer:
Nonmetals do not react with dilute acid, (exception: halogen). Chlorine reacts with dil. hydrobromic acid to form bromine and HCl.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 28

Question 20.
How do nonmetals react with hydrogen?
Answer:
Nonmetals react with hydrogen under certain conditions (such as proper temperature, pressure, catalyst, etc.)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 29

Question 21.
What is meant by an ionic compound?
Answer:
The compound formed from two units, namely cation and anion is called an ionic compound.

Question 22.
What is meant by an ionic bond?
Answer:
The cation and anion being oppositely charged, there is an electrostatic force of attraction between them, this force of attraction between cation and anion is called the ionic bond.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 23.
State the general properties of ionic compounds.
Answer:

  1. Ionic compounds are solids and hard due to strong electrostatic force of attraction between oppositely charged ions.
  2. They are generally brittle. When pressure is applied they break into pieces.
  3. They have high melting and boiling points, due to intermolecular force of attraction is high in ionic compounds.
  4. They are soluble in water and insoluble in solvents such as kerosene and petrol.
  5. Ionic compounds cannot conduct electricity when in solid state, they are electrically neutral. They conduct electricity in the molten state and also in an aqueous solution.

Question 24.
Explain the following terms:
1. Concentration of ores
2. Roasting
3. Calcination
4. Refining
Answer:
1. Concentration of ores: The process of separating gangue from the other ores is called concentration of ores.
2. Roasting: The process of heating an ore to a high temperature in excess of air and converting it into its oxide is called roasting.
Examples: ZnS (zinc blend), PbS (Galena)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 30
3. Calcination: The process of heating an ore in a limited supply of air and converting it into its oxide is called calcination.
Example: Zinc carbonate (ZnCO3)
ZnCO3 → ZnO + CO2
4. Refining: The metal obtained by chemical reduction contains impurities. The process of electrolysis method is used to obtain pure metals from impure metals is known as refining.

Question 25.
State two methods of concentration of ores in which the heavy particles of ores can be separated from the light gangue particles by the gravitational method.
Answer:

  1. Wilfley table method
  2. Hydraulic separation method are two methods of concentration of ores in which the heavy particles of ores can be separated from the light, gangue particles by the gravitational method.

Question 26.
What are the different methods used for removing gangue from ores?
(OR)
Write the five methods of concentration of ores.
Answer:

  1. Wilfley table method
  2. Hydraulic separation method
  3. Magnetic separation method
  4. Froth floatation method
  5. Leaching method.

Question 27.
Write short notes on: (1) Wilfley table method (2) Hydraulic separation method (3) Magnetic separation method (4) Froth floatation method (5) Leaching method.
Answer:
(1) Wilfley table method : (Separation based on gravitation) This method of separation uses the Wilfley table, it is made by fixing narrow and thin wooden wedges/blocks on inclined surface with low slope. The table is kept continuously vibrating.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 31
Lumps of the ore is made powdered ore by using ball mill. This powdered ore is poured on the table and a stream of water is simultaneously released from the upper side. This result in the lighter gangue particles getting carried away along with the flowing water, while the heavier particles in which proportion of minerals is more and proportion of gangue particles is less, are blocked by the wooden wedges and is collected through the slits between them.

(2) Hydraulic separation method: The hydraulic separation method is based on the working of a mill. This is a tapering vessel similar to that used in a grinding mill. It opens in a tank like a container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 32
Finely ground ore is added to the tank. A fast stream of water is released in the tank from the lower side. The lighter gangue particles flow out along with the water stream from the outlet on the upper side of the tank and are collected separately, simultaneously the heavy particles of the ore are collected at the bottom from the lower side of the tank. This method is based on the law of gravitation, wherein particles of the same size are separated by their weight with the help of water.

(3) Magnetic separation method: Electro-magnetic machine is used in this method. The main parts of this machine are two types of iron rollers and the conveyor belt continuously moving around them. One of the rollers is nonmagnetic while the other is electromagnetic. The conveyor belt moving around the rollers is made up of leather or brass (nonmagnetic). The powdered ore is poured at that end of the conveyor belt which is on the side of the nonmagnetic roller. Two collector vessels are placed below the magnetic roller.

The particles of the nonmagnetic part in the ore are not attracted towards the magnetic roller. Therefore, they are carried out further along the belt and fall in the collector vessel which is away from the magnetic roller. Simultaneously the particles of the magnetic ingredients of the ore stick to the magnetic roller and therefore fall in the collector vessel near the belt.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 33
In this way the magnetic and nonmagnetic particles in the ore are separated because of their magnetic nature. For example, cassiterite is a tin ore. It contains mainly the nonmagnetic ingredient stannic oxide (SnO2) and the magnetic ingredient ferrous tungstate (FeWO4). These are separated by the electromagnetic method.

(4) Froth floatation method: The froth floatation method is based on the two opposite properties, hydrophilic and hydrophobic, of the particles. The metal sulphides particles get wet mainly with oil due to their hydrophobic property. The gangue particles get wet with water due to the hydrophilic property.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 34
In this method the finely ground mineral is put into a big tank containing a lot of water. The finely powdered ore and vegetable oil such as pine oil, eucalyptus oil are mixed with water for formation of froth. The pressurised air is blown through the mixture. There is an agitator rotating around its axis in the centre of the floatation tank. The agitator is used as per the requirement. Bubbles are formed due to the blown air.

A foam is formed from oil, water and air bubbles together, due to the agitating. This foam rises to the surface of the water and floats. Hence this method is called froth floatation. Sulphide minerals float with the foam on water as they get and can be removed. The gangue particles are wetted by water, settles down at the bottom. This method is used for concentration of zinc blend (ZnS) and copper pyrite (CuFeS2).

(5) Leaching: Leaching is the first step in the extraction of the metals like aluminium, gold and silver from their ores. In this method the ore is soaked in a particular solution for long time. The ore dissolves in that solution due to specific chemical reaction. The gangue, however, does not react and therefore does not dissolve in that solution. It can be separated easily.

For example, concentration of bauxite, the aluminium ore, is done by leaching method. Bauxite is soaked in aqueous NaOH or aqueous Na2CO3 which dissolves the main ingredient alumina in it. This means that bauxite is leached by sodium hydroxide.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 28.
Draw a neat labelled diagram of the arrangement of the equipment used in (1) Wilfley table method (2) Hydraulic separation method.
Answer:
1. Wilfley table method:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 35

2. Hydraulic separation method:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 36

Question 29.
Complete the following flow chart and answer the questions below:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 37
(i) In which method pine oil is used?
(ii) Explain that method of concentration in brief. (Practice Activity Sheet – 2)
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 38
(i) Pine oil is used in froth floatation method.

(ii) The finely powdered ore and vegetable oil such as pine oil, eucalyptus oil are mixed with water for formation of froth. The pressurised air is blown through the mixture. The agitator is used as per the requirement. Bubbles are formed due to the blown air. A foam is formed from oil, water and air bubbles together, due to the agitating. This foam rises to the surface of the water and floats. Hence this method is called froth floatation. Sulphide minerals float with the foam on water as they get and can be removed. The gangue particles are wetted by water, settles down at the bottom. This method is used for concentration of zinc blend (ZnS) and copper pyrite (CuFeS2).

Question 30.
A tapping vessel opens in a tank like container that is tapering on the lower side. The tank has an outlet for water on the upper side and a water inlet on the lower side. Finely ground ore is released in the tank. A forceful jet of water is introduced in the tank from lower side and gangue particles and pure ore are separated by this method.
(i) The above description is of which gravitation separation method?
(ii) Draw labelled diagram of this method. (March 2019)
Answer:
(i) Hydraulic separation method.
(ii)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 39

Question 31.
How are sodium, magnesium and potassium obtained from their molten chloride salts?
Answer:
The metals sodium, calcium and magnesium are obtained by electrolysis of their molten chloride salts. In this process metal is deposited on the cathode while chlorine gas is liberated at the anode.

Question 32.
Name the main ore of aluminium.
Answer:
Bauxite (Al2O3·H2O) is the main ore of aluminium.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 33.
What is bauxite? What are the main impurities found in this ore?
Answer:
Bauxite (Al2O3·H2O) is hydrated aluminium oxide. It contains 30% to 70% Al2O3. The main impurities present in it are iron oxide (Fe2O3) and sand (SiO2).

Question 34.
From which ore is aluminium extracted? What are the stages in its extraction (give only names)?
Answer:
Aluminium is extracted from bauxite (Al2O3·nH2O). Stages id the extraction: (i) Concentration of ore, i.e., conversion of bauxite into alumina, (ii) Electrolytic reduction of alumina.

Question 35.
Describe Bayer’s process for concentration of bauxite.
Answer:
(1) Bayer’s process is used to obtain pure aluminium oxide from bauxite.
(2) Bauxite is then concentrated by chemical separation. Bauxite contains impurities like iron oxide (Fe2O3) and silica (SiO2).
(3) Bauxite ore is powdered and heated with sodium hydroxide under high pressure for 2 to 8 hours at 140 °C in the digester. The aluminium oxide being amphoteric in nature present in bauxite reacts with sodium hydroxide to form water soluble sodium aluminate. This means that bauxite leached by sodium hydroxide. Silica reacts with sodium hydroxide to form soluble sodium silicate. The basic iron oxide (Fe2O3) in the gangue remains unaffected. It is separated by filtration.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 40
(4) The filtrate containing sodium aluminate and sodium silicate is stirred with water and then cooling to 50° C. It is hydrolysed to give precipitate of aluminium hydroxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 41
(5) Aluminium hydroxide is then filtered, washed with water, dried and then calcinated by heating at 1000 °C to get pure aluminium oxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 42

Question 36.
Describe Hall’s process for concentration of bauxite.
Answer:
In Hall’s process the ore is powdered and then it is leached by heating with aqueous sodium carbonate in the digester to form water soluble sodium aluminate. Then the insoluble impurities are filtered opt. The filtrate is warmed and neutralised by passing carbon dioxide gas through it. This result in precipitation of aluminium hydroxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 43
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 44
The precipitate of Al(OH)3 obtained in this processes is filtered, washed, dried and then calcinated by heating at 1000 °C to obtain alumina.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 45

Question 37.
Describe the process of preparation of aluminium by the electrolysis of alumina.
(OR)
Draw and label the diagram of electrolysis of alumina and explain the electrolytic reduction of alumina.
Answer:
Electrolytic reduction of alumina:
(1) The electrolytic cell consists of a rectangular steel tank lined from inside with graphite.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 46
(2) The carbon lining (graphite) acts as a cathode. The anode consists of graphite rods suspended in the molten electrolyte.
(3) Alumina has very high melting point ( > 2000 °C). The electrolysis of alumina is carried out at a low temperature by dissolving it in molten cryolite (Na3AlF6). The solution of alumina in cryolite and small amount of fluorspar (CaF2) is added in the mixture to lower its melting point up to 1000 °C.
(4) On passing an electric current, alumina is electrolysed.
(5) Molten aluminium is collected at the cathode, while oxygen gas is evolved at the anode.
The electrode reactions are shown below:
Al2O3 → 2Al3+ + 3O2-
Anode reaction: 2O2- → O2(g) + 4e
Cathode reaction: Al3+ + 3e → Al
The molten aluminium is heavier than the electrolyte. Therefore, it sinks to the bottom of the electrolyte and is removed from time to time. About 99% pure aluminium is obtained by this process. The oxygen gas liberated reacts with carbon anode and forms carbon dioxide. As the anode gets oxidised during the electrolysis of alumina, it has to be replaced from time to time.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 38.
In the extraction of aluminium:
(i) Name the process of concentration of bauxite.
Answer:
The process of concentration of bauxite is known as Bayer’s process.

(ii) Write the cathode reaction in electrolytic reduction of alumina.
Answer:
At the cathode: Al3+ + 3e → Al.

(iii) Write the function and formula of cryolite in the extraction of aluminium.
Answer:
Cryolite is added to the molten mixture of alumina to reduce the melting point to about 1000 °C.
The formula of cryolite is (Na3AlF6) or AlF3, 3NaF.

(iv) write an equation for the action of heat on aluminium hydroxide.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 47

(v) Draw the diagram of extraction of aluminium.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 48

(vi) Write the anode reaction in electrolytic reduction of alumina.
Answer:
Al2O3 → 2Al3+ + 3O2-
At Anode: 2O2- → O2(g) + 4e

(vii) Write the cathode reaction in electrolytic reduction of alumina.
Answer:
Al2O3 → 2Al3+ + 3O2-
Cathode: Al3+ + 3e → Al(l)

Question 39.
What happens when aluminium ore is heated with caustic soda? Write the balanced chemical equation for the same.
Answer:
When aluminium ore is heated with caustic soda solution under high pressure for 2 to 8 hours and at 140 °C to 150 °C, aluminium oxide from aluminium ore, being amphoteric in nature, dissolves in caustic soda solution to form sodium aluminate.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 49

Question 40.
How is zinc extracted from its ore zinc sulphide or zinc carbonate?
Answer:
The crude zinc sulphide ore is heated strongly in excess of air. Zinc sulphide is converted into zinc oxide. This process is known as roasting.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 50
(OR)
The crude zinc carbonate ore is heated strongly in limited supply of air. Zinc carbonate is converted into zinc oxide. This process is known as calcination.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 51
The zinc oxide is reduced to zinc by using a reducing agent such as carbon.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 52

Question 41.
How is copper extracted from its sulphide ore?
Answer:
Copper is found as cuprous sulphide (Cu2S) in nature. When Cu2S is heated in air, copper is obtained.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 53

Question 42.
How is mercury extracted from cinnabar?
(OR)
Extraction of mercury from its ore cinnabar and write the corresponding chemical reaction.
Answer:
Cinnabar is an ore of mercury. When cinnabar is heated (roasted), it is converted into mercuric oxide (HgO). Mercuric oxide is then reduced to mercury on further heating.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 54

Question 43.
Show the steps involved in the extraction of moderately reactive metals from their sulphide ores.
Answer:
Moderately reactive elements are present as sulphides or carbonates in nature.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 55

Question 44.
In the reactivity series of metals, some metals are misplaced. Rearrange these metals in the decreasing order of their reactivity.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 56
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 57

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 45.
Complete the table, if a metal reacts with the reagent then mark ✓ and if not then ✗.

MetalFerrous
sulphate
Silver
nitrate
Copper
sulphate
Zinc
sulphate
Cu
Al

Answer:

MetalFerrous
sulphate
Silver
nitrate
Copper
sulphate
Zinc
sulphate
Cu
Al

Question 46.
Explain the term corrosion with a suitable example.
(OR)
What is corrosion ?
Answer:
The process in which a metal is destroyed gradually by the action of air, moisture or a chemical (like an acid) on its surface is called corrosion.
(OR)
Corrosion is degradation of a material due to reaction with its environment.
The major problem of corrosion occurs with iron, as it is used as a structural material in construction, bridges, shipbuilding.
Iron gets covered by reddish brown flakes when exposed to atmosphere. This is an example of corrosion.

Question 47.
Explain the different methods to prevent corrosion of metals.
Answer:
(1) Corrosion of a metal can be prevented if the contact between metal and air is cut off.
(2) Corrosion of a metal is prevented by coating with something which does not allow moisture and oxygen to react with it.
(3) A layer of oil or paint or grease is applied on the surface of a metal to prevent corrosion. The rusting or corrosion of iron can be prevented by this method.
(4) Corrosion is also prevented by coating a corrosive metal with a noncorrosive metal. Galvanising, tinning, electroplating, anodising and alloying are the different methods in which a metal is coated with a noncorrosive metal to prevent corrosion.

Question 48.
Write three methods of preventing rusting of iron.
Answer:

  1. The rusting of iron can be prevented by painting, oiling, greasing or varnishing its surface.
  2. Galvanisation is another method of protecting iron from rusting by coating iron with a thin layer of zinc.
  3. Corrosion of iron is prevented by coating iron with noncorrosive substance like carbon. This process is termed as alloying.

Question 49.
What is meant by an alloy? Give two examples with chemical composition.
Answer:
The homogeneous mixture formed by mixing a metal with other metals or nonmetals in certain proportion is called an alloy.
Examples:
1. Bronze: Bronze is an alloy formed from 90% copper and 10% tin. Bronze statues stay well in sun and rain.
2. Stainless steel: Stainless steel alloy is made from 74% iron, 18% chromium and 8% carbon. This alloy does not get stained with air or water and does not rust.

Write short notes on the following:

Question 1.
Galvanizing.
Answer:
(1) The process of coating a thin layer of zinc on iron or steel is called galvanization.
(2) In this method corrosion of zinc occurs first because zinc is more electropositive than iron. After a few years zinc layer goes away and the iron layer gets exposed and starts rusting.
(3) In galvanization an iron object is dipped into molten zinc. A thin layer of zinc is formed all over the iron object. Examples: Shiny iron nails, pin, iron pipes.

Question 2.
Tinning.
Answer:
The process of coating a thin layer of tin (molten tin) on copper or brass is called tinning. Cooking vessels made of copper and brass get a greenish coating due to corrosion. The greenish substance is copper carbonate and it is poisonous. If buttermilk or curry is placed in such a vessel it gets spoiled. Therefore, these vessels are coated with tin to prevent corrosion.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 3.
Electroplating.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 58
The process in which a less reactive metal is coated on a more reactive metal by electrolysis is called electroplating.
Examples: Silver-plated spoon, gold-plated jewellery.
(1) Which process will you study with the help of above material and solutions.
Answer:
With the help of above material and solutions, electroplating process is studied.

(2) Define the process.
Answer:
The process in which less reactive metal is coated on a more reactive metals by electrolysis is called electroplating.

(3) Write the anode and cathode reactions.
Answer:
At anode: Ag → Ag+ + e
At cathode: Ag+ + e → Ag

Question 4.
Anodizing.
(OR)
Identify the process shown in the above diagram and explain it in brief. (Practice Activity Sheet – 3)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 59
Answer:
The anodizing technique is an application of electrolysis. In this method copper or aluminium article is used as anode and it is coated with a strong film of their oxides by means of electrolysis. This oxide layer is strong and uniform all over the surface. This thin film protects the metals from corrosion. The protection can be further increased by making the oxide layer thicker during the anodization.
Examples: Kitchen articles such as ; anodized pressure cooker and anodized pan.

Question 5.
Alloying.
Answer:
A homogenous mixture of two or more metals or a metal and a nonmetal in a definite proportion is called an alloy. The physical properties of an alloy are different from those of its constituents. Alloys are corrosion resistant. Alloy decreases the intensity of corrosion of metals.

Examples: Brass is made from copper and zinc, 90 % Copper and 10 % tin are used to make an alloy called bronze, Stainless steel is made from 74% iron, 8 % carbon and 18 % chromium.

Distinguish between: (Two points of distinction)

Question 1.
Metals and Nonmetals.
Answer:
Metals:

  1. Metals have a lustre.
  2. They are generally good conductors of heat and electricity.
  3. They are generally solids at room temperature.
    Exception: Mercury and gallium are liquids.
  4. Metals form basic oxides.

Nonmetals:

  1. Nonmetals have no lustre.
    Exception: Iodine and Diamond.
  2. They are bad conductors of heat and electricity.
    Exception: Graphite.
  3. They are generally gases and solids at room temperature.
    Exception: Bromine is a liquid.
  4. Nonmetals form acidic or neutral oxides.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 2.
Roasting and Calcination.
Answer:
Roasting:

  1. In this process, the ore is heated strongly in the presence of air.
  2. In this process, sulphide ore is converted into metal oxide.
  3. During this process SO2 is given out.

Calcination:

  1. In this process, the ore is heated strongly in the limited supply of air.
  2. In this process, carbonate ore is converted into metal oxide.
  3. During this process CO2 is given out.

Give scientific reasons for the following:

Question 1.
Calcium floats on water during the reaction with water.
Answer:

  1. Calcium reacts with water less vigorously hence the heat evolved is not sufficient for hydrogen to catch fire.
  2. Instead, calcium floats on water because the bubbles of hydrogen gas formed stick to the surface of the metal.

Question 2.
Common salt has high melting and boiling points.
Answer:

  • Common salt is an ionic compound. Common salt is solid and hard due to strong electrostatic attraction between oppositely charged Na+ and Cl- ions.
  • A large amount of energy is required to break the strong intermolecular attraction (strong ionic bond). Hence, common salt has high melting and boiling points.

Question 3.
Metals are good conductors, while non-metals are poor conductors of electricity.
Answer:

  • The electrons in the outermost orbit of atoms of a metal are free to move throughout the metal.
  • When a potential difference is applied between the ends of a metal wire, the movement of the electrons constitutes an electric current. Hence, metals are good conductors of electricity.
  • Nonmetals involve covalent bonding and do not have free electrons like metals. Hence, nonmetals are poor conductors of electricity.

Question 4.
Sodium is more reactive than aluminium.
Answer:

  • If the number of electrons in the outermost orbit of an atom of a metal is less, the metal is more reactive.
  • Sodium has electronic configuration (2, 8, 1) and aluminium has electronic configuration (2, 8, 3). The number of electrons in the outermost orbit of sodium and aluminium atoms are 1 and 3, respectively. Hence, sodium is more reactive than aluminium.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 5.
When zinc granules are added to copper sulphate solution, the blue coloured solution turns colourless.
Answer:

  • Zinc is more reactive than copper.
  • When zinc granules are added to copper sulphate solution, they displace copper from the copper sulphate solution to form zinc sulphate solution. As zinc sulphate is colourless, the blue coloured solution of copper sulphate disappears.

Question 6.
When an iron nail is dipped into a copper solution, a shiny coat of copper is formed on the nail.
Answer:

  • Iron is more reactive than copper.
  • When an iron nail is dipped into copper sulphate solution, iron displaces copper from the copper sulphate solution. The copper so liberated deposits on the iron nail. As a result, a shiny coat of copper is formed on the nail.

Question 7.
Cryolite (Na3AlF6) and fluorspar (CaF2) are added to the electrolytic mixture containing pure alumina.
Answer:
(1) Alumina has very high melting point ( > 2000 °C). Cryolite (Na3AlF6) and fluorspar (CaF2) lower the fusion temperature of the mixture containing alumina from 2000 ° C to 1000 ° C, thereby saving electrical energy.
(2) They increase the conductivity and the mobility of the fused mixture. Hence, cryolite and fluorspar are added to the electrolytic mixture containing pure alumina.

Question 8.
Air is bubbled through the mixture in Froth floatation process.
Answer:
(1) In the froth floatation process, in a tank water, ore and an oil are mixed. When air is bubbled through the mixture the oil forms froth.
(2) The mineral particles are wetted by the oil and float on the surface.
(3) The gangue particles are wetted by water and settle down. Hence, the ore can be concentrated. Hence, air is bubbled through the mixture in froth floatation process.

Question 9.
Silver amalgam is used for filling dental cavities.
Answer:
(1) Silver is a soft metal and wears off on constant usage particularly due to abrasion. Silver amalgam is an alloy of silver with mercury.
(2) It is a hard substance. It is nontoxic. Besides these properties it is a lustrous shining substance. It melts at a comparatively low temperature and can therefore conveniently fill in the cavities. Hence, silver amalgam is used for filling dental cavities.

Explain the following reactions with the help of balanced equations:

Question 1.
Out of sodium and sulphur which is a metal? Explain its reaction with oxygen. (March 2019)
Answer:
Sodium is a metal. Sodium reacts with oxygen in air at room temperature to form sodium oxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 60

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 2.
Magnesium burns in air.
Answer:
When magnesium bums in air, it combines with oxygen, emitting intense light and heat to form magnesium oxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 61

Question 3.
Copper reacts with air.
Answer:
Copper tarnishes in moist air and forms black coloured oxide when strongly heated.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 62

Question 4.
Sodium reacts with water.
Answer:
When sodium reacts with water, it evolves hydrogen which immediately catches fire producing a lot of heat.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 63

Question 5.
Calcium reacts with water.
Answer:
Calcium reacts with water less vigorously to form hydrogen gas and calcium hydroxide. In this reaction, the heat evolved is not sufficient for hydrogen to catch fire.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 64

Question 6.
Steam is passed over aluminium.
Answer:
When steam is passed over aluminium, hydrogen gas and aluminium oxide are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 65

Question 7.
Steam is passed over iron.
Answer:
When steam is passed over iron, iron (III) oxide and hydrogen gas are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 66

Question 8.
Magnesium reacts with dil. hydrochloric acid.
Answer:
When magnesium reacts with dil. hydrochloric acid, magnesium chloride is formed and hydrogen gas is evolved.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 67

Question 9.
Aluminium is treated with dil. hydrochloric acid.
Answer:
When aluminium is treated with dil. hydrochloric acid, aluminium chloride and hydrogen gas are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 68

Question 10.
Zinc reacts with dil. hydrochloric acid.
Answer:
When zinc reacts with dil. hydrochloric acid, zinc chloride and hydrogen gas are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 69

Question 11.
Iron is treated with dil. hydrochloric acid.
Answer:
When iron reacts with dil. hydrochloric acid, ferrous chloride and hydrogen gas is formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 70

Question 12.
Copper is reacted with cone, nitric acid.
Answer:
When copper is reacted with cone, nitric acid, copper nitrate and reddish brown nitrogen dioxide are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 71

Question 13.
Copper is reacted with dil. nitric acid.
Answer:
When copper is reacted with dil. nitric acid, copper nitrate and nitric oxide are formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 72

Question 14.
Sodium metal is reacted chlorine gas.
Answer:
When sodium metal is reacted with chlorine, sodium chloride an ionic compound is formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 73

Question 15.
Sulphur burns in air.
Answer:
When sulphur burns in air, it combines with oxygen to form acidic sulphur dioxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 74

Question 16.
Chlorine dissolves in water.
Answer:
When chlorine dissolves in water, hypochlorous acid is formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 75

Question 17.
Chlorine is treated with hydrobromic acid.
Answer:
When chlorine is treated with hydrobromic acid, chlorine displaces bromine from hydrobromic acid.
Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(aq)

Question 18.
Hydrogen gas is passed over boiling sulphur.
Answer:
When hydrogen gas is passed over boiling sulphur, sulphur combines with hydrogen to form hydrogen sulphide which has rotten egg smell.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 76

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Question 19.
Sodium aluminate is treated with water.
Answer:
When sodium aluminate is treated with water, it is hydrolysed to give a precipitate of aluminium hydroxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 77

Question 20.
Dry aluminium hydroxide is ignited at 1000 °C.
Answer:
When dry aluminium hydroxide is ignited at 1000 °C, alumina (Al2O3) formed.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 78

Question 21.
Zinc sulphide is heated strongly in excess of air.
Answer:
When zinc sulphide is heated strongly in excess of air, it forms zinc oxide and sulphur dioxide gas.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 79

Question 22.
Zinc carbonate is heated strongly in a limited supply of air.
Answer:
When zinc carbonate is heated strongly in a limited supply of air, it gives zinc oxide and carbon dioxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 80

Question 23.
Zinc oxide is treated with carbon.
Answer:
When zinc oxide is treated with carbon, it is reduced to zinc. In this reaction, carbon acts as reducing agent.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 81

Question 24.
Manganese dioxide is heated with aluminium.
Answer:
When manganese dioxide is heated with aluminium, manganese dioxide is reduced to manganese and large amount of heat is evolved.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 82

Question 25.
Cinnabar is heated in air.
Answer:
When cinnabar is heated in air, it forms mercuric oxide and sulphur dioxide.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 83

Question 26.
Cuprous sulphide is heated in air.
Answer:
When cuprous sulphide is heated in air, cuprous oxide is formed. Cuprous oxide is reduced to copper in the presence of ore.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy 84

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 8 Metallurgy

Project: (Do it your self)

Project 1.
Which metals are used in day to day life? What are its uses?

Project 2.
Which nonmetals are used in day to day life? What are its uses?

Maharashtra State Board Class 10 Science Solutions 

Space Missions Class 10 Questions And Answers Maharashtra Board

Std 10 Science Part 1 Chapter 10 Space Missions Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 1 Chapter 10 Space Missions Question Answer Maharashtra Board

Question 1.
Fill in the blanks and explain the statements with reasoning:
a. If the height of the orbit of a satellite from the earth’s surface is increased, the tangential velocity of the satellite will ………………
Answer:
If the height of the orbit of a satellite from the earth’s surface, is increased, the tangential velocity of the satellite will decrease.
Explanation: The gravitational force (F) exerted by the earth on the satellite will decrease if the height of the orbit of the satellite from the earth’s surface is increased. Hence, the tangential velocity of the satellite will decrease.
The formula
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 1
shows that υc decreases with increasing h.

b. The initial velocity (during launching) of the Mangalyaan must be greater than ………… from the earth.
Answer:
The initial velocity (during launching) of the Mangalyaan must be greater than the escape velocity from the earth.
Explanation: If a satellite is to travel beyond the gravitational pull of the earth, its velocity must be more than the escape velocity from the earth.
[Note: The velocity must be atleast equal to the escape velocity. Refer the definition of escape velocity.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 2.
State with reasons whether the following statements are true or false.
a. If a spacecraft has to be sent away from the influence of the earth’s gravitational field, its velocity must be less than the escape velocity.
Answer:
False.
Explanation: The escape velocity of a body is the minimum velocity with which it should be projected from the earth’s surface, so that it can escape the influence of the earth’s gravitational field. This clearly shows that the given statement is false.

b. The escape velocity on the moon is less than that on the earth.
Answer:
True.
Explanation: Escape velocity of an object from the earth,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 2

c. A satellite needs a specific velocity to revolve in a specific orbit.
Answer:
True.
Explanation:
Centripetal force on the satellite \(\frac{m v_{c}^{2}}{R+h}\) = gravitational force exerted by the earth on the satellite \(\frac{G M m}{(R+h)^{2}}\)
where,
m: mass of the satellite
υc: critical velocity of the satellite
h: height of the satellite from the surface of the earth
M: mass of the earth
R: radius of the earth
G: gravitational constant
∴ \(v_{\mathrm{c}}^{2}=\frac{G M}{R+h}\)
∴ \(v_{\mathrm{c}}=\sqrt{\frac{G M}{R+h}}\)
Thus, if the value of h changes, the value of υc also changes. It means a satellite needs to be given a specific velocity (in the tangential direction) to keep it revolving in a specific orbit.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 3

d. If the height of the orbit of a satellite increases, its velocity must also increase.
Answer:
False.
Explanation:
Centripetal force on the satellite \(\frac{m v_{c}^{2}}{R+h}\) = gravitational force exerted by the earth on the satellite \(\frac{G M m}{(R+h)^{2}}\)
where,
m : mass of the satellite
υc : critical velocity of the satellite
h : height of the satellite from the surface of the earth
M : mass of the earth
R : radius of the earth
G : gravitational constant
∴ \(v_{\mathrm{c}}^{2}=\frac{G M}{R+h}\)
∴ \(v_{\mathrm{c}}=\sqrt{\frac{G M}{R+h}}\)
Thus, if the value of h changes, the value of υc also changes. It means a satellite needs to be given a specific velocity (in the tangential direction) to keep it revolving in a specific orbit.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 4
As per the formula υc = \(\sqrt{\frac{G M}{R+h}}\) , if the value of h increases, the value of υc decreases. Hence, if the height of the satellite from the surface of the earth increases, its velocity decreases.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 3.
Answer the following questions:
a. What is meant by an artificial satellite? How are the satellites classified based on their functions?
(OR)
Write the importance of artificial satellites in your words. (Practice Activity Sheet – 3)
Answer:
A manmade object orbiting the earth or any other planet is called an artificial satellite. Satellites work on solar energy and hence photovoltaic panels are attached on both sides of the satellite, which look like wings. Satellites are also installed with various transmitters and other equipment to receive and transmit signals between the earth and the satellites.

Classification of satellites depending on their functions:
(1) Weather satellites: weather satellites collect the information regarding weather conditions of the region. It records temperature, air pressure, wind direction, humidity, cloud cover, etc. this information is sent to the space research station on the earth and then with this information weather forecast is made.

(2) Communication satellites: In order to establish communication between different places on the earth through mobile phones or computer assisted internet, communication satellites are used. Many artificial satellites placed at various locations in the earth’s orbit are well interconnected and help us to have communication with any place, from anywhere, at any time and in any form including voicemail, email, photographs, audio mail, etc.

(3) Broadcasting satellites: Broadcasting satel¬lites are used to transmit various radio and television programs and even live programs from any place on the earth to any other place. As a result, one can have access to information about current incidents, events, programs, sports and other events right from his drawing room with these satellites.

(4) Navigational satellites: Navigational satel¬lites assist the surface, water and air transportation and coordinate their busy schedule. These satellites also assist the user with current live maps as well as real time traffic conditions.

(5) Military satellites: Every sovereign nation needs to keep the real time information about the borders. Satellites help to monitor all movements of neighboring countries or enemy countries. Military satellites also help to guide the missiles effectively.

(6) Earth observation satellites: These satellites observe and provide the real time information about the earth. These satellites also help us to collect information about the resources, their management, continuous observation about a natural phenomenon and the changes within it.

(7) Other satellites: Apart from these various satellites, certain satellites for specific works or purposes are also sent in the space. E.g. India has sent EDUSAT for educational purpose; CARTOSAT for surveys and map making. Similarly, satellites with telescopes, like Hubble telescope or a satellite like International Space Station help to explore the universe. In fact, ISS (International Space Station) provides a temporary residence where astronauts can stay for a certain short or long period and can undertake the research and study space activities.
The various functions listed above show the importance of artificial satellites.

b. What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
Answer:
Orbit of a satellite is its path around the earth.
Orbits of artificial satellites can be classified on various basis.
(1) On the basis of the angle of the orbital plane: Orbital plane of a satellite can be the equatorial plane of the earth or it can be at an angle to it.
(2) On the basis of the nature of the orbit: Orbital plane can be circular or elliptical in shape.
(3) On the basis of the height of the satellite: Orbit of a satellite can be HEO, MEO or LEO.

(i) High Earth Orbit (HEO) satellite: A satellite orbiting at a height equal to or greater than 35780 km above the earth’s surface is called a High Earth Orbit satellite. The critical velocity (υc) of a satellite revolving in an orbit at 35780 km above the earth surface is 3.08 km/s. Such a satellite will take about 23 hours 54 minutes to complete one revolution around the earth. The earth completes one rotation about its axis in the same time. The orbital plane of such a satellite is the equatorial plane of the earth. The satellite’s relative position appears stationary with respect to a place on the earth. This satellite is, therefore, called a geostationary satellite or geosynchronous satellite.

(ii) Medium Earth Orbit (MEO) satellite: A satellite orbiting at a height between 2000 km and 35780 km above the earth’s surface is called a Medium Earth Orbit satellite. The orbital path of such a satellite is normally elliptical and passes through the North and the South polar regions. These satellites take about 12 hours to complete one revolution around the earth.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 5

(iii) Low Earth Orbit (LEO) satellite:
A satellite orbiting at a height between 180 km and 2000 km above the earth’s surface is called a Low Earth Orbit satellite. Normally, these satellites take 90 minutes to complete one revolution around the earth. Weather satellites, space telescopes and International Space Station are Low Earth Orbit satellites.

c. Why are geostationary satellites not useful for studies of polar regions? (Practice Activity Sheet – 4)
(OR)
Explain the following statement. A geostationary satellite is not useful in the study of polar regions. (Practice Activity Sheet – 1)
Answer:
Geostationary satellites have two distinct characteristics:
(1) Geostationary satellites are HEO satellites and are placed at 35780 km above the earth’s surface.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 6

(2) A geostationary satellite revolves in the equatorial plane of the earth, and thus, it can never fly above the polar regions.
Hence, geostationary satellites are not useful for studies of polar regions.

d. What is meant by a satellite launch vehicle? Explain the satellite launch vehicle developed by ISRO with the help of a schematic diagram.
Answer:
A rocket used to carry an artificial satellite to a desired height above the earth’s surface and then project it with a proper velocity so that the satellite orbits the earth in the desired orbit is called a launch vehicle. A satellite launch vehicle needs a specific velocity as well as a thrust to reach the desired height above the earth’s surface. The velocity and the thrust of a satellite launch vehicle depend on the weight and orbital height of the satellite.

Accordingly, the structure of the launch vehicle is decided and designed. The weight of the fuel also contributes a major portion in the total weight of the launch vehicle. This also influences the structure of the launch vehicle. In order to use the fuel optimally, multiple stage launch vehicles are now designed and used.

The Polar Satellite Launch Vehicle (PSLV) developed by ISRO is shown below in a schematic diagram.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 7

e. Why is it beneficial to use a satellite j launch vehicle made up of more than one stage?
Answer:
Earlier Satellite Launch Vehicles (SLV) used to be of a single stage vehicles. Such SLVs used to be very heavy as well as expensive in terms of its fuel consumption. As a result, SLVs with multiple stages were developed.

In multistage SLVs, as the journey of the launch vehicle progresses and the vehicle achieves a specific velocity and a certain height, the fuel of the first stage is exhausted and the empty fuel tank gets detached from the main body of the launch vehicle and falls back into a sea or on unpopulated land. As the fuel in the first stage is exhausted, the engine in the second stage is Ignited. However, the weight of the launch vehicle is now less than what it was earlier and hence it can move with higher velocity, Thus, it saves fuel consumption. Hence, it is beneficial to use a multistage satellite launch vehicle.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 4.
Complete the following table:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 8
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 9

Question 5.
Solve the following problems:
a. If the mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet?
Answer:
Given:
(1) The mass of the planet (M) is eight times the mass of the earth, i.e., 8 × 6 × 1024 kg
(2) The radius of the planet (R) is twice the radius of the earth, i.e., 2 × 6.4 × 106 km
(3) G = 6.67 × 10-11 N·m2/kg2
Escape velocity for that planet
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 10
= 2.237 × 104 m/s
= 22.37 km/s

b. How much time would a satellite in an orbit at a height of 35780 km above the earth’s surface take to complete one revolution around the earth, if the mass of the earth were four times its original mass?
Answer:
Given: R (Earth) = 6400 km = 6.4 × 106 m,
M (Earth) = 6 × 1024 kg
∴ M’ = 4M = 4 × 6 × 1024 kg
h = 35780 km = 3.578 × 107 m = 35.78 × 106 m,
G = 6.67 × 10-11 N·m2/kg2, T = ?
The time that the satellite would take to complete one revolution around the earth,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 11
= Approx 4.303 × 104 s
= Approx 11.95 h
or 11 hours 57 minutes 10 seconds.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

c. If the height of a satellite completing one revolution around the earth in T seconds is h1 meters, then what would be the height of a satellite taking 2\(\sqrt{2}\) T seconds for one revolution?
Answer:
Given:
(1) Time: T seconds
(2) Height: h1
Let us assume the height of the satellite completing one revolution in 2\(\sqrt{2}\) T seconds as h2.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 12
∴ R + h2 = 2R + 2h1
∴ h2 = R + 2h1

Project:

Project 1.
Collect information about the space missions undertaken by Sunita Williams.
Hints:
The following sources can be used to get the information on the above topic:
(1) Google Search Engine
(2) YouTube
(3) E-books on Sunita Williams
(4) English and other regional language books on Sunita Williams available in your library
(5) Newspaper clippings

Based on the information you have collected from the above sources, complete the project in about 5 pages. You can do value addition to your project with the help of suitable photos, clippings, charts, graphs and sketches.

Project 2.
Assume that you are interviewing Sunita Williams. Prepare a questionnaire and also the answers.
Answer:
Points to make a list of a questionnaire for the interview of Sunita Williams :
(1) Primary and higher education
(2) The source of inspiration to become an astronaut
(3) Information about her mentor
(4) General and specific training
(5) Initial experience of being an astronaut
(6) First space mission, its nature, duration and experience
(7) Natureofresearchcarriedoutinspace
(8) Some special memories
(9) Future plans
(10) Tips and guidance for the younger generation.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Can you recall? (Text Book Page No. 135)

Question 1.
What is the difference between space and sky?
Answer:

  1. The visible portion of the atmosphere and outer space seen by simple eyes, without any equipment from the earth, is known as the sky.
  2. The infinite three-dimensional expanse in which the Solar system, stars, celestial bodies, galaxies and the endless Universe exist is known as space.
  3. Both sky and space lack a definite boundary. However, the sky is a very tiny part of space.

Question 2.
What are different objects in the Solar system?
Answer:

  1. Our Solar system is a very tiny part of a huge Galaxy-Milky Way.
  2. The Sun is at the centre of the Solar system. Sun is a star.
  3. Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune are planets in our Solar system. These planets revolve around the Sun. Some of these planets have their own natural satellites.
  4. Besides, there are asteroids, meteoroids, comets and meteors in the Solar system.

Question 3.
What is meant by a satellite?
Answer:

  1. An astronomical object orbiting any planet of our Solar system is called a satellite.
  2. Mercury and Venus have no satellites.
  3. Some planets have more than one satellite. E.g. Jupiter has 69 satellites.

Question 4.
How many natural satellites does the earth have?
Answer:
The earth has one natural satellite called the moon.

Question 5.
Which type of telescopes are orbiting around the earth? Why is it necessary to put them in space?
Answer:
(1) The following three types of telescopes are orbiting around the earth:

  • Optical Refracting Telescope.
  • Optical Reflecting Telescope.
  • Radio Telescope.

(2) Visible light and radio waves emitted by celestial bodies in space pass through the atmosphere before reaching the earth’s surface. During this journey, some light is absorbed by the atmosphere. Hence, the intensity of the light reaching the earth’s surface decreases. Besides, temperature and air pressure cause the atmospheric turbulence. Hence, light rays change their path, resulting in a change in the position of the image of a celestial body.

City lights during night, and bright sunlight during day also put limitations on usage of optical telescopes on the earth. To minimize these problems, optical telescopes are situated on mountain top, away from inhabited places. However, limitations caused by the atmosphere still persist.

To get rid of these problems scientists have successfully launched telescopes in space. Images obtained by these telescopes are brighter and clearer than those obtained by the telescopes located on the earth’s surface.

Can you recall? (Text Book Page No. 135)

Question 1.
Where does the signal in your cellphone come from?
Answer:
In nearby area of our residence, many mobile towers are installed at various places. Cellphones receive signals from one of these mobile towers.

Question 2.
Where from does it come to mobile towers?
Answer:
All mobile towers are connected to satellites. Cellphone signal reaching the nearest mobile tower in our vicinity is first transmitted to the satellite. The satellite transmits the signal to the mobile tower near the destination.

Question 3.
Where does the signal to your TV set come from?
Answer:
(1) Television Centre or Studio transmits the TV program which first reaches the satellite. The dish antenna of the cable operator in our area receives these signals. The TV programs reach our TV set through a cable connected between the cable operator’s receiving station and our TV set.

(2) Alternatively, a small portable dish antenna fixed on the rooftop is also used to receive the TV signals directly from the satellites. Finally, a cable connected to the dish antenna and TV set brings the programme to our TV set.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 4.
You may have seen photographs showing the position of monsoon clouds over the country in the newspaper. How are these images obtained?
Answer:
Weather satellites take photographs of the sky above the earth’s surface at regular intervals. Some satellites, capable of receiving radio signals, also collect the information of weather conditions and finally images of the sky are built with computers. Territorial boundaries of the states and the country are drawn later on these images. Such satellite images with imposed boundaries are printed in media or shown on the television.

Fill in the blanks:

Question 1.
A man-made object revolving around the earth in a fixed orbit is called …………..
Answer:
A man-made object revolving around the earth in a fixed orbit is called an artificial satellite.

Question 2.
Chandrayaan-I discovered the presence of ………….. on the moon.
Answer:
Chandrayaan-I discovered the presence of water on the moon.

Question 3.
Apart from launching a satellite around the earth, India has been able to launch a satellite around ……………
Answer:
Apart from launching a satellite around the earth, India has been able to launch a satellite around Mars.

Question 4.
All satellites work on …………… energy.
Answer:
All satellites work on solar energy.

Question 5.
……………. are used to carry and place a satellite in a specific orbit.
Answer:
Satellite launchers are used to carry and place a satellite in a specific orbit.

Question 6.
USA has developed ……………. as an alternative to space launch vehicles.
Answer:
USA has developed space shuttles as an alternative to space launch vehicles.

Question 7.
Hubble telescope is a ………….. satellite.
Answer:
Hubble telescope is a Low Earth Orbit (LEO) satellite.

Question 8.
……………. executed the first ever mission to the moon in the world.
Answer:
Russia executed the first ever mission to the moon in the world.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 9.
………… executed the first manned mission to the moon in the world.
Answer:
USA executed the first manned mission to the moon in the world.

Select the appropriate answer from given options:

Question 1.
Which one of the following is a Low Earth Orbit (LEO) satellite?
(a) Navigational satellite
(b) Geostationary satellite
(c) International Space Station
(d) All of the above
Answer:
(c) International Space Station

Question 2.
Which of the following satellite launchers is developed by India?
(a) INSAT
(b) IRNSS
(c) EDUSAT
(d) PSLV
Answer:
(d) PSLV

Question 3.
Which of the following astronauts travelled through space shuttle ‘Discovery’ first time? (Practice Activity Sheet – 4)
(a) Kalpana Chawla
(b) Rakesh Sharma
(c) Sunita Williams
(d) Neil Armstrong
Answer:
(c) Sunita Williams

Considering the correlation between the words of the first pair, pair the third word accordingly with proper answer. (OR) Considering the first correlation, complete the second.

Question 1.
IRNSS : Direction showing satellite :: INSAT :………… (Practice Activity Sheet – 1)
Answer:
IRNSS : Direction showing satellite :: INSAT : Weather satellite

Question 2.
Hubble telescope : 569 km high from the earth’s surface :: Revolving orbit of Hubble telescope :………. (Practice Activity Sheet – 2; March 2019)
Answer:
Hubble telescope : 569 km high from the earth’s surface :: Revolving orbit of Hubble telescope : Low Earth Orbit.

Match the column:

Question 1.

Column AColumn B
(1) Clouds over India(a) Low Earth Orbit
(2) Global communication(b) PSLV
(3) Launch vehicle made by ISRO(c) Communication satellite
(4) International Space Station(d) EDUSAT
(5) Navigational satellite(e) Weather satellite
(f) Medium Earth Orbit

Answer:
(1) Clouds over India – Weather satellite
(2) Global communication – Communication satellite
(3) Launch vehicle made by ISRO – PSLV
(4) International Space Station – Low Earth Orbit
(5) Navigational satellite – Medium Earth Orbit.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Answer the following questions in one sentence each:

Question 1.
What do you mean by the orbit of a satellite?
Answer:
Orbit of a satellite is its path around the earth.

Question 2.
Which factor decides the orbit of a satellite?
Answer:
The function of a satellite decides the orbit of the satellite.

Question 3.
What is a High Earth Orbit satellite?
Answer:
A satellite orbiting at a height equal to or greater than 35780 km above the earth’s surface is called a High Earth Orbit satellite.

Question 4.
Give two examples of Low Earth Orbit satellites.
Answer:
Weather satellite and International Space Station are Low Earth Orbit satellites.

Question 5.
What is a launch vehicle?
Answer:
A rocket used to carry an artificial satellite to a desired height above the earth’s surface and then project it with a proper velocity so that the satellite orbits the earth in the desired orbit is called a launch vehicle.

Question 6.
Name the launch vehicle developed by India.
Answer:
The launch vehicle developed by India is known as PSLV, i.e., Polar Satellite Launch Vehicle.

Answer the following questions:

Question 1.
Write the proper name of the orbits of satellites shown in the following figure with their height from the earth’s surface. (Practice Activity Sheet – 4)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 13
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 14
(a) Low earth orbits: height above the earth’s surface: 180 km to 2000 km
(b) Medium earth orbits: height above the earth’s surface: 2000 km to 35780 km
(c) High earth orbits: height from the earth’s surface > 35780 km

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 2.
Explain the need and importance of space missions.
Answer:
Man has always been curious about the sun, moon, stars and the world beyond the earth. Initially, man tried to observe space with the help of telescopes. However, later he dreamt to fly into space and finally succeeded to reach into space.

Space missions are now essential to understand the origin and evolution of our solar system as well as to study the Universe beyond the Solar system.

Space missions have given us many benefits and made our life simpler. It is because of space missions that the real-time immediate communication and exchange of information across the globe is now possible. We can receive the abundant information at the desk at our home or office. We also get information about any topic at any time and anywhere at fingertips through the Internet. Besides, the advanced alerts about some natural calamities like cyclones or storms are received through satellites sent as a part of space missions. Satellites have also helped us in entertainment. Programmes, sports events, etc., can be telecast live and can reach millions at a time throughout the world.

Satellite surveillance of the enemy, exploring the reserves of various minerals resources, access to various activities like trade, tourism and navigation, and easy global reach to make world a global village is all possible due to the space missions. Thus, space missions are extremely important in defence, communication, weather forecast, observation, direction determination, etc.

Question 3.
What are space expeditions? Explain their need and importance in your words. (Practice Activity Sheet – 2)
Answer:
A mission planned (i) for establishing artificial satellites in the earth’s orbit, using them for research or for the benefit of life, or (ii) for sending a spacecraft to the various components of the solar system or outside is called a space expedition.

Man has always been curious about the sun, moon, stars and the world beyond the earth. Initially, man tried to observe space with the help of telescopes. However, later he dreamt to fly into space and finally succeeded to reach into space. Space missions are now essential to understand the origin and evolution of our solar system as well as to study the Universe beyond the Solar system.

Space missions have given us many benefits and made our life simpler. It is because of space missions that the real-time immediate communication and exchange of information across the globe is now possible. We can receive the abundant information at the desk at our home or office. We also get information about any topic at any time and anywhere at fingertips through the Internet. Besides, the advanced alerts about some natural calamities like cyclones or storms are received through satellites sent as a part of space missions. Satellites have also helped us in entertainment. Programmes, sports events, etc., can be telecast live and can reach millions at a time throughout the world.

Satellite surveillance of the enemy, exploring the reserves of various minerals resources, access to various activities like trade, tourism and navigation, and easy global reach to make world a global village is all possible due to the space missions. Thus, space missions are extremely important in defence, communication, weather forecast, observation, direction determination, etc.

Question 4.
What are the objectives of the space mission?
Answer:
Man initially tried to satisfy his curiosity to know the world and universe beyond the earth with the help of telescopes. However, it has some obvious limitations and to overcome these limitations, man later ventured into space missions.

Space missions carried out by man were aimed at four specific objectives:

  1. To launch artificial satellites in the earth’s orbit for study and research.
  2. To launch artificial satellites in the earth’s orbit for various purposes like telecommunication, weather forecast, radio and TV programme transmission, etc.
  3. To send artificial satellites beyond the earth’s orbit to observe, study and collect the information from other planets, meteors, meteoroids, asteroids and comets.
  4. To sense and understand space beyond the solar system.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 5.
Write on significant space missions carried out by man.
Answer:
Man has carried out many space missions within and beyond the earth’s orbit. Significant space missions are as follows:
(1) Space missions within the earth’s orbit: Man has so far sent many artificial satellites of various types in the earth’s orbit. These satellites have made the life of man simpler. Besides, it has also helped us in resource management, communication, disaster management, etc.

(2) Moon missions : Moon is the natural satellite , of the earth and it is the nearest celestial body to us. Naturally, our initial space missions were directed to the moon. As of now, only Russia, USA, European Union, China, Japan and India have successfully undertaken . moon missions. Russia executed 15 moon missions between 1959 and 1976. Of these, last 4 missions brought the stone samples from the moon for study and analysis. However all these missions were unmanned. USA executed moon missions between 1962 and 1972. Some of these missions were unmanned.

However, the historic moon mission took place on 20th July, 1969, when American astronaut Neil Armstrong became the first human to step on the moon. India has undertaken the moon mission. Indian Space Research Organisation (ISRO) successfully launched Chandrayaan-I and placed it in orbit of the moon. It sent useful information to the earth for about a year. The most important discovery made during the mission was the presence of water on moon’s surface. India was the first country to discover this.

(3) Mars mission: The second nearest celestial object to the earth is Mars and many nations sent spacecraft towards it. But only few of them have been successful. However, the performance of Mangalyaan, the Indian spacecraft sent by ISRO towards Mars, was remarkable. Mangalyaan was launched in November 2013 and was placed in the orbit of Mars successfully in September 2014. It has obtained useful information about the surface and atmosphere of Mars.

(4) Space missions to other planets: Other than moon and Mars missions, many other space missions were undertaken for studying other planets. Some spacecraft orbited the planets, some landed on some planets, and some just observed the planets, passed near them and went further to study other celestial bodies. Some spacecraft were sent specifically to study asteroids and comets. Some spacecraft’s have brought dust and stone samples from asteroids for the study.

All these space missions are very useful in getting information and helping us in clarifying our concepts about the origin of the earth and the Solar system.

Question 6.
Bring out the contribution of India’s space missions.
Answer:
Successful space missions as well as scientific and technological accomplishments by India in space technology have made a significant contribution in the national and social development of our country.
India has indigenously built various launchers and these launchers can put the satellites having the mass up to 2500 kg in orbit.

Indian Space Research Organisation (ISRO) has designed and built two important launchers: Polar Satellite Launch Vehicle (PSLV) and Geosynchronous Satellite Launch Vehicle (GSLV).

Many satellites in INS AT and GSAT series are active in telecommunication, television broadcasting, meteorological services, disaster management and in monitoring and management of natural resources. EDUSAT is used specifically for education while satellites in IRNSS series are used for navigation. Thumba, Sriharikota and Chandipur are Indian satellite launch centers.

Vikram Sarabhai Space Centre at Thiruvananthapuram, Satish Dhawan Space Research Centre at Sriharikota and Space Application Centre at Ahmedabad are space research organizations of India.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Question 7.
What is meant by space debris? Why is there need to manage the debris? (March 2019)
Answer:
In a space nonessential objects such as the parts of launchers and satellites, revolving around the earth are called the debris in space.

The debris can be harmful to the artificial satellites. It can collide with the satellite or spacecrafts and damage them. Therefore the future of artificial satellites or spacecrafts are in danger.
Hence, it is necessary to manage the debris.

Solve the following examples/numerical problems:
[Note: See the textbook for the relevant data.]

Problem 1.
If the mass of a planet is 8 times that of the earth and its radius is twice the radius of the earth, what will be the escape velocity for that planet? (Escape velocity for the earth = 11.2 km/s) (Practice Activity Sheet – 2)
Answer:
Given:
Mass of the planet = 8ME radius of the planet, Rp = 2RE,
escape velocity for the earth, υescE = 11.2 km/s
escape velocity for the planet, υescP = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 15

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Problem 2.
Calculate the critical velocity (υc) of the satellite to be located at 35780 km above the surface of the earth.
Answer:
Given:
G: 6.67 × 10-11 N·m2/kg2,
M(Earth): 6 × 1024 kg,
R(Earth): 6.4 × 106 m,
h: 35780 km = 35780 × 103 m,
υc = ?
Critical velocity of the satellite
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 16
= 3.08 × 103 m/s
= 3.08 Km/s.

Problem 3.
In the above example (2) how much time will the satellite take to complete one revolution around the earth?
Answer:
Given:
R: 6400mkm = 6.4 × 106 m
h: 35780 km = 3.5780 × 107 m
v: 3.08 km/s = 3.08 × 103 m/s
T = ?
The time required for the satellite to complete one revolution around the earth,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions 17
= Approx 86060 s
= 23 hours 54 minutes 20 seconds

Problem 4.
Calculate the critical velocity (υc) of the satellite to be located at 2000 km above the surface of the earth.
Answer:
Refer to the example (2) above.
Here,h = 2 × 106 m
υc = 6902 m/s

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 10 Space Missions

Problem 5.
In the above example (4), how much time will the satellite take to complete one revolution around the earth?
Answer:
Refer to example (3) above.
Approx 7647 s
= 2 hours 7 minutes 27 seconds.
[Note: For more solved problems and problems for practice, refer Chapter 1 (Gravitation)]

Maharashtra State Board Class 10 Science Solutions 

Life Processes in Living Organisms Part – 2 Class 10 Questions And Answers Maharashtra Board

Std 10 Science Part 2 Chapter 3 Life Processes in Living Organisms Part – 2 Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part – 2 Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 2 Chapter 3 Life Processes in Living Organisms Part – 2 Question Answer Maharashtra Board

Question 1.
Complete the following chart.

Asexual reproductionSexual reproduction
1. Reproduction that occurs with the help of somatic cells is called asexual reproduction.1. __________________________________________
____________________________________________
2. __________________________________________
____________________________________________
2. Male and female parent are necessary for sexual reproduction.
3. This reproduction occurs with the help of mitosis only.3. __________________________________________
____________________________________________
4. __________________________________________
____________________________________________
4. New individual formed by this method is genetically different from parents.
5. Asexual reproduction occurs in different individuals by various methods like binary fission, multiple fission, budding, fragmentation, regeneration, vegetative propagation, spore production, etc.5. __________________________________________
_____________________________________________
_____________________________________________
_____________________________________________

Answer:

Asexual reproductionSexual reproduction
1. Reproduction that occurs with the help of somatic cells is called asexual reproduction.1. Reproduction that occurs due to fertilization of gametes is called sexual reproduction.
2. For asexual reproduction only one parent is necessary.2. Male and female parents are necessary for sexual reproduction.
3. This reproduction occurs with the help of mitosis only.3. This reproduction occurs with the help of both mitosis and meiosis.
4. New individual formed by this method is genetically identical with parents.4. New individual formed by this method is genetically different from parents.
5. Asexual reproduction occurs in different individuals by various methods like binary fission, multiple fission, budding, fragmentation, regeneration, vegetative propagation, spore production, etc.5. Sexual reproduction occurs in two steps: First formation of haploid gametes by meiosis and then fertilization of these haploid gametes to form diploid zygote. There are no subtypes in the sexual reproduction.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 2.
Fill in the blanks.
a. In humans, sperm production occurs in the organ …………
(a) prostate gland
(b) testis
(c) ovaries
(d) Cowper’s gland
Answer:
(b) testis

b. In humans, …………. chromosome is responsible for maleness.
(a) X
(b) Y
(c) Z
(d) O
Answer:
(b) Y

c. In male and female reproductive system of human, …………. gland is same.
Answer:
There is no similar gland in male and female reproductive system. There may be some homologies but there is no similarity.

d. Implantation of embryo occurs in …………
(a) ovaries
(b) fallopian duct
(c) uterus
(d) vagina
Answer:
(c) uterus

e. …………type of reproduction occurs without fusion of gametes.
(a) Asexual
(b) sexual
(c) Fertilization
(d) Gamete formation
Answer:
(a) Asexual

f. Body breaks up into several fragments and each fragment begins to live as a new individual.
This is ………. type of reproduction.
(a) regeneration
(b) fragmentation
(c) binary fission
(d) budding
Answer:
(b) fragmentation

g. Pollen grains are formed by division in locules of anthers.
(a) meiosis
(b) mitosis
(c) amitosis
(d) binary
Answer:
(a) meiosis

Question 3.
Complete the paragraph with the help of words given in the bracket:
(Luteinizing hormone, endometrium of uterus, follicle stimulating hormone, estrogen, progesterone, corpus luteum)
Growth of follicles present in the ovary occurs under the effect of ………….. This follicle secretes estrogen. Ovarian follicle along with oocyte grows/regenerates under the effect of estrogen. Under the effect of ………….., fully grown up follicle bursts, ovulation occurs and …………….. is formed from remaining part of follicle. It secretes ……………. and ………. Under the effect of these hormones, glands of ……….. are activated and it becomes ready for implantation.
Answer:
Growth of follicles present in the ovary occurs under the effect of follicle stimulating hormone. This follicle secretes estrogen. Ovarian follicle along with oocyte grows/regenerates under the effect of estrogen. Under the effect of Luteinizing hormone, fully grown up follicle bursts, ovulation occurs and corpus luteum is formed from remaining part of follicle. It secretes estrogen and progesterone. Under the effect of these hormones, glands of endometrium of uterus are activated and it becomes ready for implantation.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 4.
Answer the following questions short.
a. Explain with examples types of asexual reproduction in unicellular organism.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 1
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 2
There are different methods of asexual reproduction in different unicellular animals.
(1) Binary fission: The process in which the parent cell divides to form two similar daughter cells
is binary fission. It takes place either by mitosis or amitosis. When there are favourable conditions and abundant food supply then the organisms undergo binary fission. Prokaryotes, Protists and eukaryotic 5 cell-organelle like mitochondria and chloroplasts perform binary fission.

Based on axis of fission there are three subtypes of binary fission.
(a) Simple binary fission: The plane of division is not definite, it can be in any direction due to lack of specific shape as in Amoeba.
(b) Transverse binary fission: The plane of J division is transverse, as in Paramoecium.
(c) Longitudinal binary fission: The plane of division is in length-wise direction as in Euglena.

(2) Multiple fission:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 3
During unfavourable conditions when there is lack of food, multiple fission is shown by amoeba. Amoeba forms protective covering and becomes encysted. Inside the cyst, amoeba undergoes repeated nuclear division. This is followed by cytoplasmic divisions. Many amoebulae are formed which remain dormant inside the cyst. When favourable conditions reappear, they come out by breaking the cyst.

(3) Budding in yeast:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 4
Yeast is unicellular fungus that performs budding. The parent cell produces two daughter nuclei by mitotic division. This results in a small bulgingbud on the surface of parent cell. One daughter nucleus enters the bud. It then grows and upon becoming big it separates from the parent cell to have independent life as new yeast cell.

b. Explain the concept of IVF.
Answer:
(1) IVF means In Vitro Fertilization (IVF)
(2) This is the technique in the modern medical field where childless couples can be blessed by their own child.
(3) IVF technique is used for childless couples who are faced with problems such as less sperm count, obstacles in oviduct, etc.
(4) The IVF technique is done by removing the oocyte from the mother and artificially fertilizing by the sperms collected from father. This fertilization is done in a test-tube. Thus it is also called test tube baby. The embryo formed is implanted in uterus of real mother or a surrogate mother at appropriate time.

c. Which precautions will you follow to maintain the reproductive health?
Answer:
About reproductive health one should have scientific and authentic information. The cleanliness of body is very essential but keeping the mind clean is also important to maintain good reproductive health. One should be careful about sexual relationships. These things should not be experimented in young age. Mistakes committed like these can change the sexual health forever. The cleanliness and hygiene during menstruation, the cleanliness of genitals and other private parts are the aspects of personal hygiene. When living in a society, one should always be away from cross-infections of venereal type.

d. What is menstrual cycle? Describe it in brief.
Answer:

  • Menstrual cycle is the events of cyclic changes that takes place with the interval of 28 to 30 days in mature woman.
  • Hormones from pituitary, FSH (Follicle Stimulating Hormone) and LH (Luteinizing Hormone) and hormones from ovary, estrogen and progesterone control the menstrual cycle.
  • Due to influence of FSH, the ovarian follicle grows along with the oocyte that is present in it.
  • This growing follicle produces estrogen.
  • Under the influence of estrogen, the uterine inner layer called endometrium grows or regenerates. In the meantime the development of follicle is completed.
  • LH from pituitary stimulates the bursting of ovarian follicle and releases the mature oocyte out of the follicle and the ovarian wall. This process is called ovulation.
  • The empty ovarian follicle after the ovulation becomes corpus luteum. Corpus luteum produces hormone progesterone.
  • Under the influence of progesterone, the glands from uterine endometrium start secreting. The oocyte if fertilized is implanted over this endometrium.
  • If oocyte is not fertilized, the corpus luteum becomes a degenerate body called corpus albicans. The corpus albicans cannot secrete estrogen and progesterone.
  • Due to lack of these hormones, the endometrial layer of the uterus collapses. The tissue debris, along with unfertilized egg is given out through the vagina as menstrual flow. This results in bleeding for about 5 days.
  • If woman is not pregnant, then this menstrual cycle keeps on repeating with regularity.

Question 5.
In case of sexual reproduction, newborn show similarities about characters. Explain this statement with suitable examples.
Answer:
(1) Sexual reproduction occurs due to two different gametes. One male gamete is from father while the other female gamete is from mother.
(2) Both the gametes are produced by meiosis.
(3) When the gametes unite it is called process of fertilization which produces diploid zygote.
(4) Due to the chromosomes of parents, their DNA pass to the next generation through such fertilization. Therefore, the characters of newborn show similarities with parents.

Question 6.
Sketch the labelled diagrams.
a. Human male reproductive system.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 5

b. Human female reproductive system.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 6

c. Flower with its reproductive organs.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 7

d. Menstrual cycle.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 8

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 7.
Give the names.
a. Hormones related with male reproductive system.
Answer:
Follicle stimulating hormone and ICSH or Luteinizing hormone secreted by pituitary gland, testosterone secreted by testis.

b. Hormones secreted by ovary of female reproductive system.
Answer:
Estrogen and progesterone.

c. Types of twins.
Answer:
Monozygotic twins, Siamese twins and Dizygotic twins.

d. Any two sexual diseases.
Answer:
Gonorrhea and Syphilis.

e. Methods of family planning.
Answer:
Copper T, condoms, oral contraceptive pills.

Question 8.
Gender of child is determined by the male? partner of couple. Explain with reasons whether this statement is true or false.
(OR)
“A couple shall have a male child or female child totally depends upon husband”. Prove truthfulness of this statement with scientific reason.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 9
Sex determination in Human being
(1) The statement Gender of child is determined by the male partner of couple is true.
(2) It is clearly seen from the diagram that there are two types of sperms produced by males. One sperm has a X chromosome while the other has a Y chromosome, apart from autosomes. The mother on the other hand has all X bearing oocytes. Thus the sperm that fertilizes the oocyte decides the sex of the child.
(3) If X bearing sperm fertilizes the oocyte, daughter is born and when Y bearing sperm fertilizes the oocyte, son is born.
(4) Thus father or male partner is responsible for the determination of the sex.

Question 9.
Explain asexual reproduction in plants.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 10

  • Vegetative propagation is the method of asexual reproduction in plants.
  • It takes place with the help of vegetative parts like root, stem, leaf and bud.
  • Potato, suran (Amorphophallus) and other tubes propagate with the help of ‘eyes’ which are buds. These eyes are present on the stem tubers.
  • In case of plants like sugarcane and grasses, buds present on nodes perform vegetative propagation.
  • Plants like Bryophyllum performs vegetative propagation with the help of buds present on leaf margin.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 10.
Modern techniques like surrogate mother, sperm bank and IVF technique will help the human beings. Justify this statement.
OR
Despite various diagnostic tests, a couple could not have a child. In this situation, which remedies will you suggest? (July 2019)
Answer:
(1) Some couples want a child but they are not able to bear one due to various problems either in mother or in father. In such cases modern techniques such as IVF, surrogacy and sperm bank are useful in conceiving a child.

(2) These methods are as follows:
(i) Surrogacy: In woman if there is problem regarding the implantation of embryo in uterus, then help of another women is taken. This women is called surrogate mother.

Oocyte from real mother is taken out and fertilized with sperms collected from her husband. These gametes are fertilized outside in a test-tube and then the fertilized zygote is implanted in the surrogate mother.

(ii) In Vitro Fertilization (IVF) is done when there are problems like less sperm count or obstacles in oviduct. In IVF, fertilization is done in the test-tube. The embryo formed is implanted in uterus of woman for further growth.

(iii) Sperm bank: If man has problems with the sperm production, then the sperms are collected from the sperm bank. Sperm bank is the place where the donor’s donate the sperms and such sperms are kept stored. The donor’s identity is kept secret and he should also be physically and medically fit person.

Question 11.
Explain sexual reproduction in plants.
(OR)
Explain double fertilization in plants.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 11

  • Plants reproduce sexually with the help of flowers.
  • Androecium and gynoecium are male and female parts of the flowers respectively.
  • In the carpel, the ovule undergoes meiosis and forms embryo sac.
  • A haploid egg cell and two haploid polar nuclei are present in each embryo sac.
  • The pollen grains from the anther reach the stigma of flower by the process of pollination. They germinate here on the stigma.
  • As a result of germination, long pollen tube and two male gametes are formed.
  • The pollen tube travels through the style of flower and the male gametes present in the pollen tube are transferred till the embryo sac in ovary. Upon reaching there, tip of the pollen tube bursts releasing two male gametes in embryo sac.
  • One male gamete unites with the egg cell and forms zygote. While other male gamete unites with two polar nuclei forming the endosperm.
  • Because there are two nuclei participating in this process, therefore it is called double fertilization.
  • After fertilization ovule develops into seed and ovary forms a fruit. When the seed again gets favourable conditions, it can produce a new plant.

Activities: (Do it your self)

Question 1.
Collect the official data about present and a decade old population of various Asian countries and plot a graph of that data. With the help of it, draw your conclusions about demographic changes.

Question 2.
With the help of your teacher, compose and present a road show to increase the awareness about prenatal gender detection and gender bias.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Can you recall? (Text Book Page No. 22)

Question 1.
Which are the important life processes in living organisms?
Answer:
The important life processes in living organisms are respiration, circulation, nutrition, excretion, sensation and response through nervous system.

Question 2.
Which life processes are essential for production of energy required by body?
Answer:
The oxidation of nutrients that are absorbed in body is done because of oxygen supplied to cells by respiratory and circulatory system. This helps in liberation of energy. Thus respiration, circulation and nutrition are the life processes that are essential for production of energy required by body.

Question 3.
Which are main types of cell division? What are the differences?
Answer:
The main types of cell division are mitosis and meiosis. In mitosis, the chromosome number remains the same. 2 daughter cells are obtained from one cell. In meiosis, the chromosome number is reduced to half. From one cell, four daughter cells are obtained.

Question 4.
What is the role of chromosomes in cell division?
Answer:
Due to chromosomes, the DNA from parental cells enter into daughter cells. The hereditary, characters are transmitted to next generation by cell division.

Can you recall? (Text Book Page No. 22)

Question 1.
What do we mean by maintenance of species?
Answer:
Maintenance of species means a species undertakes successful reproduction and produces individuals of its own kind. This keeps the species existing on the earth.

Question 2.
Whether the new organism is genetically exactly similar to earlier one that has produced it?
Answer:
No. The new organism produced from the old one is not genetically exactly similar to the parents. In meiotic cell division there is crossing over in the homologous chromosomes. This produces genetic recombination. Thus the new organism is different from the earlier one. However, if the reproduction is of asexual type, then the young one is exactly similar to the parents.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 3.
Who determines whether the two organisms of a species will be exactly similar or not?
Answer:
The type of reproduction, whether it is asexual or sexual, the type of crossing over, the extent of genetic recombination, etc. determine the similarity among the parent organisms and their, offspring. Based on this genetic recombination the two organisms of a species do not show exact similarity. However, in case of monozygotic twins there is exact similarity. In asexual reproduction to there is similarity.

Question 4.
What is the relationship between the cell division and formation of new organism of same species by earlier existing organism?
Answer:
In the process of reproduction, there is division of chromosomes. Due to cell division, the gametes are formed. The union of gametes produce new offspring. In sexual reproduction, all these processes take place due to cell division. In asexual reproduction too there is cell division. Growth of new organism also occurs due to cell division.

Let’s Think: (Text Book Page No. 26)

Question 1.
What would have been happened if the male and female gametes had been diploid?
Answer:
Diploid (2n) gametes if united, they will form 4n, i.e. tetraploid variety. Such zygote will show severe abnormality. The chromosome number will not be maintained.

Question 2.
What would have been happened if any of the cells in nature had not been divided by meiosis?
Answer:
If meiosis does not happen the gametes produced will be diploid. This will create abnormality.

Can you recall? (Text Book Page No. 28)

Question 1.
Which different hormones control the functions of human reproductive system through chemical coordination?
Answer:
Pituitary gland secretes FSH and LH. LH is known as ICSH in males, as its function in the male body is different. From the gonads of male and female, hormones are secreted which are essential for male and female reproductive functions respectively. These hormones are testosterone secreted from testis in males and estrogen and progesterone secreted from the ovaries in females. Testosterone is essential for masculinity as well as for sperm production while female hormones are essential for changes in the female body leading to motherhood.

Question 2.
Which hormones are responsible for changes in human body occurring during onset of sexual maturity?
Answer:
Testosterone in male body and estrogen in female body are responsible for maturity onset changes in human body.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 3.
Why has the Government of India enacted the law to fix the minimum age of marriage as 18 in girls and 21 in boys?
Answer:
The full growth of female body is not completed till the age of 18. Till 18 years of age the physical and emotional maturity is not attained. Therefore, she is not suitable for marriage, sexual relationship and pregnancy. Similarly, boy attains complete growth only the age of 21. Therefore, to keep individuals and their progeny safe and healthy the Government of India enacted the law to fix the minimum age of marriage as 18 in girls and 21 in boys.

Can you recall? (Text Book Page No. 31)

Question 1.
Which hormone is released from pituitary of mother once the foetal development is completed?
Answer:
The hormone oxytocin is released from the posterior pituitary of mother once the foetal development is completed.

Question 2.
Under the effect of that hormone, which organ of the female reproductive system starts to contract and thereby birth process (parturition) is facilitated?
Answer:
Due to oxytocin, uterus contracts involuntarily and the baby is expelled out. Thus initiation of birth process is possible due to contractions of uterus.

Use your brain power. (Text Book Page No. 24)

Question 1.
Does the parent cell exist after asexual reproduction-fission?
Answer:
In fission, the parent cell divides into two. This nucleus and cytoplasm, both are divided. Thus, parent cell does not exist any longer, it is converted into new cells.

Choose the correct alternative and write that alphabet against the sub-question number:

Question 1.
Pranav and Pritee are twins in your class. They belong to ……….. twins type.
(a) monozygotic
(b) dizyotic
(c) siamese
(d) none of the above
Answer:
(b) dizyotic

Question 2.
Longitudinal binary fission is seen in …………..
(a) Paramoecium
(b) Euglena
(c) Amoeba
(d) Spirogyra
Answer:
(b) Euglena

Question 3.
Yeast cell performs asexual reproduction by ……………..
(a) fragmentation
(b) budding
(c) binary fission
(d) regeneration
Answer:
(b) budding

Question 4.
Carrot and raddish undergoes …………. with the help of their roots.
(a) vegetative propagation
(b) fragmentation
(c) budding
(d) regeneration
Answer:
(a) vegetative propagation

Question 5.
Androecium and gynoecium are ……….. whorl of the flower.
(a) accessory
(b) essential
(c) external
(d) internal
Answer:
(b) essential

Question 6.
Flowers without stalk are called ……….. flowers.
(a) stalkless
(b) sessile
(c) incomplete
(d) complete
Answer:
(b) sessile

Question 7.
………….. on the inner surface of fallopian ducts (oviducts) push the egg towards uterus.
(a) Cilia
(b) Tentacles
(c) Flagella
(d) Fibres
Answer:
(a) Cilia

Question 8.
Pregnant mother supplies nourishment to her foetus through …………..
(a) breasts
(b) uterus
(c) placenta
(d) ovaries
Answer:
(c) placenta

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 9.
The length of a sperm is about …………. micrometers.
(a) 400
(b) 5
(c) 60
(d) 600 (July ’19)
Answer:
(c) 60

Question 10.
Vegetative propagation is performed with the help of ……….. in sweet potato.
(a) root
(b) stem
(c) leaf
(d) flower
Answer:
(a) root

Question 11.
Which of the following is not a unisexual flower?
(a) Coconut
(b) Papaya
(c) Gulmohar
(d) Maize
Answer:
(c) Gulmohar

Write whether the following statements are true or false, with the suitable reason:

Question 1.
Absence of genetic recombination is an advantage whereas fast process is drawback of asexual reproductive method.
Answer:
False. (Absence of genetic recombination is a drawback whereas fast process is advantage of asexual reproductive method.)

Question 2.
Prokaryotes show fission which occurs either by mitosis or amitosis.
Answer:
True. (Prokaryotes show fission by both the methods, i.e. mitosis and amitosis.)

Question 3.
During favourable conditions multiple fission is performed by amoeba.
Answer:
False. (During unfavourable conditions multiple fission is performed by amoeba.)

Question 4.
Any encysted Amoeba or any other protist is called ‘Cyst’.
Answer:
True. (Cyst is the tough capsule like structure which keeps the protists dormant inside it. This helps the organisms to tide over unfavourable conditions.)

Question 5.
If the body of Sycon breaks up accidentally into only large and few fragments, then only each fragment develops into new Sycon.
Answer:
False. (If the body of Sycon breaks up accidentally into many fragments, each fragment develops into new Sycon. Because the capacity to regenerate is very strong in poriferan Sycon, even a small piece of parent Sycon can give rise to entire new individual.)

Question 6.
Pollen tube reaches the zygote via style.
Answer:
False. (Pollen tube reaches the embryo sac via style. Later, double fertilization takes place and the zygote and endosperm are formed.)

Question 7.
There is glucose sugar in the semen.
Answer:
False. (There is fructose sugar in the semen. Glucose is not present in semen.)

Question 8.
Out of 2 – 4 million ova, approximately only 400 oocytes are released up to the age of menopause.
Answer:
True. (During the reproductive span of the woman, from menarche to menopause only one oocyte per one month is released in the span of 30 to 35 years.)

Question 9.
If the oocyte is fertilized, secretion of estrogen and progesterone stops completely.
Answer:
False. (If the oocyte is not fertilized, there is no need of corpus luteum which secretes progesterone. In absence of conception, the progesterone is not needed, thus corpus luteum degenerates and forms corpus albicans.)

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 10.
During menstruation there is need of rest along with special personal hygiene.
Answer:
True. (During phase of menstruation there is pain and bleeding in woman. Her body is also susceptible for infections. There is weakness and hence she needs rest along with special personal hygiene.)

Find the odd one out:

Question 1.
Circulation, Excretion, Sensation, Reproduction.
Answer:
Reproduction. (All others are processes necessary for survival of the individual.)

Question 2.
Budding in hydra, Regeneration, Binary fission, Fragmentation
Answer:
Binary fission. (All others are processes of asexual reproduction in multicellular organisms.)

Question 3.
Carrot, Radish, Potato, Sweet potato.
Answer:
Potato. (All others are edible roots.)

Question 4.
Vas eferens, vas deferens, prostate gland, epididymis.
Answer:
Prostate gland. (All others are duct systems in male reproductive system.)

Question 5.
Prostate gland, Bartholin’s gland, Cowper’s gland, Epididymis.
Answer:
Bartholin’s glands. (All others are parts of male reproductive system.)

Question 6.
Stigma, style, pollen, ovary.
Answer:
Pollen. (All others are parts of gynoecium.)

Identify the correlation between the first two words and suggest the suitable words in the fourth place:

Question 1.
Amoeba : Fission : : Hydra : ………….
Answer:
Amoeba : Fission : : Hydra : Budding

Question 2.
Transverse binary fission : Paramoecium : : Longitudinal binary fission : ………… (July ‘19)
Answer:
Transverse binary fission : Paramoecium : : Longitudinal binary fission : Euglena

Question 3.
Calyx : Sepals : : Corolla : ………….
Answer:
Calyx : Sepals : : Corolla : Petals

Question 4.
Accessory whorls : Calyx and corolla : : Essential whorls : ………..
Answer:
Accessory whorls : Calyx and corolla : : Essential whorls : Androecium and gynoeciuin

Question 5.
Bisexual flower : Hibiscus : : Unisexual flower : ………….
Answer:
Bisexual flower : Hibiscus : : Unisexual flower : Papaya

Question 6.
FSH : Development of qocyte : : LH : ………….
Answer:
FSH : Development of qocyte : : LH : Ovulation

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Define the following/Give meanings of the following:

Question 1.
Budding in yeast.
Answer:
Budding in yeast: Budding is the asexual reproductive process in which a! small bulge or bud appears on the surface of parent cell as seen in unicellular yeast.

Question 2.
Budding in hydra.
Answer:
Budding in hydra: Budding in hydra is asexual reproductive process in which an outgrowth is formed by repeated divisions of regenerative cells of body wall called bud.

Question 3.
Regeneration.
Answer:
Regeneration: Regeneration is the asexual reproduction in Planaria in which the body is broken up into two parts and resulting each part regenerates remaining part of the body.

Question 4.
Fragmentation.
Answer:
Fragmentation: Fragmentation is the asexual type of reproduction in which the body of parent organism breaks up into many fragments. Each fragment can start living independently.

Question 5.
Vegetative propagation.
Answer:
Vegetative propagation: Vegetative propagation is a type of asexual reproduction in plants that takes place with the help of vegetative parts like root, stem, leaf and bud.

Question 6.
Fertilization.
Answer:
Fertilization: The process by which two haploid gametes unite to form a diploid zygote is called fertilization.

Question 7.
Pedicel.
Answer:
Pedicel: The stalk of the flower which is for the support is called pedicel.

Question 8.
Pollination.
Answer:
Pollination: Transfer of pollen grains from anther to the stigma is called pollination.

Question 9.
Self-Pollination.
Answer:
Self-Pollination: Pollination involving only one flower or two flowers borne on same plant is called self-pollination.

Question 10.
Cross-Pollination.
Answer:
Cross-Pollination: Pollination involving two flowers borne on two plants of same species is cross-pollination.

Question 11.
Endosperm.
Answer:
Endosperm: Endosperm is the nourishing substance formed by the union of second male gamete with two polar nuclei at the time of fertilization in plants.

Question 12.
Embryo sac.
Answer:
Embryo sac: There are many ovules in the ovary, the structure formed in each of the ovule by meiosis is called embryo sac.

Question 13.
Menopause.
Answer:
Menopause: Stoppage of functioning of female reproductive system due to lack of synthesis of hormones due to advancing age is called menopause.

Question 14.
Placenta.
Answer:
Placenta: An organ developed in the uterus of the pregnant mother, through which the embryo is given nourishment is called placenta.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 15.
Menstrual cycle.
Answer:
Menstrual cycle: The repetitive changes at the interval of every 28-30 days in female reproductive system that take place after puberty, form menstrual cycle.

Question 16.
Corpus luteum.
Answer:
Corpus luteum: Corpus luteum is the secondary structure that is formed from empty ovarian follicle after ovulation. This corpus luteum produces progesterone and thereby maintains pregnancy.

Question 17.
Corpus albicans.
Answer:
Corpus albicans: Corpus albicans is the degenerate body which is formed from corpus luteum, if the ovum is not fertilized.

Question 18.
Ovulation.
Answer:
Ovulation: Bursting of mature ovarian follicle under the influence of hormones to release the oocyte is called ovulation.

Question 19.
IVF.
Answer:
IVF: In Vitro Fertilization is the technique in which fertilization is brought about outside the female body but in the test-tube and the embryo is implanted in uterus of woman.

Question 20.
Sperm bank.
Answer:
Sperm bank: Sperm bank is the place where semen donated by the desired men is collected after their thorough physical and medical check-up and stored at sub-zero temperatures in sterile conditions.

Name the following/Give the names:

Question 1.
Different glands associated with male reproductive system.
Answer:
Seminal vesicles, Prostate gland, Cowper’s or bulbourethral glands.

Question 2.
Agents of pollination.
Answer:
Biotic: Insects, birds, few animals.
Abiotic: Water and wind.

Question 3.
Components of semen.
Answer:
Secretion of prostate gland seminal vesicles and Cowper’s glands along with sperms.

Question 4.
Two accessory whorls in flower.
Answer:
Calyx and corolla.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 5.
Two essential whorls in flower.
Answer:
Androecium and gynoecium.

Question 6.
The modern techniques in reproduction.
Answer:
In Vitro Fertilization, Surrogate mother, Sperm bank.

Question 7.
Symptoms of gonorrhea.
Answer:
Painful and burning sensation during urination, oozing of pus through penis and vagina, inflammation of urinary tract, anus, throat, eyes, etc.

Question 8.
Symptoms of syphilis.
Answer:
Occurrence of chancre (patches) on various parts of body including genitals, rash, fever, inflammation of joints, alopecia, etc.

Write the functions of the following organs:

Question 1.
Sporangium.
Answer:
Storing the spores and releasing them by bursting.

Question 2.
Calyx.
Answer:
Protection of inner whorls of the flower.

Question 3.
Corolla.
Answer:
Attracting insects for pollination. Protecting inner whorls.

Question 4.
Androecium.
Answer:
Production of pollen grains, the male gametes of flower.

Question 5.
Gynoecium.
Answer:
Production of female gametes of flower. Participating in production of fruits.

Question 6.
Endosperm.
Answer:
Nourishment of the growing embryo.

Question 7.
Testis.
Answer:
Production of sperms and male hormone-testosterone.

Question 8.
Scrotum.
Answer:
Protection and temperature control of testis.

Question 9.
Seminal vesicles.
Answer:
Secretion of seminal fluid which forms major portion of semen. Nourishment of sperms.

Question 10.
Penis.
Answer:
Transferring of sperms to vagina at the time of intercourse. Release of urine at the time of urination.

Question 11.
Ovary.
Answer:
Production of oocytes and female hormones – estrogen and progesterone.

Question 12.
Uterus.
Answer:
Growth and development of foetus during pregnancy. Helping in parturition (childbirth) by contractions.

Question 13.
Fallopian tubes/ducts.
Answer:
Transporting the released oocyte after ovulation to the uterus. Providing space for fertilization of oocyte by sperm. Conception is possible only when sperm and oocyte meet in the fallopian tube.

Question 14.
Vagina.
Answer:
Passage for copulation/intercourse. Birth canal. Passage for menstrual flow.

Question 15.
Placenta.
Answer:
Supplying nourishment to the growing foetus.

Distinguish between:

Question 1.
Binary fission and Multiple fission.
Answer:
Binary fission:

  1. Two new individuals are formed from one old individual at one time.
  2. The division of nucleus and cytoplasm takes place initially.
  3. The axis of division can be transverse, longitudinal or any one axis as it is in simple binary fission.
  4. Formation of protective cyst does not take place.
  5. Binary fission can be done during favourable period.

Multiple fission:

  1. Many new individuals are formed from one old individual at one time.
  2. Only nucleus divides initially followed by division of cytoplasm.
  3. There is no exact axis for the fission.
  4. Protective covering is formed around dividing amoebulae which is called cyst.
  5. Multiple fission takes place only at the time of unfavourable period.

Question 2.
Human male and Human female reproduction system.
Answer:
Human male reproductive system:

  1. Testis are essential organs which are located outside the abdomen in the scrotal sacs.
  2. There is common urethra through which urine and semen, are passed out.
  3. Reproductive system of male continues to work even in old age.
  4. Sperms or male gametes are produced by meiosis in the testis.
  5. Sperms are produced in millions at one time.
  6. Three accessory glands are associated with the male reproductive system.
  7. Testis secrete testosterone which is essential male hormone.

Human female reproductive system:

  1. Ovaries are essential organs which are located along with all other organs inside the lower abdomen.
  2. Urethra and vagina are two separate openings that open to outside.
  3. Reproductive system works only till menopause.
  4. Oocytes or ova are produced by meiosis in the ovaries.
  5. Only single oocyte is produced per month.
  6. Only one gland is associated with female reproductive system.
  7. Ovaries produce estrogen and progesterone which are essential female hormones.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 3.
Monozygotic twins and Dizygotic twins.
Answer:
Monozygotic twins:

  1. Two children developing from only one zygote are called monozygotic twins.
  2. Monozygotic twins develop from same oocyte.
  3. Gender of both the twins is same.
  4. The monozygotic twins are genetically exactly alike.

Dizygotic twins:

  1. Two children developing from two different zygotes are called dizygotic twins.
  2. Dizygotic twins develop from two different oocytes.
  3. Gender of both the twins can be same or can be different.
  4. The dizygotic twins are genetically not exactly alike.

Give scientific reasons:

Question 1.
Individual developed by sexual reproduction always carry recombined genes of both the parents.
Answer:

  • In sexual reproduction, the haploid male and female gametes are united to form diploid zygote.
  • The zygote thus carries chromosomes of both parents which are transferred via male and female gametes.
  • While producing gametes, there is meiosis in which genetic recombination takes place.
  • Therefore, the individual developed by sexual reproduction always carry recombined genes of both the parents.

Question 2.
Flower is the structural unit of sexual reproduction in plant.
Answer:

  • Flower produces male and female gametes.
  • For this purpose there are essential whorls of androecium and gynoecium.
  • The double fertilization also takes place in flower.
  • Therefore, flower is called the structural unit of sexual reproduction in plants.

Question 3.
Fertilization in plants is called double fertilization.
Answer:

  • After pollination the pollen grains drop on the sticky stigma of the flower.
  • They germinate here producing two male gametes and a long pollen tube.
  • The male gametes travel through the pollen tube till they reach the embryo sac.
  • Here the male gametes are released by bursting the pollen tube. One male gamete unites with the egg cell to form zygote while the second male gamete unites with two polar nuclei forming endosperm.
  • In this way because two nuclei participate in the fertilization process, therefore it is called double fertilization.

Question 4.
By the age of 45 – 50 women gets menopause.
Answer:

  • By the age of 45-50, the secretion of hormones which control the functioning of the reproductive system is reduced gradually and then it stops.
  • This causes end of menstrual cycle. This results into menopause.

Question 5.
Older mothers have greater chance of conceiving abnormal children.
Answer:

  • In older women the menopausal age approaches.
  • The oocytes, released from ovaries during this phase are not normal.
  • Their meiotic cell division is abnormal and thus oocyte becomes abnormal too.
  • If such abnormal oocytes are fertilized, the baby is born with many genetic problems, e.g. Down’s syndrome or Turner’s syndrome.

Question 6.
Indians should follow family planning for controlling the population.
Answer:

  • There is severe population explosion in India. It has almost reached to 121 crores.
  • This results into unemployment, decreasing per capita income and increasing loan, stress on natural resources, etc.
  • Only by controlling population, the quality of life can be restored.
  • Therefore, Indians should follow family planning for controlling the population.

Answer the following questions in short:

Question 1.
How does reproduction take place in fungus Mucor?
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 12

  • Mucor reproduces asexually by spore formation.
  • It has filamentous body that possess sporangia.
  • When the spores are formed, the sporangia burst. The spores are released which settle down at suitable Places.
  • They germinate in moist and warm place forming a new fungal colony.

Question 2.
What is the type of asexual reproduction shown in the diagram below? (Board’s Model Activity Sheet)
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 13
Type of asexual reproduction shown in the diagram above is fragmentation in Spirogyra.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 3.
Describe the structure of a flower.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 14
Answer:
(1) The structural unit of sexual reproduction in plants is flower. There are total four floral whorls. Of these, two are accessory floral whorls while two are essential floral whorls.
(2) Calyx and corolla are accessory whorls. They are protective in nature.
(3) Members of calyx are known as sepals. They are usually green in colour. They protect the inner whorls.
(4) The members of corolla are called petals. They can be of different colours.
(5) Androecium and gynoecium are essential whorls as they participate in sexual reproduction.
(6) The male whorl androecium is made up of stamens. Each stamen has a filament with anther located at the upper end. In the anther there are four locules. Inside the locules the meiosis takes place forming pollen grains. During suitable time, the pollen grains are released from anther lobes.
(7) Gynoecium is made up of carpels, either in separate form or are united. Each carpel is formed of ovary at the basal end hollow ‘style’ and the stigma at the tip of style. There are one or many ovules inside the ovary.
(8) In bisexual flowers both androecium and gynoecium are located in the same flower, e.g. Hibiscus.
(9) In unisexual flowers, androecium is present in male flowers and gynoecium is present in the female flowers, e.g. Papaya.

Question 4.
Describe the human male reproductive system.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 15
In human male reproductive system, the reproductive organs are as follows:

  • Testes, different types of duct systems and glands.
  • Testes are in pair. Each testis lies in the scrotum which lies outside to abdominal cavity.
  • Testes -consist of numerous seminiferous tubules. The germinal epithelium of seminiferous tubules form sperms by undergoing meiosis.
  • These sperm cells are immature.
  • They are pushed gradually through various duct systems till the penis.
  • This path is as follows:
    Rete testis → vas efferentia → epididymis → vasa deferentia → Ejaculatory duct → urethra
  • As the sperms are travelling, they gradually become mature. They are made capable to perform process of fertilization.
  • Seminal vesicles (in pairs), Single prostate gland and a pair of Cowper’s glands secrete their secretions. These secretions and the sperms together form semen.
  • This semen is deposited in the vagina with help of penis.

Question 5.
Describe the human female reproductive system.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 16

  • All the organs of the human female reproductive system are located inside the lower abdomen.
  • There are pair of ovaries, pair of fallopian ducts and a single median uterus.
  • The uterus opens out by vagina. In vaginal walls there are Bartholin’s glands.
  • The urethra in female body is separate and not a common passage as in male body.
  • The free end of fallopian duct is funnel-like having an opening in the centre. The oocyte released from the ovary due to ovulation is picked up by this funnel.
  • The other end of fallopian duct opens into uterus. There are cilia on inner surface of oviduct. With the help of the cilia the oocyte is pushed to the uterus through the fallopian duct.
  • The fertilization of oocyte can take place only in the middle’part of the fallopian duct.
  • The lower end of uterus opens into vagina. The contractions of uterus help in the process of parturition.
  • Vagina is the birth canal as well as copulatory passage. It is also a passage for menstrual flow.

Question 6.
What problems cause infertility in couple?
Answer:

  • In woman if there are problems like irregularity in menstrual cycle, difficulties in oocyte production or implantation in uterus, obstacles in the oviduct, etc.
  • In man if there are no sperms in the semen, slow movement of sperms, or anomalies in the sperms then he becomes sterile.
  • But now with the help of advanced medical techniques these problems can be overcome and a childless couple can be parents.

Question 7.
Answer the following questions: (July 2019)
(a) In our country, there seems to be lack of awareness regarding reproductive health. Why?
(b) Write the symptoms of disease gonorrhea.
(c) What precautions will you take to maintain reproductive health?
Answer:
(a) There is lack of awareness about reproductive health among majority of people of our country. This is due to social customs, traditions, illiteracy, social taboo and shyness.

(b) Symptoms of gonorrhea are as follows:

  1. Painful burning during urination.
  2. Oozing of pus through penis or vagina.
  3. Inflammation of urinary tract, anus, throat, eyes, etc.

(c) Precautions to maintain reproductive health are cleanliness and personal hygiene. Guarding against any sexual infections.

Question 8.
If a piece of bread is kept in a container in moist place for 2-3 days, (1) What will you see? (2) Write scientific name and a character of the organism you may observe.
Answer:
(1) If a piece of bread is kept in moist container we can see growth of fungus on it.
(2) Fungi belonging to species Mucor is seen. It has filamentous body and sporangia. Sporangia burst open to spread spores. It has saprophytic mode of nutrition as it devoid of chlorophyll.

Write short notes on:

Question 1.
Multiple fission.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 17
During unfavourable conditions when there is lack of food. multiple fission is shown by amoeba. Amoeba forms protective covering and becomes encysted. Inside the cyst, amoeba undergoes repeated nuclear division. This is followed by cytoplasmic divisions. Many amoebulae are formed which remain dormant inside the cyst. When favourable conditions reappear, they come out by breaking the cyst.

Question 2.
Regeneration.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 18
In developed animals like wall lizard the process of regeneration is used to restore the lost parts like tail or limbs. As the reproductive system is one of the fullfledged system in the body, the process of regeneration cannot be called type of reproduction.

But some primitive organisms such as Plarfaria use this method for procreation. Planaria breaks up its body into two parts. Each part has the ability to develop the lost part by process of regeneration. This forms two new Planaria.

Question 3.
Seed germination.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 19
Seed germination is the process in which the seed develops into a new plantlet. In the plants, after fertilization the ovule develops into seed and ovary turns into fruit. Seeds fallen on the ground due to bursting of the fruits start germinating. Only under favourable conditions in the soil, this germination takes place. The zygote present inside the seed uses food stored in endosperm of seed and hence develops further to produce a new plantlet.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 4.
Budding in hydra.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 20
In multicellular organisms asexual reproduction by budding is shown by hydra. In fully grown Hydra, at specific part of its body there is development of bud.

This development is only during favourable period. The bud is an outgrowth developed due to repeated divisions of regenerative cells of body wall. It grows up gradually to form a small hydra. Parent hydra’s dermal layers and digestive cavity are in continuity with those of the budding hydra. It receives all the nutrition from parent hydra.When the budding hydra grows sufficiently, it detaches from parent hydra. Then it leads an independent life.

Question 5.
Fragmantation.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 21
Fragmentation is one of the type of asexual reproduction in multicellular organisms. During fragmentation, the body of parent organism breaks up into many fragments. All the resulting fragments start to develop as an independent new organism. In alga Spirogyra, and sponge like Sycon asexual reproduction takes place by fragmentation. Spirogyra grow up very fast and break up into many small fragments when there are favourable conditions. Each newly formed fragment lives independently as a new Spirogyra. Similarly the body of Sycon if accidentally broken into many fragments, develops into new Sycon from each old fragment.

Question 6.
Monozygotic twins.
Answer:
The twins developed from a single embryo are called monozygotic twins. If within 8 days of zygote formation i.e. in early embryonic development cells of that embryo divide into two groups. Each one develop as two separate embryos forming two monozygotic twins. Monozygotic twins are genetically exactly similar to each other. The gender of the twins is also same.

The Siamese twins develop from monozygotic twins, if the embryonic cells are divided into two groups 8 days after the zygote formation. These are conjoined twins where some parts of body are joined to each other. Also some organs are common in Siamese twins.

Read the paragraph and answer the questions given below:

Reproduction is the process by which the living species continues its existence. Lower organisms carry out asexual reproduction while higher plants and animals always show sexual reproduction. Plants reproduce asexually by methods such as fragmentation, vegetative propagation, budding, spore formation. For sexual reproduction they form gametes. In animal kingdom, budding, fission of different types and parthenogenesis are some of the methods that do not require both the sexes. Though regeneration also forms new individual, it is not considered to be a reproductive process because, basically it is a repair process. The ability to regenerate is lost in higher phyla. In human beings | it is restricted only to wound healing. Sexual reproduction is also undergoing lots of experimentation such as cloning which may make females capable of producing their own baby without intervention of any male.

Questions and Answers:

Question 1.
How do living species continue their existence?
Answer:
Through the process of reproduction, living species continue their existence.

Question 2.
Which are asexual methods of reproduction in kingdom Animalia?
Answer:
Fission, budding and parthenogenesis are the asexual methods of reproduction in Kingdom Animalia.

Question 3.
Why is regeneration not true method of reproduction?
Answer:
Regeneration is the repair process than a reproductive process. It is not done with the intention of producing offspring, but is for healing or repairing the lost part.

Question 4.
What are methods of reproduction in plants?
Answer:
Plants reproduce by asexual as well as sexual methods. Asexual reproduction is by fragmentation, vegetative propagation, budding, spore formation, while by formation of gametes, sexual reproduction is done.

Question 5.
What is the modern method of reproduction aimed at in higher organisms?
Answer:
Cloning is the modern method of reproduction by which production of young one can be aimed at.

Diagram-based Questions:

Question 1.
Observe the figure 3.18 and answer the questions below: (Board’s Model Activity Sheet)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 22
(a) What does the figure 3.18 indicate?
Answer:
The figure indicates the menstrual cycle in human female.

(b) Which human organs are involved in this process?
Answer:
The ovary and uterus are primarily involved in this process. But the pituitary gland also controls this cycle.

(c) Which hormones take part in this process?
Answer:
Following hormones regulate this menstrual cycle.
Pituitary hormones: Follicle Stimulating Hormone (FSH) and Luteinizing Hormone (LH).
Ovarian hormones: Estrogen and progesterone.

(d) What is the periodicity for these changes?
Answer:
The menstrual cycle shows repetitive changes every 28 to 30 days.

(e) The body of woman undergoing this process is impure, she should remain away from other people. What is your opinion about this statement? Give justification for your opinion.
Answer:
A menstruating woman is not at all with impure body. It is a natural process in which the endometrium of the uterus is sloughed off and repaired.

She should get enough rest and nutrition during this period. It is painful period in which there is a possibility of infections. Therefore, she should take ! hygienic care and rest till the bleeding persists. But blind faith and superstition to keep her away from others should not be followed.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Question 2.
Observe the diagram (Fig. 3.19) of menstrual cycle and answer the following questions:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 23
(1) What is the period of menstruation?
Answer:
1 to 5 days is the period of menstruation.

(2) On which day does ovulation occur during menstrual cycle?
Answer:
Ovulation occurs on 14th or 15th day.

(3) During which period is corpus luteum active during menstrual cycle? Which hormones are secreted by corpus luteum ?
Answer:
Corpus luteum is active till the 28th day of menstrual cycle. During this time if there is no union of sperm and ovum, then corpus luteum degenerates. Corpus luteum secretes estrogen and progesterone.

(4) In menstrual cycle which reproductive organs undergo changes?
Answer:
Ovary and uterus undergo changes during menstrual cycle.

(5) Which period is said to be period of regeneration of endometrium?
Answer:
In menstrual cycle, days 5 to 14 are period of regeneration of endometrium.

(6) Which period is said to be period of secretions of glands in endometrium?
Answer:
Period of secretions of glands in endometrium is 15 to 28 days.

Question 3.
Observe the following picture and describe the type of reproduction shown in.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 24
Answer:
In developed animals like wall lizard the process of regeneration is used to restore the lost parts like tail or limbs. As the reproductive system is one of the fullfledged system in the body. the process of regeneration cannot be called type of reproduction.

But some primitive organisms such as Planaria use this method for procreation. Planaria breaks up its body into two parts. Each part has the ability to develop the lost part by process of regeneration. This forms two new Planaria.

Question 4.
Answer the following questions: (March 2019)
(a) “Gender of child is determined by the male partner of couple.” Draw a diagram explaining the above statement.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 25

(b) Prepare a slogan for campaign against female foeticide.
Answer:

  • Save the girl child.
  • Daughters give lot of joy, it is not only the boy.

(c) In the following figure, explain how new fungal colonies of mucor are formed:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 26
Answer:
Mucor is a fungus having filamentous body. The filaments bear sporangia. Mature sporangia burst and release spores. Spores germinate to form new hyphae upon getting favourable moist and warm place.

(d) Identify and state the type of reproduction represented in the above figure.
Ans. The spore formation is asexual type of reproduction seen in Mucor.

Question 5.
Write the type of asexual reproduction shown in the figure.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 2 Life Processes in Living Organisms Part - 2, 27
Answer:
The figure shows budding in yeast. Budding is the type of asexual reproduction.

Experiments:
(Try this: Text Book Pages 23 and 24)

(1) Observation of Paramoecia.
(2) Observation of yeast.
(3) Study of Hibiscus.
[For detailed information on practicals, refer to Vikas Science and Technology Experiment Book: Standard X]

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 3 Life Processes in Living Organisms Part - 2

Projects:

Project 1.
Use of ICT. (Textbook page no. 27)
Make an video album of pollination and show it in the class.

Project 2.
Internet is my friend. (Textbook page no. 33)
You may have read that sometimes a woman may deliver more than two offspring at a time. Collect more information from internet about reasons for such incidences.

Project 3.
Get information. (Textbook page no. 34)
Visit a public health centre nearby your place and collect the information through an interview of health officer about meaning and various methods of family planning.

Maharashtra State Board Class 10 Science Solutions 

12th Biology Chapter 1 Exercise Reproduction in Lower and Higher Plants Solutions Maharashtra Board

Reproduction in Lower and Higher Plants Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 1 Reproduction in Lower and Higher Plants Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 1 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 1 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Insect pollinated flowers usually possess ………………
(a) sticky pollen with a rough surface
(b) large quantities of pollens
(c) dry pollen with a smooth surface
(d) light-colored pollens
Answer:
(a) sticky pollen with a rough surface

Question 2.
In ovule, meiosis occurs in ………………
(a) Integument
(b) Nucellus
(c) Megaspore
(d) Megaspore mother cell
Answer:
(d) Megaspore mother cell

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 3.
The ploidy level is NOT the same in ………………
(a) Integuments and nucellus
(b) Root tip and shoot tip
(c) Secondary nucleus and endosperm
(d) Antipodals and synergids
Answer:
(c) Secondary nucleus and endosperm

Question 4.
Which of the following types require pollination but result is genetically similar to autogamy?
(a) Geitonogamy
(b) Xenogamy
(c) Apogamy
(d) Cleistogamy
Answer:
(a) Geitonogamy

Question 5.
If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?
(a) Endosperm
(b) Leaf cells
(c) Cotyledons
(d) Synergids
Answer:
(d) Synergids

Question 6.
In angiosperms, endosperm is formed by/ due to ………………
(a) free nuclear divisions of megaspore
(b) polar nuclei
(c) polar nuclei and male gamete
(d) synergids and male gametes
Answer:
(c) polar nuclei and male gamete

Question 7.
Point out the odd one.
(a) Nucellus
(b) Embryo sac
(c) Micropyle
(d) Pollen grain
Answer:
(d) Pollen grain

2. Very Short Answer Questions

Question 1.
The part of gynoecium that determines the compatible nature of pollen grain.
Answer:
Stigmatic surface.

Question 2.
How many haploid cells are present in a mature embryo sac?
Answer:
6 cells, 2 synergids, 1 egg cell, 3 antipodals.

Question 3.
Even though each pollen grain has 2 male gametes why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?
Answer:
Angiosperms have phenomenon of double fertilization in which both the male gametes are utilized, one for fusion with egg cell to form zygote and other for fusion with secondary nucleus to form endosperm.

Question 4.
Megasporogenesis
Answer:
It is the process of formation of haploid megaspores from diploid megaspore mother cell.

Question 5.
What is hydrophily?
Answer:
Transfer of pollen grains in pollination process through agency of water is known as hydrophily.

Question 6.
The layer which supplies nourishment to the developing pollen grains.
Answer:
Tapetum

Question 7.
Parthenocarpy
Answer:
The condition in which fruit is developed without the process of fertilization is called parthenocarpy.

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 8.
Are pollination and fertilization necessary in apomixis?
Answer:
Apomixis is formation of embryos without formation of gametes hence there is no need of pollination and fertilization.

Question 9.
The part of pistil which develops into fruit and seed.
Answer:
Ovary develops into fruit and ovules into seed.

Question 10.
What is the function of filiform apparatus ?
Answer:
Filiform apparatus guides the pollen tube towards egg cell.

3. Short Answer Questions

Question 1.
How polyembryony can be commercially exploited?
Answer:

  1. Polyembryony is the development of more than one embryo inside the seed.
  2. When such polyembryonic seed germinate we get multiple seedlings from it.
  3. This condition increases the chances of survival of new plants.
  4. Nucellar embryos are genetically identical to parent plants hence we get uniform plants.
  5. In horticulture we can utilize these as rootstock for grafting, hence they have significant role in fruit breeding programmes e.g. Citrus, Mango.

Question 2.
Pollination and seeds formation are very crucial for the fruit formation, Justify.
Answer:

  1. After fertilization, ovary is transformed into fruit, where ovary wall becomes fruit wall, i.e pericarp.
  2. Mature ovules are transformed into seeds after fertilization.
  3. Fertilization is a process where male gametes unites with female gamete to form zygote which develops into embryo.
  4. In pollination process pollen grains carrying non-motile male gamete are transferred on stigma.
  5. Seeds have embryo which germinate into new plant hence the goal of reproduction to create offspring for next generation is achieved. Hence these are the crucial events for fruit formation.

Question 3.
Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?
Answer:

  1. Self incompatibility or self-sterility is a genetic mechanism that prevents germination of pollen on stigma of same flower. This favours cross pollination. E.g. Tobacco.
  2. In pollen-pistil interaction, when pollen grain is deposited on stigma, pistil has the ability to recognize and allow germination of right type of pollen.
  3. Special type of proteins on stigmatic surface determine compatibility or incompatibility.
  4. A physiological mechanism operates to ensure successful germination of compatible pollen.
  5. Compatible pollen absorbs water and nutrients from stigmatic surface that are absent in pollen and then pollen tube emerges which grow-s through style.

Question 4.
Describe three devices by which cross pollination is encouraged in Angiosperms by avoiding self-pollination?
Answer:

  1. Unisexuality, dichogamy, prepotency, heteromorphy and herkogamy are the outbreeding devices.
  2. Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.
  3. Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.
  4. Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. E.g. Calotropis where pollinia are situated below the stigma.

4. Long Answer Questions

Question 1.
Describe the process of double fertilization.
Answer:
Double fertilization:
(1) Out of the two male gametes produced by the male gametophyte in angiosperms, one unites with the female gamete and the other with the secondary nucleus. Since both the male gametes take part in fertilization and fertilization occurs twice, it is called double fertilization.

(2) During double fertilization, the pollen tube on reaching the ovule enters the embryo sac through micropyle and bursts in one of the synergids. Owing to this, the two male gametes contained in the pollen tube, are set free.
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 1

(3) Out of the two male gametes, one unites with the egg or female gamete and the other unites with the secondary nucleus of the embryo sac, forming a triploid or triple fusion nucleus, called the primary endosperm nucleus. The process involving the fusion of one of the male gametes with the egg nucleus, resulting in the formation of a diploid zygote is called syngamy.

(4) The reproductive process in which non-motile male nuclei are carried to the egg cell through a pollen tube is called siphonogamy.

(5) After fertilization, zygote develops into an embryo. Certain changes take place in the ovule leading to the development of a seed.

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 2.
Explain the stages involved in the maturation of microspore into male gametophyte.
OR
Describe the development of male gametophyte before pollination in angiosperms.
OR
Sketch and label male gametophyte in angiosperm.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 2

  1. Microspore or pollen grain is first cell of male gametophyte.
  2. The protoplast of pollen grain divides mitotically to form two unequal cells – a small thin walled generative cell and a large naked vegetative or tube cell.
  3. The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.
  4. The vegetative, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.
  5. In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.
  6. The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

Question 3.
Explain the development of dicot embryo.
Answer:
Development of embryo (dicot) in angio- sperm:
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 3
The oospore undergoes a transverse division to form a large basal cell towards the micropyle and a small apical or terminal cell towards the chalaza of the embryo sac. This two celled structure is called proembryo. The basal cell or suspensor initial undergoes repeated transverse divisions to form a multicellular structure called suspensor. The suspensor pushes the embryo towards the endosperm to draw its nutrition.

  1. The development of embryo from a zygote is called embryogenesis.
  2. The fusion of male gamete and an egg cell during fertilization results in the formation of a diploid zygote. The zygote develops a wall around it and is converted into oospore.
  3. The apical cell or embryonal initial of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
  4. From octant, the lower four cells form hypocotyl and radicle while four cells of upper side form plumule with two cotyledons.
  5. The lowermost cell of suspensor is hypophysis and by its further division forms part of radicle and root cap.
  6. The cells from upper side of octant divide repeatedly to form heart shaped which elongated further to form two lateral cotyledons.
  7. Enlargement of hypocotyl and cotyledon results into curved embryo which appears horse shoe shaped.

Question 4.
Draw a diagram of the L.S of anatropous ovule and list the components of embryo sac and mention their fate after fertilization.
Answer:
Components of Embryo sac.
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 4

  1. Mature embryo sac is 7-celled and 8 nucleate.
  2. Egg apparatus at micropylar end – with 2 synergids and egg cell.
  3. Central cell with secondary nucleus formed by 2 polar nuclei
  4. Antipodal cells at chalazal end – 3 cells.
  5. Pollen tube enters the synergids, Synergids guide the growth of pollen tube towards egg.
  6. Male gamete fuses with female gamete, i.e. syngamy to form zygote which develops into embryo.
  7. One male gamete fuses with secondarynucleus to form primary endosperm nucleus (PEN) which forms endosperm, nutritive tissue for embryo.

5. Fill in the Blanks

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 5
Question 1.
The ……………… collects the pollen grains.
Answer:
biotic agents

Question 2.
The male whorl, called the ……………… produces ………………
Answer:
androecium, pollen grains

Question 3.
The pollen grains represent the ………………
Answer:
male

Question 4.
The ……………… contains the egg or ovum.
Answer:
embryo sac

Question 5.
…………….. takes place when one male gamete and the egg fuse together. The fertilized egg grows vs into seed from which the new plants can grow.
Answer:
Fertilization

Question 6.
The ……………… is the base of the flower to which other floral parts are attached.
Answer:
thalamus

Question 7.
……………… is the transfer of pollen grains from anther of the flower to the stigma of the same or a different flower.
Answer:
Pollination

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

Question 8.
Once the pollen reaches the stigma, pollen tube traverses down the ……………… to the ovary where fertilization occurs.
Answer:
style

Question 9.
The ……………… are coloured to attract the insects that carry the pollen. Some flowers also produce ……………… or ……………… that attracts insects.
Answer:
petals, fragrance, nectar

Question 10.
The whorl ……………… is green that protects the flower until it opens.
Answer:
Calyx.

6. Label the Parts of seed.

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 6
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants 7

7. Match the following

Column I (Structure Before seed formation)Column II (Structure After seed formation)
A. Funiculusi. Hilum
B. Scar of Ovuleii. Tegmen
C. Zygoteiii. Testa
D. Inner Integumentiv. Stalk of Seed
v. Embryo

Maharashtra Board Class 12 Biology Solutions Chapter 1 Reproduction in Lower and Higher Plants

(a) A-v, B-i, C-ii, D-iv
(b) A-iii, B-iv, C-i, D-v
(c) A-iv, B-i, C-v D-ii
(d) A-iv, B-v C-iii, D-ii
Answer:
(c) A-iv, B-i, C-v D-ii

Maharashtra State Board 12th Std Biology Textbook Solutions

Elements, Compounds and Mixtures Class 7 Questions And Answers Maharashtra Board

Std 7 Science Chapter 14 Elements, Compounds and Mixtures Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 14 Elements, Compounds and Mixtures Question Answer Maharashtra Board

1. Who are my companions?

Question a.

Column ‘A’Column B’
1. Stainless steela. Non-metal
2. Silverb. Compound
3. Bhajani mixture for millingc. Mixture
4. Saltd. Element
5. Coale. Alloy
6. Hydrogenf. Metal

Answer:

Column ‘A’Column B’
1. Stainless steele. Alloy
2. Silverf. Metal
3. Bhajani mixture for millingc. Mixture
4. Saltb. Compound
5. Coala. Non-metal
6. Hydrogend. Element

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

2. Write the names of elements from the following symbols:
Zn, Cd, Xe, Br, Ti, Cu, Fe, Si, Ir, Pt.

Question a.
Write the names of elements from the following symbols:
Zn, Cd, Xe, Br, Ti, Cu, Fe, Si, Ir, Pt.
Answer:

SymbolElement
ZnZinc
CdCadmium
XeXenon
BrBromine
TiTitanium
CuCopper
FeIron
SiSilicon
IrIridium
PtPlatinum

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

3. What are the molecular formulae of the following compounds?
(Hydrochloric acid, Sulphuric acid, Sodium chloride, Glucose, Methane)

Question a.
What are the molecular formulae of the following compounds?
(Hydrochloric acid, Sulphuric acid, Sodium chloride, Glucose, Methane)
Answer:

  • Hydrochloric acid → HCl
  • Sulphuric acid → H2SO4
  • Sodium Chloride → NaCl
  • Glucose → C6H12O6
  • Methane → CH4
  • Water → H2O
  • Carbon dioxide → CO2
  • Sucrose (sugar) → C12H22O11

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

4. Give scientific reasons:

Question a.
Buttermilk is churned to get butter.
Answer:

  1. Churning is the process of shaking up buttermilk to make butter.
  2. Butter is essentially the fat of milk.
  3. Churning physically agitates the cream until it ruptures the membrane surrounding milk fat
  4. Fat droplets can join with each other to form clumps of fat.

Question b.
In Chromatography the ingredients of a mixture rise up to a limited height when water rises up to the upper end of the paper.
Answer:
In Chromatography two properties of substances are used, (a) They are the stability of the substances in the solvent that moves up. The ability of the substance to stick to the stationary filter paper, (b) So all the components of the mixture do not rise all the way to the upper end of the filter paper but remain behind at limited heights.

Question c.
A wet cloth is wrapped around a water storage container in summer.
Answer:
A wet cloth will absorb the heat from the surroundings and it will keep the water in the container, cool for a longer time.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

5. Explain the difference.

Question a.
Metals and Non-metals
Answer:

MetalsNon-metals
1. Metals are good conductors of heat and electricity.1. Non-metals are poor conductors of heat and electricity.
2. Metals are solids at room temperature except for mercury.2. Non-metals exists in all three states.
3. Metals are lustrous, (shiny)3. Non-metals are not lustrous, (dull appearance) except graphite.
4. Metals are malleable (can be hammered into sheets)4. Non-metals are brittle, not malleable.
5. Metals are ductile, can be drawn into wire.5. Non-metals are not ductile.
6. Metals are very hard and strong.6. Non-metals are brittle, will break down into pieces except diamond.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question b.
Mixture and Compound
Answer:

MixtureCompound
1. The different substances are not chemically joined together to form a mixture but mixed physically.1. A compound is formed when two or more elements chemically combine together e.g. H2O → chemical formula of water.
2. Each substance in the mixture retains its own properties2. Compound has fixed properties.
3. Mixtures are impure substances.3. Compounds are pure substances.
4. Substances from the mixture can easily be separated by physical methods.4. The constituents of a compound can be separated only by chemical methods.
5. The constituents of a mixture are present in varying proportions.5. The constituents of a compound are present in fixed proportion.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question c.
Atoms and Molecules
Answer:

AtomsMolecules
1. An atom is the fundamental part of matter.1. A molecule is a group of chemically bonded atoms.
2. Atoms are basic building blocks of matter.2. Molecule is the smallest unit of a chemical compound
3. It is foundation of molecules.3. It is foundation of chemical compound.

Question d.
Separation by distillation and Separation by separating funnel
Answer:

Separation by distillationSeparation by separating funnel
1. Distillation is used for purification of impure liquids, for separating liquid and solids (Separation of mixture containing two miscible liquids) e.g. to separate salt and water from saltwater.1. This procedure is used for separating two liquids e.g. separating oil and water, (two immiscible liquids)

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

6. Write answers to the following questions in your own words.

Question a.
How are the components of mixture separated by simple methods?
Answer:
1. Component of mixtures are separated by straining, filtering, sifting, picking, sorting, winnowing, combing with a magnet and sublimation.
2. In a mixture, the constituent substances do not lose their identity, they can be separated easily by physical methods.
a. Sedimentation: (i) It is a process of separating an insoluble solid from a liquid in which it is suspended, by allowing it to settle to the bottom of the container, e.g. Muddy water contains heavier particles like sand and soil, (ii) Leave this muddy water undisturbed for some time, (iii) The heavier soil and sand particles settle down and the clear upper water is poured out by decantation.

b. Filtration: It is used for separating insoluble solids from a liquid, e.g. mixture of chalk and water is poured through a filter paper in a funnel while the water gets

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 1.1

collected in the beaker below, chalk is retained in the filter paper.

c. Evaporation: It is used for recovering dissolved solid substances from solutions by evaporation e.g. sugar can be recovered from sugar-water.

d. Distillation: It is a process of heating a solution containing soluble solids to form vapours of the liquid and then cooling the vapours to get the liquid back. e.g. A mixture of common salt and water is taken in distillation flask and heated. Steam rises up and comes out into condenser.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 2

e. Sublimation: It is a process in which some solids on heating are transformed directly to vapour without passing through the liquid state, (i) It is used to separate a mixture of solids. The vapours are cooled separately, (ii) Used to separate ammonium chloride, iodine, camphor and sulphur from any mixture.

f. Magnetic separation: A mixture with iron fillings as one of the components can be separated using magnet to attract the iron particles away from the mixture.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question b.
Which elements (metals, non-metals) compounds and mixtures do we use in our day to day life?
Answer:
Elements – non-metal

  • Oxygen, nitrogen → present in air
  • Hydrogen → present in water
  • Silicon → in memory card
  • Lithium → to make batteries
  • Neon → in neon signs/lights
  • Sulphur → used in water treatment, agricultural pesticides

Elements – metals

  • Calcium → in milk
  • Silver, gold → used in jewellery
  • Aluminium, copper, iron → Kitchen vessels
  • Mercury → in thermometer
  • Copper → electric wires

Compounds:

  • Sodium chloride → table salt
  • Sodium carbonate → washing soda
  • Sodium bicarbonate → baking soda
  • Sodium hypochloride → bleaching powder
  • Sodium hydroxide, Potassium hydroxide → in making soaps
  • Calcium oxide, Calcium hydroxide → in white washing the buildings.
  • Hydrochloric acid → in cleaning toilets
  • Sucrose → sugar used in cooking and baking.

Mixtures:

  • Bhel → mixture of puffed rice, sev, groundnuts, lemon juice
  • Concrete → mixture of cement, sand and rocks
  • Salad → mixture of onion, cucumber, tomato, lettuce, etc.
    Sherbet and saltwater are also mixtures.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question c.
In everyday life, where and for what purpose do we use centrifugation?
Answer:
Centrifugation: It is a process which involves application of centrifugal force.
1. It is used in industrial and laboratory settings. It is used for separation of fluids, gas or liquid based on density. In centrifugation mixture is separated through spinning and the solid settles to the bottom and the solution is clear.

2. Best example of centrifugal force is working of washing machine when it dries the clothes. The wet clothes are moving in circular path and a force acts on water particles in clothes and this force pulls water all outer side. Force involved is centrifugal force which removes the water from the clothes and clothes are dried in this way.

3. Cream separation: When the machine moves, the rod set inside milk moves and milk is pressurised and solid part from this milk goes outside due to centrifugal force and this solid part is cream. In this way the cream is separated from the milk.

Question d.
Where are methods of separation by distillation and by separating funnel used?
Answer:
1. Distillation: It is the process of separating the components or substances from a compound. It is a process of heating a solution containing soluble solids to form vapours of the liquid and then cooling the vapours to get the liquid back.
Distillation method is used in the following:

  • to separate components of air into oxygen, nitrogen, argon, etc.
  • to separate salt from saltwater.
  • to purify impure liquids.
  • to prepare distilled water.

2. Separating funnel: When a mixture of two immiscible liquids is left undisturbed, two layers are clearly seen to have formed, (a) The heavier of the liquids remain below and the lighter liquid floats on it. (b) So two liquids in a mixture can be separated by making use of this property, e.g.

  • to separate kerosene and water.
  • to separate oil and water.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question e.
Which precaution will you take while using the methods of distillation and separation by separating funnel?
Answer:
In using distillation:

  1. Allow sufficient space to work in, working area should be well lit and well ventilated to prevent the accumulation of alcoholic vapours.
  2. Keep a fire extinguisher handy.
  3. There should be no obstruction in the piping that could cause pressure build-up in the stills and cause it to burst.

In using separating funnel:

  1. Stopper should be tightly fitted so that the solution does not leak out when the separating funnel is inverted.
  2. Never throw any layer away until you are absolutely sure that you isolated the final product.

Project:

Question a.
Visit a jaggery or a sugar factory. Obtain information about the methods that are used to separate the components of the mixture while making jaggery or sugar. Present it in the class.

Class 7 Science Chapter 14 Elements, Compounds and Mixtures Important Questions and Answers

Fill in the blanks and rewrite the sentences:

Question 1.
………………, ……………….. and are three states of matter.
Answer:
Solid, liquid, gaseous

Question 2.
Molecules in ……………….. phase of matter are tightly packed together.
Answer:
solid

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 3.
A ……………….. has a definite volume, but not a definite shape, it takes the shape of its container.
Answer:
liquid

Question 4.
The temperature at which a liquid becomes a solid is called ……………….. .
Answer:
freezing point

Question 5.
Anything that takes up space and has mass is called ……………….. .
Answer:
matter

Question 6.
A ……………….. has no definite shape and no definite volume.
Answer:
Gas

Question 7.
The amount of space that matter occupies is called ……………….. .
Answer:
volume

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 8.
A ……………….. has a definite shape and definite volume.
Answer:
solid

Question 9.
When a liquid is heated it will ……………….. .
Answer:
expand

Question 10.
A ……………….. is formed by mixing different elements or compounds.
Answer:
mixture

Question 11.
The substance formed by a chemical combination of two or more elements is called a ……………….. .
Answer:
compound

Question 12.
The elements that show some properties of metals and non-metals are called ……………….. .
Answer:
metalloids

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 13.
Elements are generally classified into ……………….. and ……………….. .
Answer:
metal, non-metals

Question 14.
The smallest particles of substances are called ……………….. .
Answer:
molecules

Question 15.
To date, scientists have discovered ……………….. elements and of these ……………….. elements occur in nature.
Answer:
118, 92

Question 16.
……………….. was the first scientist to use symbols for elements.
Answer:
Berzelius

Question 17.
Symbol for Tungsten is and it is ……………….. and it is derived from its German name ……………….. .
Answer:
W, Wolfram

Question 18.
22-carat gold is an ……………….. of gold.
Answer:
alloy

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 19.
……………….. of a compound is a short form of its name written using symbols of its constituent elements.
Answer:
Molecular formula

Question 20.
For separating insoluble particles of blood (blood cells) from its liquid part (plasma), ……………….. method is used.
Answer:
centrifugation

Give scientific reason.

Question 1.
Saltwater is a mixture.
Answer:

  1. The salt and water are mixed to give saltwater and can both be separated.
  2. It can be separated by boiling saltwater.
  3. Water can be obtained by condensation whereas the salt will be leftover.
  4. As the salt and water do not react chemically and they are separated by simple processes therefore saltwater is called a mixture.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 2.
Classify the following substances according to their properties
(water, thermocol, soil, iron, coal, paper, rubber, copper, coir, plastic)
Answer:

MetalsMixturesCompounds
Copper, IronSoil, ThermocolPaper, Coir, Plastic, Rubber, Water

Can you tell?

Answer the following questions:

Question 1.
What are objects made of?
Answer:
An object is made of a certain substance, (i) The term matter is also used as a synonym of substance, (ii) Object is made of matter.

Question 2.
What are these articles of everyday use made of?
Answer:

  • Electric wire → copper-metal
  • Kitchen utensils → stainless steel, Brass-alloy (aluminium-metal).
  • Nails → iron-metal
  • Tables, chairs → wood, plastic
  • Sugar → sucrose
  • Window panes → glass
  • Salt → NaCl (Sodium chloride) it is a compound.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 3.
What do the short-forms Dr, H.M., AC, Adv., C.M., DC stand for?
Answer:
Short forms of Dr. – Doctor, H.M. – Headmaster, AC – Air conditioner, Adv – Advance, C.M. – Chief Minister, DC – Direct current

Question 4.
Which metals do we use in day-to-day life?
Answer:

  • Gold and silver → used in jewellery, computer and solar cells
  • Iron, copper, aluminium → used to make utensils
  • Copper → used in electrical gadgets
  • Tungsten → used in electric bulbs.
  • Zinc → for coating on iron to prevent rusting
  • Tin → for coating on copper and brass vessels
  • Iron → nails, agricultural equipments, in construction of buildings, to make steel

Question 5.
Are metals elements?
Answer:
Yes, metals are elements.

Question 6.
Which element helps combustion?
Answer:
Oxygen helps combustion.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 7.
Does water help combustion?
Answer:

  1. Water is formed by a combination of hydrogen and oxygen but does not help combustion.
  2. It is used to extinguish a fire.

Question 8.
What are the mixtures used in everyday life?
Answer:
Oil and water, lemon juice and sherbet, bhel, honey and tea, milk and chocolate, coffee and cream, salt and water, smog (smoke + fog) air (oxygen + nitrogen), salad, milk.

Question 9.
Are all mixtures useful to us?
Answer:
No, adulterated foodstuff is also a kind of mixture and it is not useful. It is harmful, because when an unwanted and harmful substance is mixed with another substance the resulting mixture no longer remains useful.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 10.
How will you separate each component from a mixture of semolina, salt and iron filings?
Answer:

  1. If a magnet is moved through this mixture all iron filings will stick to the magnet, remove them.
  2. Add water to the remaining mixture, all salt will dissolve and then filter through filter paper.
  3. Samolina will settle in filter paper.
  4. The filtrate will be saltwater and salt can be separated by evaporation or distillation.

Use your brainpower!

Answer the following questions:

Question 1.
In day to day life we come across many things in our surroundings. We touch them, we study their properties. Are all these things made from only one kind of matter or from more than one kind of matters?
Answer:

  1. Things are made from only one kind of matter or may be made from more than one kind of matter.
  2. When a matter is made up of only one type of atoms it is called an element.
  3. We find many elements in our surroundings.
  4. But we also see many substances which are made up of two or more elements they are called compounds.
  5. Also we see many substances which are formed by physically mixing two or more elements, they are called mixtures.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 2.
Classify the following according to the nature of matter in them – whether it is made from one kind of matter or from more than one kind of matter, and whether it is in solid, liquid or gaseous state: an engraved idol, gold, milk, water, a plank, concrete, salt, soil, coal, smoke, sherbet, cooked khichadi, steam.
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 3

Question 3.
Which of the following are mixtures?
(water, sharbet, iron, steel, coal, air, salt, copper, brass, soil)
Answer:
Mixtures: sherbet, steel, air, brass and soil

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 4.
Which elements are present in air?
Answer:
Nearly all of the earth’s atmosphere is made up of gases → Nitrogen, Oxygen, Carbon dioxide, Argon and Neon, Helium and Krypton.

Question 5.
Is carbon dioxide an element?
Answer:
No, carbon dioxide is a compound. A molecule of the compound CO2 contains one atom of the element carbon and two atoms of the element oxygen.

Question 6.
What are the properties of elements due to?
Answer:
Properties of elements are due to molecules present in it. The molecules are made of one or more atoms which are exactly alike. The mass and volume of atoms of different elements are different.

Question 7.
Are the atoms of different elements similar or dissimilar?
Answer:
The atoms of different elements are dissimilar.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 8.
Is the water that falls from clouds naturally pure?
Answer:

  1. Any form of water that falls from the clouds is known as precipitation. Several forms → rain, snow, hail stones.
  2. Yes, it is pure as water vapours condense to form rain.
  3. The same process is used in distillation to obtain the purest water.

Question 9.
Which properties of a liquid are seen in the distillation method?
Answer:
Liquid when boiled turns into gaseous state (e.g. water vapours) and on cooling it condenses to liquid.

Question 10.
For what purposes is distilled water used?
Answer:
Distilled water is used in pharmacy as a solvent, in photography to wash the negatives, in laboratory experiments and in cleaning medical tools.

Answer the following questions:

Question 1.a.
Take a piece of chalk and keep on dividing it in to smaller pieces, what will happen?
Answer:
We will get very tiny particles of chalk.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question b.
Wipe a drop of ink with a handkerchief what effect does it have on the cloth of the handkerchief?
Answer:
The cloth will absorb the ink drop and cloth will have a stain mark on it.

Question c.
What happens when the lid of a bottle of perfume is opened?
Answer:
The molecules of perfume move out of the bottle and collide with other molecules in the air and eventually perfume spreads throughout the room.

Question 2.
Fill water in a spray pump, spray the water and observe the spray.
Answer:
When we spray the water, spray is composed of small particles of water. They are very tiny that we can not see them.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 3.
Take water in a tea pot and cover it, heat the water to a boil. What do you see on the inside of the lid?
Answer:

  1.  When we boil the water in a tea pot it boils and changes into gaseous state.
  2. We find vapours (steam) but when we cover it with a lid, the water vapour cools down, condenses into liquid. So we find water droplets on the inside of the lid.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 4

Question 4.a.
Take sugar in a test tube and heat the test tube. Observe what happens. What remains behind?
Answer:
When a test tube containing sugar is heated, the sugar melts and then it loses water leaving behind a black substance. This black substance is carbon.
Sugar is sucrose, a molecule of sugar is composed of 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen. (C12H22O11).

  1. Sugar is a compound made from these three elements carbon, hydrogen, oxygen.
  2. It is a Carbohydrate.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question b.
Using tongs hold a magnesium ribbon in a flame and observe what changes took place?
Answer:
When a magnesium ribbon is ignited, light and heat are produced. Magnesium reacts chemically with oxygen in the air
2Mg(s) + O2(g) → 2MgO(s)
Magnesium oxide is produced.

Question c.
What does the name carbon dioxide imply – how many and which elements is this substance made of?
Answer:

  1. Carbon dioxide is a compound. It is a tasteless, odourless gas. It is a natural product of respiration.
  2. Plants use this to produce food.
  3. Humans breath out the CO2.

[A molecule of the compound carbon dioxide contains one atom of the element carbon and two atoms of the element oxygen.]

Question d.
Which of these are compound, which are elements?
Answer:

  • Water → compound H2O (hydrogen + oxygen), Oxygen → Element
  • Carbon dioxide → compound CO2 (Carbon + Oxygen)

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question e.
What is the smallest particle of a compound called?
Answer:
The smallest particle of a compound is called molecule.

Question 5.
From the internet or reference books obtain information about elements and prepare a table according to the format given below:
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 5 Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 6

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 6.
Some methods of separating the components of a mixture.
Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 7
Answer:

  1. Winnowing
  2. Filteration
  3. Hand picking
  4. Churning
  5. Sieving

Question 7.
What is a molecular formula?
Answer:
Like an element, a compound is also written in an abridged form. A molecule of a compound is formed by a chemical combination of atoms of two or more elements. Therefore a molecular formula is used to represent a compound. Molecular formula of a compound is a short form of its name written with the help of the symbols of constituent elements and the number of their respective atoms, e.g.: Nael – Sodium chloride.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 8.
Collect information and prepare table:
Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 8 Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 9

Write short notes on:

Question a.
Atoms
Answer:
Atoms are basic building blocks of matter. Desk, chair, air are made up of atoms.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question b.
Compounds
Answer:
1. A compound is a pure substance that contains two or more elements. Compounds are chemical combination of elements with properties that are different from the elements that formed them.
2. Substance is a compound only if its molecules are made up of atoms of different types e.g. Water, one molecule of water is made of two atoms of hydrogen and one atom of oxygen.

Question c.
Mixtures
Answer:

  1. A mixture is two or more elements or compounds which are combined physically but no chemical reaction occurs.
  2. This means you can separate them again, e.g. air, brass.

Question 9.
Write short notes on:
(1) Distillation (2) Sublimation (3) Magnetic separation (4) Sedimentation (5) Filtration (6) Evaporation
Answer:
1. Distillation: It is a process of heating a solution containing soluble solids to form vapours of the liquid and then cooling the vapours to get the liquid back. e.g. A mixture of common salt and water is taken in distillation flask and heated. Steam rises up and comes out into condenser.

2. Sublimation: It is a process in which some solids on heating are transformed directly to vapour without passing through the liquid state, (i) It is used to separate a mixture of solids. The vapours are cooled separately, (ii) Used to separate ammonium chloride, iodine, camphor and sulphur from any mixture.

3. Magnetic separation: A mixture with iron fillings as one of the components can be separated using magnet to attract the iron particles away from the mixture.

4. Sedimentation: (i) It is a process of separating an insoluble solid from a liquid in which it is suspended, by allowing it to settle to the bottom of the container, e.g. Muddy water contains heavier particles like sand and soil, (ii) Leave this muddy water undisturbed for some time, (iii) The heavier soil and sand particles settle down and the clear upper water is poured out by decantation.

5. Filtration: It is used for separating insoluble solids from a liquid, e.g. mixture of chalk and water is poured through a filter paper in a funnel while the water gets collected in the beaker below, chalk is retained in the filter paper.

6. Evaporation: It is used for recovering dissolved solid substances from solutions by evaporation e.g. sugar can be recovered from sugar-water.

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 10.
Name the method you will use to separate the following mixtures:

  1. Cream from buttermilk.
  2. Mud from muddy water.
  3. Stones from grains.
  4. Tea leaves from boiled tea.
  5. Salt from water.
  6. Kerosene and water.
  7. Blood cells and plasma.
  8. Ink from water

Answer:

  1. Churning
  2. Sedimentation and decantation
  3. hand picking
  4. filtration
  5. distillation
  6. separating funnel
  7. Centrifugation
  8. Chromatography

Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures

Question 11.
Draw the experimental setup of weperation using separating funnel:
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 14 Elements, Compounds and Mixtures 10

Maharashtra State Board Class 7 Science Textbook Solutions

Properties of a Magnetic Field Class 7 Questions And Answers Maharashtra Board

Std 7 Science Chapter 19 Properties of a Magnetic Field Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 19 Properties of a Magnetic Field Question Answer Maharashtra Board

1. Write the appropriate terms in the blanks:

Question a.
The alloys called …………. and …………. are used for making industrial magnets.
Answer:
alnico, Nipermag

Question b.
A magnetic field can pass through ………… and …………. .
Answer:
cardboard, bottle

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question c.
The intensity of a magnetic field is indicated by the lines of …………. .
Answer:
force

Question d.
The real test of a magnet is ……………… .
Answer:
repulsion

2. With whom should I pair up?

Question a.

Column ‘A’Column ‘B’
1. Compassa. The highest magnetic force
2. Door of a cupboardb. Like poles
3. Repulsionc. A magnet
4. Magnetic poled. A magnetic needle

Answer:

Column ‘A’Column ‘B’
1. Compassd. A magnetic needle
2. Door of a cupboardc. A magnet
3. Repulsionb. Like poles
4. Magnetic polea. The highest magnetic force

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

3. Write answers to the following questions:

Question a.
Distinguish between the two methods of making artificial magnets
Answer:

Single touch methodDouble touch method
1. In this method, one magnet is used to magnetise the steel bar.1. In this method, two magnets are used to magnetise the steel bar.
2. In this method, the magnet is dragged from one end of the steel bar to another end.2. In this method, two magnets are dragged from the middle of the steel bar to the opposite directions.
3. As only one magnet is used to induce magnetism, this method is called single touch method.3. As two bar magnets are used to induce magnetism, this method is called double-touch method.
4. The magnetism created by this method is of low strength and lasts for a short time.4. The magnetism created by this method is of high strength and lasts for a longer time.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question b.
Which substances are used for making electromagnets?
Answer:
The substances used for making electromagnets are, an iron nail, insulated copper wire, one or more batteries and a pair of wire strippers.

Question c.
Write a note on magnetic field
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field 1

  1. The region around a magnet where the magnetic force acts on an object is called a magnetic field.
  2. The magnetic field around a magnet can be shown by means of magnetic lines of force.
  3. The intensity of the magnetic field can be obtained from the number of lines of force.
  4. The intensity of a magnetic field is low where the lines of force are sparse and the intensity is high where the lines of force are concentrated.
  5. A magnetic field is the magnetic effect of electric currents and magnetic materials.
  6. The magnetic field at any given point is specified by both the direction and the magnitude.
  7. Magnetic fields are produced by electric current.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question d.
Why is a magnetic needle used in a compass?
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field 2

  1. The magnetic compass is the most familiar compass type.
  2. It functions as a pointer to magnetic ‘north’ the local magnetic meridian, because the magnetized needle at its heart aligns itself with the horizontal component of the earth’s magnetic field.
  3. A compass is an instrument used for navigation and orientation that shows direction relative to the geographic cardinal direction.

Question e.
Explain with the help of a diagram how the intensity and direction of the magnetic field of a bar magnet can be determined.
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field 1

  1. The region around a magnet where the magnetic force acts on an object is called a magnetic field.
  2. The magnetic field around a magnet can be shown by means of magnetic lines of force.
  3. The intensity of the magnetic field can be obtained from the number of lines of force.
  4. The intensity of a magnetic field is low where the lines of force are sparse and the intensity is high where the lines of force are concentrated.
  5. A magnetic field is the magnetic effect of electric currents and magnetic materials.
  6. The magnetic field at any given point is specified by both the direction and the magnitude.
  7. Magnetic fields are produced by electric current.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

4. Give detailed information about how the merchants of olden times used a magnet while travelling.

Question a.
Give detailed information about how the merchants of olden times used a magnet while travelling.
Answer:

  1. The merchants of older times used a magnet whose north-south side was known.
  2. When the magnet is freely suspended, the north pole of the magnet settles in the direction of the geographic north pole of the earth.
  3. This is how the merchants come to know the directions while travelling.
  4. The idea for using loadstone in a compass first appeared in China.
  5. The first compass were made of loadstone, a naturally magnetised ore of iron.
  6. The Chinese used “south pointing spoon” which when thrown upon the ground comes to rest pointing at the south.
  7. Then the transition was made from loadstone spoons to iron needles.
  8. In the medieval period a magnetic needle in the shape of a fish, floating in the bowl of water was used as a compass, pointing south.
  9. It was called floating compass.
  10. In the twentieth century, liquid filled magnetic compass began to appear.
  11. There is a lesser known compass called dry mariners compass.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Project:

Question a.
Obtain information about the function of metal detectors.

Class 7 Science Chapter 19 Properties of a Magnetic Field Important Questions and Answers

Write the appropriate terms in the blanks:

Question 1.
The magnetic lines of force always run from the north pole to the ……………. .
Answer:
south pole

Question 2.
The unit of magnetic field strength is …………….. .
Answer:
A/M (Amperes per metre)

Question 3.
The earth behaves like a giant ………… .
Answer:
barmagnet

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 4.
The ……………. properties of magnetic field was proposed by scientist
Answer:
Michael Faraday

Question 5.
The region around a magnet where the magnetic force acts on an object is called the …………. .
Answer:
magnetic field

Question 6.
The ……………. of the magnetic field can be obtained from the number of lines of force.
Answer:
strength

Question 7.
The function of the metal detector is based on ………… .
Answer:
electromagnets

Question 8.
The magnetism is ………….. in the case of an electromagnet.
Answer:
temporary

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Match the following:

Question 1.

Column ‘A’Column ‘B’
1. Iron, nickel, cobalta. Electromagnet
2. Doorbell magnetb. Permanent magnet
3. Nickel, cobalt, aluminiumc. Iron, nickel, titanium, aluminium
4. Cupboard magnetd. Magnetic metals
5. Lodestonee. Mariner’s compass
6. Nipermagf. Alnico

Answer:

Column ‘A’Column ‘B’
1. Iron, nickel, cobaltd. Magnetic metals
2. Doorbell magneta. Electromagnet
3. Nickel, cobalt, aluminiumf. Alnico
4. Cupboard magnetb. Permanent magnet
5. Lodestonee. Mariner’s compass
6. Nipermagc. Iron, nickel, titanium, aluminium

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Write True or False correct and rewrite the false statements:

Question 1.
Material alnico is a mixture of aluminium, nickel and iron.
Answer:
False. Alnico is an alloy of aluminium, nickel and cobalt

Question 2.
Nipermag is alloy of iron, nickel, cobalt.
Answer:
False. Nipermag is an alloy of iron, nickel, aluminium and titanium

Question 3.
The magnetic force is concentrated at the centre of the magnet.
Answer:
False. The magnetic force is concentrated at the poles of the magnet.

Question 4.
Magnetism of electromagnet is permanent.
Answer:
False. Magnetism of electromagnet is temporary

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 5.
Like poles attract each other and unlike poles repel each other.
Answer:
False. Like poles repel each other and unlike poles attract each other

Question 6.
Two poles of a magnet cannot be separated from each other.
Answer:
True.

Answer in one or two sentences:

Question 1.
In which direction does a freely suspended magnet settle?
Answer:
The north pole of a freely suspended magnet settles in the direction of the geographic north pole of the earth. It settles in north-south direction.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 2.
Write any two functions of metal detectors.
Answer:

  1. Metal detectors are used in very important places like an airport, bus station, certain temples and buildings to inspect those who are entering these places.
  2. To detect very precious articles.

Question 3.
Which magnetic poles attract each other?
Answer:
Opposite magnetic poles attract each other.

Question 4.
What metals are used for making magnets?
Answer:

  1. Magnets are made from alloys of iron, cobalt and nickel.
  2. Nipermag is an alloy made up of iron, nickel, aluminium and titanium whereas alnico is an alloy made up of aluminium, nickel and cobalt.

Give scientific reasons:

Question 1.
Earth is called as a gigantic magnet.
Answer:

  1. The earth is called as a gigantic magnet because north pole of a freely suspended magnet always settles in the direction of the geographic north pole of the earth.
  2. It means that the South pole of some gigantic magnet must be near the geographic north pole of that magnet, near the geographic south pole of the earth.
  3. Gilbert inferred from this that the earth itself is a gigantic magnet.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 2.
A bar magnet kept on the cardboard arranges the iron fillings in line.
Answer:

  1. A bar magnet kept on the cardboard arranges the iron fillings in line because of the ‘magnetic lines of force’.
  2. These magnetic lines gauge the intensity of the bar magnet.
  3. The intensity of the magnetic field is low where the lines of force are sparse and the intensity is high where the lines of force are concentrated.
  4. Therefore, the iron fillings arrange themselves according to intensity of the magnetic lines of force.

Write short note on:

Question 1.
Functions of metal detector.
Answer:

  1. A metal detector is an electronic instrument which detects the presence of metals nearby.
  2. The function of these machines is based on electromagnets.
  3. Metal detectors are useful for finding metal inclusions hidden within objects, or metal objects buried underground.
  4. They often consist of a handheld unit with a sensor probe which can be swept over the ground or other objects.
  5. Metal detectors in the food manufacturing industry are used to detect and eliminate threats of metal contaminants in food.
  6. Metal detectors are used in very important places like an airport, bus station, temples, hospitals, railway stations, theaters, malls.
  7. They are used for inspection of persons entering these places.
  8. In geology, these machines are used to detect the presence and quantity of metals.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 2.
Write down the characteristics of magnets.
Answer:

  1. A freely suspended magnet settles in the north- south direction always.
  2. Magnets attract objects of iron, cobalt and nickel.
  3. The force of attraction of a magnet is greater at its poles than in the middle.
  4. Like poles of two magnets repel each other.
  5. Opposite poles of two magnets attract each other.
  6. If a bar magnet is suspended by a thread and if it is free to rotate its south pole will move towards the north pole of the earth.

Question 3.
Characteristic of magnetic lines of force
Answer:

  1. Magnetic lines of force start from the north pole and end at the south pole.
  2. They are continuous through the body of magnet. These are imaginary lines.
  3. Magnetic lines of force can pass through iron more easily than air
  4. The number of magnetic lines force determines the strength of the magnetic field at a particular point.
  5. Two magnetic lines of force do not intersect each other.
  6. They contract longitudinally and expand laterally.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Can you tell?

Answer the following questions:

Question 1.
Why does freely suspended magnet always settle in the north-south direction?
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field 3

  1. Gilbert gave a scientific explanation that earth itself is a gigantic magnet.
  2. The north pole of a freely suspended magnet settles in the direction of the geographic north pole of the earth since unlike poles attract.
  3. It means that the south pole of some gigantic magnet must be near the geographic north pole of the earth.
  4. In other words, the thing we call Earth’s geographic north pole is actually the south pole of the magnet inside the earth.
  5. That is why a freely suspended magnet always settles in geographic north-south attracted by opposite poles of the earth as a great magnet.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 2.
Which magnetic poles attract each other?
Answer:
Opposite magnet poles attract each other.

Question 3.
Which pole of a spherical magnet will get attracted towards the south pole of the bar magnet?
Answer:
The north pole of the spherical magnet will get attracted towards the south pole of the bar magnet.

Question 4.
Why does a bar magnet kept at a distance doesn’t attract pins?
Answer:

  1. The region around the magnet where the magnetic force acts is known as the magnetic field.
  2. It has its influence over a certain distance where these magnetic lines of force act.
  3. As these pins are outside this area of influence, a bar magnet does not attract those pins which are at a distance.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 5.
What is an electromagnet?
Answer:

  1. A soft metal core made into a magnet by the passage of electric current through a coil surrounding it is called an electromagnet.
  2. An electromagnet is a magnet in which the magnetic field is produced by electricity.
  3. The strength of an electromagnet can easily be changed by changing the amount of electric current that flows through it.
  4. The poles of an electromagnet can even be reversed by reversing the flow of electricity.
  5. An electromagnet works because an electric current produces a magnetic field.
  6. The magnetic field produced by electric current forms circles around the electric current.
  7. The magnetic field disappears when the current is turned off.

Question 6.
How can an electromagnet be made?
Answer:
Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field 4

  1. It is easy to make an electromagnet. To make an electromagnet we needs an iron nail of 10 cm length 1 metre long insulated copper wire, a battery cell, pins.
  2. Neatly wrap some insulated copper wire around an iron nail.
  3. The more wires you wrap around the nail, the stronger will be the electromagnet.
  4. Attach a battery to the wire, an electric current will begin to flow and the iron nail will become magnetised.
  5. When the battery is disconnected the iron nail will lose its magnetism.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Use your brainpower!

Answer the following questions:

Question 1.
Which direction will a magnetic needle show on the geographic north pole?
Answer:
North, all magnetic needle will show north direction.

Question 2.
Is magnetic force a vector or a scalar quantity?
Answer:

  1. It is a vector quantity. Because a magnetic field is the magnetic effect of electric current and magnetic materials.
  2. The magnetic field at any given point is specified by both direction and magnitude as such it is represented by a vector field.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 3.
What is meant by magnetic force?
Answer:

  1. The magnetic force, is a consequence of the electromagnet force, one of the four fundamental forces of nature.
  2. It occurs whenever objects interact in which there is charge in motion, eg. compass, a motor.
  3. Magnetic force is an attraction or repulsion that arises between electrically charged particles because of their motion.

Question 4.
How does a magnetic force act without direct contact?
Answer:
Magnetic force acts without even direct contact because of the penetrating ability of the magnetic field.

Question 5.
What is the difference between the gravitational force and magnetic force?
Answer:

Gravitational ForceMagnetic Force
1. Gravitational force acts between any objects with mass.1. Magnetic force acts between objects with specific properties.
2. It depends on the mass of the objects.2. It does not depend on the mass of objects.
3. It acts over long-range as a gravitational field is large.3. It acts over short-range as magnetic field is short.
4. Object are always attracted to each other.4. Objects are either attracted or repelled.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 6.
Why is repulsion the real test for identifying a magnet?
Answer:
A magnet can attract the opposite pole of the magnet as well as magnetic substances (iron, cobalt, nickel).
So if a magnet attracts another object, we cannot say that other object is a magnet or a magnetic substance.
But if the magnet repels the object then we can surely say that the other object is a magnet because like poles of magnets repel each other.

Question 7.
How will you find a magnet from among the various articles given to you?
Answer:

  1. Magnet has the property to attract things made of iron, cobalt, nickel.
  2. A magnet has magnetic field which is invisible but it has the property of attracting, pulling other
  3. ferromagnetic material such as iron, cobalt and nickle.
  4. So if we bring any iron, nickle objects near it, they will be pulled toward the magnet.
  5. The real test is repulsion.
  6. The article which shows repulsion when one of the poles of a magnet is brought near is surely a magnet.

Find out:

Answer the following questions:

Question 1.
Why does the magnetic needle of a compass not settle parallel to the ground but at an angle to it?
Answer:

  1. Freely suspended magnet does not point to north, but aligns to earth’s magnetic field axially ie – north-south.
  2. The magnetic pole of the earth is different from true north (geographic north) because the magnetic north is constantly changing.
  3. It leads to a margin of error called declination in which there is an angle of difference (11.51)° between Magnetic North and True North.
  4. That is the reason the magnetic needle does not settle parallel to the ground.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Observe the diagrams and answer the questions given below:
Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field 5

Question 1.
Name the methods shown.
Answer:
Single touch and Double touch methods of magnetising a steel bar.

Maharashtra Board Class 7 Science Solutions Chapter 19 Properties of a Magnetic Field

Question 2.
Give 2 points of difference between the methods.
Answer:

Single touch methodDouble touch method
1. In this method, one magnet is used to magnetise the steel bar.1. In this method, two magnets are used to magnetise the steel bar.
2. In this method, the magnet is dragged from one end of the steel bar to another end.2. In this method, two magnets are dragged from the middle of the steel bar to the opposite directions.
3. As only one magnet is used to induce magnetism, this method is called single touch method.3. As two bar magnets are used to induce magnetism, this method is called double-touch method.
4. The magnetism created by this method is of low strength and lasts for a short time.4. The magnetism created by this method is of high strength and lasts for a longer time.

Maharashtra State Board Class 7 Science Textbook Solutions

Towards Green Energy Class 10 Questions And Answers Maharashtra Board

Std 10 Science Part 2 Chapter 5 Towards Green Energy Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 2 Chapter 5 Towards Green Energy Question Answer Maharashtra Board

Question 1.
Remake the table taking into account relation between entries in three columns.

IIIIII
CoalPotential energyWind electricity plant
UraniumKinetic energyHydroelectric plant
Water reservoirNuclear energyThermal plant
WindThermal energyNuclear power plant

Answer:

IIIIII
CoalThermal energyThermal plant
UraniumNuclear energyNuclear power plant
Water reservoirPotential energyHydroelectric plant
WindKinetic energyWind electricity plant

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 2.
Which fuel is used in thermal power plant? what are the problems associated with this type of power generation?
Answer:
(1) The fuel used in the thermal power plant is coal. Coal contains chemical energy. Upon burning it releases heat energy. This heat is used for generation of electricity in the thermal power plants.

(2) Problems associated with power 8enerations by thermal power plant:
(a) Air pollution: Due to burning of coal, there is emission of carbon dioxide, carbon monoxide, sulphur dioxide and nitrogen dioxide gases. These are harmful and toxic to health.
(b) Soot particles emitted during combustion can cause severe respiratory problems such as asthma.

Question 3.
Other than thermnl power plant. which power plants use thermal energy for power generation? In what different ways is the thermal energy obtained?
Answer:
(1) The power plant based on natural gas and the nuclear power plants also used thermal energy for the power generation. Apart from these, solar energy is also used to produce heat and thereby create the power.
(2) In nuclear power plant, the energy is released by carrying out fission of nuclei of atoms like Uranium or Plutonium. This energy is used to generate the steam of high temperature and high pressure. The steam rotates the turbine. The kinetic energy in steam drives the turbine and turbine in turn drives the generator.
(3) The combustion of natural gas produces gas, which is used to run the turbine. This gas is under high pressure and high temperature. This is used to produce thermal energy.
(4) In solar thermal power plant, thermal energy is generated with the help of solar radiation. For this reflectors and absorbers are used for concentrating solar radiation and converting it into thermal energy.

Question 4.
Which type/types of power generation involve maximum number of steps of energy conversion? In which power generation is the number minimum?
Answer:
The steps of energy conversion are maximum in the thermal power generation. They are minimum in wind energy generation.

Question 5.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 1
a. Maximum energy generation in India is done using …………… energy.
Answer:
Maximum energy generation in India is done using thermal energy.

b. ………. energy is a renewable source of energy.
Answer:
wind energy is a renewable source of energy.

c.Solar energy can be called ……. energy.
Answer:
Solar energy can be called clean energy.

d. ……. energy of wind is used in wind mills.
Answer:
kinetic energy of wind is used in wind mills.

e. ………. energy of water in darns is used for generation of electricity.
Answer:
Potential energy of water in darns is used for generation of electricity.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 6.
Explain the difference.
a. Conventional and Non-conventional Sources of energy.
Answer:
Conventional Sources of energy:

  1. Conventional sources of energy are largely polluting, they release lot of carbon through its emissions.
  2. Conventional sources of energy are not eco¬friendly.
  3. The fuels produced from the conventional sources of energy are comparatively costlier.
  4. Conventional energy power plants require less area and its management cost is also less.
  5. Conventional source of energy are non-renewable.
  6. Conventional sources of energy are in the form of limited reserves. After few years they will be completely over. e.g. Fossil fuels, coal, crude oil, diesel, petrol, natural gas, etc.

Non-conventional Sources of energy:

  1. Non-conventional sources of energy are not polluting, They do not release carbon or other toxic gases.
  2. Non-conventional sources of energy are eco-friendly.
  3. The energy obtained from the non-conventional sources of energy are comparatively cheaper.
  4. Non-conventional energy power plants require more area and its management cost is also more.
  5. Non-conventional source of energy are renewable.
  6. Non-conventional energy sources qre in abundance on the earth. They are persistent and sustainable. Thus they will not get over. e.g. Solar energy, wind energy, etc.

b. Thermal electricity generation and Solar thermal electricity generation.
Answer:
Thermal electricity generation:

  1. After burning the coal, the heat that is produced is used in the generation of thermal electricity.
  2. For producing heat, the coal is burnt in the boilers.
  3. The combustion of coal produces heat. This heat converts water into steam, which is under very high temperature and pressure. By its force the turbines move. The turbines in turn are connected to generator which rotates and produces energy.
  4. Thermal energy is polluting and not eco-friendly.
  5. The fuel here is coal, its reserves are limited.

Solar thermal electricity generation:

  1. Solar radiations are used in solar thermal electricity production.
  2. For production of heat, many reflectors are used which reflect the radiations of the sun into the absorbent.
  3. Sun’s heat convert the water into steam that rotates the turbine. The turbines then rotate the generators. This generates the electricity.
  4. Solar energy is not polluting, it is eco-friendly.
  5. The solar radiations are in abundance and are sustainable and persistent.

Question 7.
What is meant by green energy? Which energy sources can be called green energy sources and why? Give examples.
Answer:
(1) Green energy means eco-friendly form of energy which does not cause environmental problems and are non-exhaustible, perpetual and sustainable.
(2) These sources of energy do not produce toxic gases or other pollutants, therefore they are safe.
(3) Examples of green energy: (i) Hydroelectric energy (ii) Wind energy (iii) Solar energy (iv) Energy obtained biofuels.

Question 8.
Explain the following sentences.
a. Energy obtained from fossil fuels is not green energy.
Answer:
Fossil fuels like petrol, diesel or natural gas when burnt, emit toxic gases and soot particles. Thus, fossil fuels cause air pollution. Burning of fossil fuels cause increased levels of carbon dioxide, carbon monoxide and nitrogen dioxide. The increased carbon dioxide emission results in global warming. Nitrogen oxide results later in acid-rain. Soot particles generated through burning of fuels cause respiratory problems iike asthma.

Moreover, the fossil fuels are non-renewable and exhaustible fuels. They have to be explored from the
deeper layers of the earth causing lots of environmental problems. Green energy is sustainable, renewable and abundant. It never creates any environmental problems and is non-polluting. Thus, energy obtained from fossil fuels is not at all a green energy.

b. Saving energy is the need of the hour.
Answer:
In modern civilization, continuous energy supply is needed for the technology and development. The energy has become a basic need for man. Most of the energy used in India is obtained from thermal power plant. For this energy generation, various fuels are used. The coal and fossil fuels are limited. Due to over-exploitation, these reserves are getting fast depleted. Use of fossil fuels is also resulting in pollution and climate change.

Nuclear energy can be very hazardous. Lot of research is being done in the field of green energy, but the tremendous human population always is in need of more energy. Therefore, each and every person should save the energy, as saving energy is the need of the hour.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 9.
Answer the following questions.
a. How can we get the required amount of energy by connecting solar panels?
Answer:

  • The photovoltaic solar cells can be connected in a series or in parallel to make a solar panel.
  • When solar cells are connected in a series, the potential difference of individual cells are added in the combination, however the currents from individual cells are not added.
  • When solar cells are connected in parallel, the currents of the individual cells are added in the combination, but the potential differences from individual cells are not added.
  • Through such connections the required potential difference and current can be obtained.
  • Many such solar panels are connected in series and in parallel to generate required current and potential difference.
  • When many solar panels connected in series they form a solar string. Many solar strings connected in parallel make a solar array. In such manner we can get the required amount of energy by connecting solar panels.

b. What are the advantages and limitations of solar energy?
Answer:
I. Advantages:

  • While generating the power through solar radiations, no fuel is burnt.
  • Solar energy generation thus does not create any type of pollution. The technology can be completely utilized in regions with abundant sunlight.
  • Solar energy is eco-friendly, green energy.

II. Limitations:

  • Sunlight is available only during day time. Thus solar cells can generate power only during day.
  • In rainy season and in cloudy conditions, solar power generation suffers.
  • The power present in the solar cells is DC while most of the domestic equipments work on AC.

Question 10.
Explain with diagram step-by-step energy conversion in
a. Thermal power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 2a
In thermal power plant the turbines are rotated using steam. Here the coal is burnt. The heat energy liberated from this burning is used to heat the water in the boiler. This water produces steam of very high temperature and pressure. The kinetic energy in the steam rotates the turbines. The rotation of turbines produces its own mechanical kinetic energy.

The generators connected to turbines produce electrical energy. The steam is condensed in a condenser and converted back into water. In this way in thermal power plant, thermal energy to kinetic energy, kinetic energy into mechanical energy and mechanical energy to electrical energy, are the conversions that take place.

b. Nuclear power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 3a
In nuclear power plant, the energy is releasdd by fission of nuclei of atbms like Uranium or Plutonium. This energy is used to generate the steam or high temperature ind high pressure. The kinetic energy in the steam rotates the turbine. The turbine in turn drives the generator to produce electricity.

c. Solar thermal power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 4a
Solar radiation is used to produce thermal energy. For this purpose, many reflectors are used which concentrate the solar radiation on absorbers. The heat energy created due to solar radiations is used to make steam. The steam possesses kinetic energy. This kinetic energy drives turbine and generator. The electrical energy is thus created from this kinetic energy.

d. Hydroelectric power plant:
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 5a
In hydroelectric plant the water stored in the reservoir is used as a source or potential energy. This water is made to fall at a great speed and hence there is production of kinetic energy in flowing water. This fast flowing water ralling down from the reservoir is brought to the turbine at the lower levels. The kinetic energy of the flowing water in turn drives the turbine, The turbine then drives the generator and electrical energy is produced.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 11.
Give scientific reasons:
a. The construction of turbine is different for different types of power plants.
Answer:

  • Generators work on the principles of electromagnetic induction.
  • For this the generator must be rotated.
  • For this purpose, there is a turbine for each generator.
  • This rotation needs energy. The turbines are different according to the type of energy source that is used for its rotation.
  • Therefore, the construction of turbine is different for each power plant.

b. It is absolutely necessary to control the fission reaction in nuclear power plants.
Answer:

  • Nuclear fission reaction is a type of chain reaction.
  • In nuclear power plants these reactions are closely controlled.
  • If these reactions are not managed properly, there can be more production of neutrons in an uncontrolled way.
  • Each released neutron further causes fission of 3 Uranium (U-235) atoms, such uncontrolled reactions can cause hazardous accidents, hence ft is absolutely necessary to control the fission reaction in nuclear power plants.

c. Hydroelectric energy, solar energy and wind energy are called renewable energies. (July ’19, Board’s Model Activity Sheet)
Answer:

  • Hydroelectric energy, solar energy and wind energy is obtained respectively from flowing water, solar radiations and flowing wind.
  • These sources, i.e. water reservoirs, sun and the wind are inexhaustible and sustainable. They will not be finished.
  • On the contrary, the conventional energy sources such as coal and fossil fuels have limited reserves.
  • They cannot be renewed and may get exhausted in future. Hydroelectric energy, solar energy and wind energy can be replenished and hence they are called renewable.

d. It is possible to produce energy from mW to MW using solar photovoltaic cells.
Answer:

  1. Solar panels can be constructed by connecting solar photovoltaic cells in either series or in parallels.
  2. The combinations are done in such a way that it can give the desired potential difference and the current.
  3. Solar strings are then made by joining solar panels in a series.
  4. When solar strings are joined in parallel; they form solar array.
  5. Therefore, by proper combinations, it becomes possible to produce energy from mW to MW using solar photovoltaic cells.

Question 12.
Draw a Schematic diagram of Solar thermal electric energy generation.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 6

Question 13.
Give your opinion about whether hydroelectric plants are environment-friendly or not?
Answer:

  • Hydroelectric plants are advantageous in some respect while in some aspects it does create problems.
  • Hydroelectric power generation does not need burning of fuels. Therefore, there is no problem regarding combustion of fuels and release of toxic pollutants.
  • Electricity can be obtained as and when required if there is enough water in the reservoir.
  • Water is replenished every time when there is sufficient rainfall.
  • All the above facts give an impression that hydroelectric power generation is eco-friendly but it is not.
  • Many villages and settlements are submerged when a dam and reservoir is constructed. The displaced people are given re-settlement, but it causes lot of emotional trauma to people.
  • Biodiversity is affected as forest lands is submerged. The river flow is obstructed by the dam which affects the aquatic organisms residing in such water.
  • Due to excessive pressure of water on land, it is said that the region gets prone to earthquakes.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 14.
Draw a neat labelled diagrams.
a. Energy transformation in solar thermal electric energy generation.
Answer:

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 7a

b. One solar panel produces a potential difference of 18 V and current of 3A. Describe how you can obtain a potential difference of 72 Volts and current of 9 A with a solar array using solar panels. You can use sign of a battery for a solar panel.
Answer:
Given Potential difference is 18 V and current is 3A. The requirement is potential difference of 72 V and current is 9A Voltage remains the same if connected in parallel and gets added it they are connected in series. Current remains the same if connected in series but adds if connected in parallel.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 8

Question 15.
Write a short note on Electrical energy generation and environment.
Answer:
The energy obtained through the fossil fuels as well as nuclear energy can cause degradation of the environment. If such energy sources are used, they can cause harm to the environment.

(1) The burning of fossil fuels cause air pollution. The incomplete combustion of fossil fuels cause release of carbon monoxide. Some more toxic gases and soot particles cause various respiratory diseases. The carbon dioxide produced is creating global warming and climate change. The nitrogen dioxide released through burning is responsible for acid rains.

(2) Fossil fuels are limited. They are getting fast depleted. It has taken millions of years for the fossil fuels to form. The exploration of such fuels also cause environmental degradation and marine pollution too.

(3) In production of nuclear energy, there is a great risk of accidents. The safe disposal of nuclear waste is also a problem.

(4) Hydroelectric power from water reservoirs, wind power from wind, solar energy from sun and electricity from biofuels are eco-friendly alternatives.

Projects: (Do it your self)

Project 1.
Gather information about solar light, solar water heating system and solar cooker.

Project2.
Gather information about a power plant near your locality by visiting the plant.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Can you recall? (Text Book Page No. 47)

Question 1.
What is Energy?
Answer:
The capacity to do work is called energy.

Question 2.
What are different types of Energy?
Answer:
Potential energy and kinetic energy are the two types of energy.

Question 3.
What are different forms of Energy?
Answer:
Heat, light, electric energy, solar energy, chemical energy, nuclear energy, mechanical energy, etc. are different forms of energy.

Use your brain power! (Text Book Page No. 54)

Question 1.
The schematic of hydroelectric plant is shown in Figure 5.17 on text book page no. 54. Water from about middle of the total height of the dam is taken to the turbine, as shown by point B in the diagram.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 9
(i) With reference to point B, potential energy of how much water reservoir in the dam will be converted into kinetic energy?
Answer:
When the sluice gate at point B is opened, the water from reservoir will start flowing. The potential energy of the stored water will become kinetic energy of the quantity of water that is let out through the sluice gates.

(ii) What will be the effect on electricity generation, if the channel taking water to turbine starts at point A?
Answer:
If the channel taking water to turbine starts at point A, then the water will flow with a greater speed. Since point A is at hSight, water will acquire speed. This will result into more efficient rotation of the blades of turbine. The electricity generation will thus become more efficient.

(iii) What will be the effect on electricity generation, if the channel taking water to turbine starts at point C?
Answer:
If the channel taking water to turbine starts at point C, it will affect the electricity generation adversely. Point C is on the lower height as compared to the channel that carries water to the turbine. The flow of the water thus will be affected resulting into improper rotation of blades of turbine. This will certainly affect the electricity generation.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Large ……….. are used in commercial power generation plants.
(a) machines
(b) generators
(c) turbines
(d) pannels
Answer:
(b) generators

Question 2.
The principle of electromagnetic ……….. was invented by Michael Faraday.
(a) induction
(b) attraction
(c) repulsion
(d) expulsion
Answer:
(a) induction

Question 3.
………… is used to rotate the magnet in the generator.
(a) fan
(b) Generator
(c) Turbine
(d) Panels
Answer:
(c) Turbine

Question 4.
In thermal power plants, the ………… energy in the coal is converted into electrical energy through several steps.
(a) physical
(b) biological
(c) kinetic
(d) chemical
Answer:
(d) chemical

Question 5.
At ………. in Andhra Pradesh power plant based on natural gas has been installed.
(a) Hyderabad
(b) Vishakhapatnam
(c) Samaralkota
(d) Kakinada
Answer:
(c) Samaralkota

Question 6.
Burning of coal may cause serious health problems related to ……….. system.
(a) digestive
(b) respiratory
(c) nervous
(d) excretory
Answer:
(b) respiratory

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 7.
Incomplete combustion of fuels leads to formation of ……….
(a) carbon dioxide
(b) carbon monoxide
(c) carbon tetrachloride
(d) All the above
Answer:
(b) carbon monoxide

Question 8.
Solar cells are made of a special type of material called semiconductor such as ………..
(a) silicon
(b) uranium
(c) borosilicate
(d) hydrogen
Answer:
(a) silicon

Question 9.
……….. of the following is eco-friendly energy resource. (Board’s Model Activity Sheet)
(a) Coal
(b) Hydroelectric power
(c) Fossil fuel
(d) Atomic energy
Answer:
(b) Hydroelectric power

Question 10.
Which is the most abundant and renewable energy?
(a) Thermal power
(b) Solar energy
(c) Fossil fuels
(d) Atomic power
Answer:
(b) Solar energy

Question 11.
What are the two technologies for harnessing solar energy?
(a) Solar photovoltaics and solar thermal
(b) Solar cooker and solar lamp
(c) Heat capturing and Heat conversation
(d) Active and passive technologies
Answer:
(a) Solar photovoltaics and solar thermal

Question 12.
Which of the following is used in solar cooker to harvest the solar energy?
(a) Solar panels
(b) Silicon cell
(c) Mirrors
(d) Glass lid
Answer:
(c) Mirrors

Question 13.
Which of the following is not the source of green energy?
(a) Wind
(b) Natural gas
(c) Sunlight
(d) Fossil fuel
Answer:
(d) Fossil fuel

Question 14.
The solar lamp uses the energy.
(a) Heat
(b) Wind
(c) Light
(d) Sound
Answer:
(c) Light

State whether the following statements are true or false with proper explanation:

Question 1.
In thermal power plants, the turbines work on solar energy.
Answer:
False. (In thermal power plant, the turbines work on steam. The turbines working on solar energy are not used.)

Question 2.
How to dispose the nuclear waste safely is a big challenge before the scientists.
Answer:
True. (Nuclear waste disposal is the greatest problem. It produces highly toxic effects in any ecosystem. Therefore, disposing such radioactive substances becomes a major challenge.)

Question 3.
The efficiency of power generation using coal plant is higher than that of power generation plant based on natural gas.
Answer:
False. (The efficiency of power generation using natural gas plant is higher than that of power generation plant based on coal.)

Question 4.
Energy obtained from nuclear fission is eco-friendly.
Answer:
False. (Energy obtained from nuclear fission is not eco-friendly, because if accidents happen it leads to hazardous accidents.)

Question 5.
In hydroelectric power plant, the kinetic energy in water stored in dam is converted into potential energy of water.
Answer:
False. (In hydroelectric power plant, the potential energy in water stored in dam is converted into kinetic energy of water. The forceful downpour of flowing water causes this kinetic energy.)

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 6.
The turbine is connected to electric generator, therefore the magnet rotates and electric energy is thus produced.
Answer:
True. (The rotating wheels of turbine cause mechanical energy. This energy helps to produce electrical energy.)

Question 7.
Use of energy is unavoidable in our daily life, but we must use it carefully and only in the required amount.
Answer:
True. (The energy Supply for everyday use results into lot of pollution. This causes harmful effects in the surrounding environment. Therefore, energy should be used in minimal amount and with great care.)

Question 8.
The machine which converts the potential energy of wind to electrical energy is called wind-turbine.
Answer:
False. (When wind blows, the kinetic energy is present in it. This kinetic energy is converted into electricity. The flowing wind never has a potential energy.)

Question 9.
The potential difference available from a solar cell is independent of its area.
Answer:
True. (The potential difference available from a solar cell is independent of its area. However, it is dependent on the way in which solar cells are connected.)

Question 10.
The power available from the solar cells is AC.
Answer:
False. (The power available from solar cells is always DC while the domestic appliances that we use work on AC.)

Match the columns:

Question 1.

Column IColumn II
(1) Polluting energy(a) Soot particles
(2) Eco-friendly energy(b) Thermal energy
(c) Nuclear energy
(d) Wind energy

Answer:
(1) Polluting energy – Thermal energy
(2) Eco-friendly energy – Wind energy

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 2.

Column IColumn II
(1) Pollutants(a) Soot particles
(2) Hazard to ecosystem(b) Thermal energy
(c) Nuclear energy
(d) Wind energy

Answer:
(1) Pollutants – Soot particles
(2) Hazard to ecosystem – Nuclear energy

Question 3.

Type of energyProblem
(1) Nuclear energy(a) Rehabilitation of displaced people
(2) Natural gas(b) Rainy season and darkness
(c) Limited reserves
(d) Disposal of wastes

Answer:
(1) Nuclear energy – Disposal of wastes
(2) Natural gas – Limited reserves

Question 4.

Type of energyProblem
(1) Solar energy(a) Rehabilitation of displaced people
(2) Hydroelectric  energy(b) Rainy season and darkness
(c) Limited reserves
(d) Disposal of wastes

Answer:
(1) Solar energy – Rainy season and darkness
(2) Hydroelectric energy – Rehabilitation of displaced people

Find the odd one out:

Question 1.
Kudankulam, Tarapur, Ravatabhata, Anjanvel
Answer:
Anjanvel. (All others are places having nuclear power plants.)

Question 2.
Samaralkota, Kudankulam, Bavanaa, Kondapalli
Answer:
Kudankulam. (All others are places having power plants based on natural gas.)

Question 3.
Tehari, Koyana, Srishailam, Tarapur
Answer:
Tarapur. (All others are places having hydroelectric projects.)

Question 4.
Edible oil, crude oil, LPG, CNG
Answer:
Edible oil. (All others are fossil fuels.)

Question 5.
Hydroelectric energy, Solar energy, Nuclear energy, Wind energy
Answer:
Nuclear energy. (All others are eco-friendly green energy types.)

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Explain with diagram step-by-step energy conversion in:

Question 1.
Power plant based on natural gas.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 10a
In a power plant based on natural gas, there are three main sections of the plant. There is combustion chamber with compressor in which the steam under pressure is introduced. The natural gas burns in the presence of air in this combustion chamber. This results in a production of a gas which is at very high temperature and pressure. This generated gas from the chamber runs the turbine. The kinetic energy of the turbine drives the generator. The generator produces electrical energy.

Question 2.
Power plant based on wind energy.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 11a
Wind energy is used for moving turbines. The wind with specific speed is used to rotate the large fins of wind turbine. The kinetic energy in these fins is transferred to generator which then produces electrical energy.

Explain the following questions in detail:

Question 1.
What are the advantages of hydroelectric power generation? (March 2019)
Answer:

  1. Hydroelectric energy does not cause pollution.
  2. Generation of hydroelectric energy does not involve burning of fossil fuel.
  3. If sufficient water storage is available then electricity generation can be done as per requirement.
  4. Rainwater can replenish the water storage and power generation can thus be done uninterrupted.

Question 2.
How is nuclear fission reaction carried out in nuclear power plants?
Answer:

  • In nuclear power plants neutrons are bombarded on atom of Uranium – 235.
  • This causes conversion of Uranium – 235 into its isotope U – 236.
  • U-236 is very unstable and thus forms atoms of Barium and Krypton by nuclear fission. This forms 3 neutrons and 200 MeV energy.
  • In a similar way three more Uranium – 235 atoms are subjected to nuclear fission which then releases energy.
  • The neutrons released are again used for further nuclear fission reactions. In this way nuclear fission reactions are carried out in controlled manner in nuclear power plants.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 3.
Draw schematic of power plant based on natural gas and answer the following questions: (July 2019)
(a) At which place natural gas power plant is situated in Maharashtra?
(b) How is pollution reduced in natural gas based power plant?
(c) Give two examples of eco-friendly electricity process.
Answer:
(a) Natural gas power plant is situated at Anjanvel in Maharashtra.
(b) Natural gas does not contain sulfur. Burning of such natural gas does not produce pollution.
(c) Solar energy and wind energy are two examples of eco-friendly energy.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 30

Complete the paragraph by choosing the appropriate words given in the brackets:

Question 1.
(marginal, array, cell, panel, string, current, power station, potential difference).
Many solar panels are connected in series and in parallel to generate required ………… and ……… Solar …………. is the basic unit in solar electric plant. Many solar cells come together to form a solar …………… Many solar panels connected in series form a solar ………., and many solar strings connected in parallel form a solar …………. As we can obtain as much electrical power as needed, they are used in applications which need ……….. power (e.g. calculators that run on solar energy) to ……….. of MW capacity.
Answer:
Many solar panels are connected in series and in parallel to generate required current and potential difference. Solar cell is the basic unit in solar electric plant. Many solar cells come together to form a solar panel. Many solar panels connected in series form a solar string, and many solar strings connected in parallel form a solar array. As we can obtain as much electrical power as needed, they are used in applications which need marginal power (e.g. calculators that run on solar energy) to power station of MW capacity.

Read the paragraph and answer the questions given below:

1. Renewable energy is, energy produced from sources that do not deplete or can be replenished within a human’s life time. The most common examples include wind, solar, geothermal, biorhass, and hydroelectric power. This is in contrast to non-renewable sources such as fossil fuels. Most renewable energy is derived directly or indirectly from the sun. Sunlight can be captured directly using solar technologies. The sun’s heat drives winds, whose energy is captured with turbines. Plants also rely on the sun to grow and their stored energy can be utilized for bioenergy. Not all renewable energy sources rely on the sun. For example, geothermal energy utilizes the Earth’s internal heat, tidal energy relies on the gravitational pull of the moon, and hydroelectric power relies on the flow of water.

Questions and Answers :
Question1.
What is renewable energy?
Answer:
Renewable energy is energy that is produced from sources which will not get exhausted within a human’s life time.

Question 2.
Give the examples of renewable energy.
Answer:
Wind, solar, geothermal, biomass and hydroelectric power are some examples of renewable energy.

Question 3.
Why will energy from fossil fuel be over soon?
Answer:
Fossil fuels are exhaustible in their amount. We have been using these extensively in the past 100 years and hence it may get over soon. It is a non-renewable resource.

Question 4.
Name the renewable sources of energy which are not dependent on sun. What are they dependent upon?
Answer:
Geothermal energy, tidal energy and hydroelectric power are renewable energy resources which are not dependent on sun. Geothermal energy utilizes the Earth’s internal heat, tidal energy relies on the gravitational pull of the moon, and hydropower relies on the flow of water.

Question 5.
Which type of energy do we mostly use in India?
Answer:
The most used energy resource is coal, i.e. fossil fuel based energy followed by hydroelectric energy.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

2. Read the information given below and solve the questions based on it.
Electric energy is produced in various ways like hydroelectric, wind power, solar energy, bio-fuel, etc. These energy sources are inexhaustible, sustainable. Besides, it does not cause any environmental problem.

Questions and Answers:
Question 1.
Above information is about which type of energy?
Answer:
From the above information, we understand about green energy.

Question 2.
Whether the fossil fuel is an example of this energy?
Answer:
Fossil fuels are not green energy.

Question3.
Draw the flow chart of production of electric energy.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 12

Diagram based questions:

Question 1.
Observe the connections of cells shown in the following images.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 13
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 14
(i) Which connection will give maximum potential difference?
Answer:
The solar cells shown in the diagram 5.19 (a) are connected in series. This gives maximum potential difference.

(ii) Give one advantage and one disadvantage of this energy.
Answer:
Advantage of Solar energy: Solar energy is eco-friendly which does not create pollution. It is boundless source.
Disadvantage of solar energy: Solar energy is available only when sun is in the sky. Therefore, it has to be stored in batteries.

Question 2.
Answer the following questions:
(a) Write the name of the device shown in the above diagram.
Answer:
Steam turbine is the device shown in the above diagram.

(b) Write briefly the work of this device.
Answer:
Turbine is a device with the blades. When the flow of liquid or gases is directed on the blades of the turbine, they rotate. The rotation produces kinetic energy. This turbine is then used to rotate the magnet in the electric generator. For this purpose, turbines are connected with the generators. The magnets rotate and produce electric energy by electromagnetic induction. The turbines working on steam are used in large commercial power generation plants.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 3.
Label the given diagram of Electromagnetic induction.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 15

Question 4.
Answer the questions with the help of picture.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 16
(a) Which type of energy is produced?
(b) This power plant is based on which energy source.
(c) Is this power plant eco-friendly? How?
Answer:
(a) In the picture, it is shown that using wind energy electricity is produced.
(b) The power plant shown here is based on kinetic energy of wind which is converted to electric energy by utilizing kinetic energy from rotating turbines.
(c) This power plant is eco-friendly because it does not cause pollution. Wind energy is green energy which is non-exhaustible and perpetual.

Question 5.
Observe the figure and answer the questions given below.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 17
Answer:
(a) Name the reaction.
Answer:
The reaction shown in nuclear fission or chain reaction.

(b) Where is this reaction used?
Answer:
This reaction is used in nuclear power plants where electricity is generated.

(c) Which element is used in it?
Answer:
Uranium-235 is used in the nuclear fission reactions.

(OR)

Identify the process shown in figure and name it. (March 2019)
Answer:
The above figure shows nuclear fission chain reaction of Uranium – 236.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 6.
Observe the diagram and answer the questions : (March 2019)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 18
(a) Which energy is generated from the power plant?
Answer:
The diagram shows electricity generated from natural gas.

(b) State its source.
Answer:
The energy is generated from natural gas.

(c) Which is more eco-friendly – Power generation from coal or Power generation from natural gas? Why?
Answer:
Power generation from natural gas is more eco-friendly. Natural gas does not contain sulfur and hence its burning does not cause major pollution by forming sulphur dioxide. The efficiency of power generation by natural gas is also high.

Question 7.
Write the names of apparatus that is used in thermal power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 19

Question 8.
Label correctly the diagram of Nuclear power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 20

Question 9.
Label correctly the diagram of power plant baded on natural gas.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 21

Question 10.
Label correctly the structures seen in Windmill.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 22

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 11.
Sketch two ways in which solar cells can be connected. Also draw the diagrams to show the arrangement of solar cells to form solar? panel and solar array.
a. Solar cells in series.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 23

(b) Solar cells in parallel.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 24

(c) Solar panel.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 25

(d) Solar array.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 26

Question 12.
Observe the figure given below and answer the given questions:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 27
(a) Identify the type of energy generation process shown in this picture.
(b) Name any four equipments which use this type of energy. (Board’s Model Activity Sheet)
Answer:
(a) In this figure solar energy is converted into electrical energy. Solar energy is also called clean energy.
(b) Solar energy is used in following equipment:

  • Solar cooker
  • Solar heater
  • Calculator
  • Solar Photovoltaic cell.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Activity based questions.

Question 1.
Make a table: (Text Book Page No. 47)
Make a table based on forms of energy and corresponding devices.
Answer:

Forms of energyDevices based on this type of energy
(1) ElectricElectric iron, Geyser, Heater, Oven, Refrigerator, Fans, Lights, Elevator.
(2) MechanicalSewing machine, Car, Bicycle, Different machines.
(3) ThermalChulha, Furnace, Steam engine
(4) SolarSolar cooker, Solar heater.

Question 2.
Let’s Think: (Text Book Page No. 52)
Which electricity generation process is eco-friendly and which not?
Answer:
Electricity generated through solar energy and wind energy are truly eco-friendly. Though it is said that hydroelectricity is non-polluting and eco-friendly, it is not true. Hydroelectric project cause destruction of biodiversity and displacement of the local people. Thermal energy, nuclear energy and energy obtained through natural gas are not at all eco-friendly.

Question 3.
Find out: (Text Book Page No. 55)
What is lake tapping? Why it takes place?
Answer:
A lake tap involves excavating a tunnel at the bottom of the lake. Dynamites are planted therein and blasted carefully. The water flows with greater force through the tunnel after such blasting is done. This increased flow of water is then driven to the hydroelectric power generation plant for increased electricity production. This technique is done to establish waterways for hydropower, for making drinking water available, for irrigation water purposes and also for the landing of oil and gas pipes from offshore fields.

Question 4.
Get information: (Text Book Page No. 56)
Get information about major wind-power stations in India and their capacity. Make a table of their location, state and their power generation capacity in MW.
Answer:

LocationStatePower generation capacity in MW
Muppandal, KanyakumariTamil Nadu7,684.31
Dhule, Satara, Sangli, DhalgaonMaharashtra4,664.08
BhujGujarat4,227.31
Dangiri Wind Farm Jaisalmer Wind ParkRajasthan4,123.35
Jogmatti BSESKarnataka3,082.45
Bhopal at Nagda Hills near DewasMadhya Pradesh2,288.60
Tirumala hillsAndhra Pradesh1,866.35
Telangana98.70
Kanjikode in PalakkadKerala43.50
Others4.30
Total28, 082.95

Question 5.
Find out: (Text Book Page No. 58)
Gather information about major solar photovoltaic power generating plants and their capacity in India.
Answer:
List of solar power stations:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 28a
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 29

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Project:

Project 1.
Let’s Discuss: (Text Book Page No. 47)
Make a list of the work that we do in our day-to-day life using energy. Which forms of energy do we use to do this work? Discuss with your friends.

Project 2.
Compare: (Text Book Page No. 51)
Observe the schematic of thermal power plant and the nuclear power plant. Discuss what are the similarities and differences between the two.

Project 3.
Use of ICT: (Text Book Page No. 49)
Prepare a presentation about thermal power plant using computerized presentation, animation, video, pictures, etc. Send it to others and upload on YouTube.

Project 4.
Internet is my friend: (Text Book Page No. 51)
Complete the following table for some important nuclear power plants in India.

Maharashtra State Board Class 10 Science Solutions 

Introduction to Microbiology Class 10 Questions And Answers Maharashtra Board

Std 10 Science Part 2 Chapter 7 Introduction to Microbiology Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 2 Chapter 7 Introduction to Microbiology Question Answer Maharashtra Board

Question 1.
Rewrite the following statements using correct of the options and explain the completed statements.
(gluconic acid, coagulation, amino acid, 4% acetic acid, clostridium, lactobacilli)

a. Process of ……. of milk proteins occurs due to lactic acid.
Answer:
Process of Coagulation of milk proteins occurs due to lactic acid.
Explanation: The lactobacilli are the bacteria carrying out fermentation of the milk. In this process, the lactose sugar in the milk is converted into lactic acid. This lactic acid causes coagulation of the proteins present in the milk.

b. Harmful bacteria like ………. in the intestine are destroyed due to probiotics.
Answer:
Harmful bacteria like Clostridium in the intestine are destroyed due to probiotics.
Explanation: In probiotics, there are lactobacilli which are useful. They control other bacteria present in the alimentary canal and also their metabolism. These bacteria thus stop the action of Clostridium which is a harmful bacteria.

c. Chemically, vinegar is …………
Answer:
Chemically, vinegar is 4% Acetic acid.
Explanation: Chemically vinegar is 4% acetic acid. It is a good preservative of the food and thus while using it as additive to the food, it is called vinegar.

d. Salts which can be used as supplement of calcium and iron are obtained from ……………. acid.
Answer:
Salts which can be used as supplement of calcium and iron are obtained from Gluconic acid.
Explanation: The microbe Aspergillus niger is used on the source material of glucose and corn steep liquor to produce amino acid called Gluconic acid. Gluconic acid is used for the production of minerals used as supplement for calcium and iron.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 2.
Match the pairs.

‘A’ group‘B’ group
(1) Xylitol(a) Pigment
(2) Citric acid(b) To impart sweetness
(3) Lycopene(c) Microbial restrictor
(4) Nycin(d) Protein binding emulsifier
(e) To impart acidity

[Note: In examination match the column question will ham 2 components in Column ‘A’ with 4 alternatives in Column ‘B’.]
Answer:
(1) Xylitol – To impart sweetness
(2) Citric acid – To impart acidity
(3) Lycopene – Pigment
(4) Nycin – Microbial restrictor.

Question 3.
Answer the following:
a. Which fuels can be obtained by microbial processes? Why is it necessary to increase the use of such fuels?
Answer:

  • Microbial anaerobic decomposition of urban agricultural and industrial waste forms the gaseous fuel in the form of methane gas.
  • Alcohol is another clean form of energy which is used in the form of ethanol. It is obtained by the fermentation of molasses by treating it with Saccharomyces-yeast.
  • By photoreduction of water with the help of bacteria, hydrogen gas is released in the process of bio-photolysis of water. This hydrogen gas is said to be the fuel of the future.
  • The conventional fuels are exhaustible. After few hundred years, they will be over completely. Moreover, these fossil fuels cause lot of air pollution due to emission of carbon dioxide. The fuels obtained by the microbial processes are not polluting. Therefore, it is necessary to increase the use of eco-friendly fuels.

b. How can the oil spills of rivers and oceans be cleaned?
Answer:

  • The oil spills in rivers or oceans are caused by crude oil or petroleum hydrocarbons.
  • This crude oil is highly toxic to the flora and fauna of the aquatic environment.
  • By using mechanical means the oil spill can be removed, but this is very difficult.
  • The biological way to remove this pollution is done by using culture of microbes like Pseudomonas spp. and Alcanovorax borkumensis.
  • They have the ability to destroy the pyridines and other chemicals present in the hydrocarbons.
  • These bacteria are called as hydrocarbono-clastic bacteria (HCB) which decompose the hydrocarbons and bring about the reaction of carbon with oxygen.
  • In the process CO2 and water are formed. In this way the oil spills are cleaned, by releasing HCB at the place of oil spills.

c. How can the soil polluted by acid rain be made fertile again?
Answer:

  • The soil polluted by the acid rain is made fertile again by using bacteria.
  • Acidophillium spp. and Acidobacillus ferroxidens are the bacteria which have the capacity to use sulphuric acid as their energy source.
  • Since this sulphuric acid present in the acid rain, can be controlled by these bacteria.
  • In this way, bacteria can control the soil pollution occurring due to acid rain, making the soil fertile again.

d. Explain the importance of bio pesticides in organic farming.
Answer:

  • By using bio pesticides, soil pollution is minimized. Otherwise by using chemical pesticides and fertilizers there is large scale soil pollution.
  • When chemical pesticides are used in agriculture, there is contamination of soil by fluoroacetamide – like chemicals.
  • These are harmful to other plants, animals as well as for-human beings. They may cause skin diseases in humans.
  • By using bacterial and fungal toxins the pests and pathogens can be destroyed. Such toxins are directly incorporated in the plant materials.
    E.g. Spinosad is a biopesticide produced as a by-product of fermentation.

e. What are the reasons for increasing the popularity of probiotic products?
Answer:

  • Probiotic substances are mostly milk products containing live bacteria. Such probiotics are very good for health.
  • The useful colonies of bacteria are produced in the alimentary canal of human beings due to the probiotics.
  • Probiotics decrease the population of harmful microbes like. Clostridium from our digestive tract.
  • The immunity is enhanced due to regular intake of probiotics in the diet.
  • The ill-effects of harmful substances formed during metabolic activities are reduced by the probiotics.
  • If someone takes the antibiotic treatment, then his or her useful intestinal bacterial flora becomes inactive or is eradicated. In such cases, probiotics restore the bacterial flora and make the person well again.

All these facts have made probiotics a popular choice for people.

f. How the bread and other products produced using baker’s yeast are nutritious?
Answer:

  • In order to make the bread the baker’s yeast – Saccharomyces cerevisiae is added to the flour for the fermentation process.
  • In commercial bakery, compressed yeast is used while in domestic settings dry, granular form of yeast is used.
  • The flour prepared by using commercial yeast contains various useful contents like carbohydrates, fats, proteins, various vitamins, and minerals.
  • The anaerobic fermentation also increases the nutritive content of the flour.
  • Due to this, bread and other products produced with the help of yeast become nutritive.

g. Which precautions are necessary for proper decomposition of domestic waste?
Answer:
The domestic waste should be properly segregated into biodegradable (wet waste) and non-biodegradable (dry waste). After segregation, these wastes should be stored separately into two different containers. The non-biodegradable substances should he either reused or sent for recycling. The biodegradable substances are decomposed naturally.

The decomposition process can be done at house-hold level too in a pot or a tank. This decomposition will yield a rich manure. The pot should be covered by a thin layer of soil and it should be kept in a dark but airy place.

The non-biodegradable things such as plastic articles, glass pieces, metal objects, unused 5 medicines, e-waste should never be thrown in wet wastes. The toxic substances and the insecticides if added to wet waste, will never allow the natural decomposition process. Therefore, only after taking proper precautions we can aim at proper decomposition of domestic wastes.

h. Why is it necessary to ban the use of plastic bags?
Answer:
Plastic is a non-biodegradable substance. It cannot be degraded back into its original constituents. It remains just like that for many hundreds of years. It causes solid waste pollution in any environment wherever it is thrown indiscriminately. If burnt, it releases very toxic gases. If dumped in landfills it obstructs the other decomposition processes.

If thrown in water bodies, it causes harm to aquatic life. Cattle graze on plastic unknowingly and are killed by it as it clogs inside their alimentary canal. The gutters and rain water drains get clogged due to plastic bags and this causes cities to submerge in water during heavy rains. Nowadays, the fishermen get more than half of plastic if they cast their net in the sea.

People use the plastic bags indiscriminately without any thought towards their environmental impact. There are better alternatives for plastic bags such as cloth bags which can be reused again and again. Therefore, it is absolutely necessary to ban the use of plastic bag.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 4.
Complete the following conceptual picture.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 1
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 2

Question 5.
Give scientific reasons.
a. Use of mutant strains has been increased in industrial microbiology.
Answer:

  • By using industrial microbiology, the commercial use of microbes is done.
  • In such experiments, various economic, social and environment related processes and products are included.
  • In this, fermentation processes are used to make bread, cheese, wines, enzymes, nutrients, etc.
  • Different types of antibiotics are also made by using processes of industrial microbiology.
  • In pollution control and solid waste management, the industrial microbiology becomes helpful.
  • In farming too biotechnology is used to produce BT crops.

b. Enzymes obtained by microbial process are mixed with detergents.
Answer:

  • When detergents are mixed with microbial enzymes, they start working more efficiently.
  • The cleaning process takes place at lesser temperatures.
  • Therefore, for better results, enzymes obtained by microbial process are mixed with detergents.

c. Microbial enzymes are used instead of chemical catalysts in chemical industry. (March 2019)
(OR)
Microbial enzymes are said to be eco-friendly.
Answer:

  • Microbial enzymes are active at low temperature, pH and pressure.
  • Due to this property, the energy is saved. The costlier erosion-proof instruments need not be used.
  • In enzymatic reactions, the unnecessary byproducts are not formed as the reactions are highly specific.
  • The expenses on purification of the product are minimized as no unnecessary products are formed.
  • The elimination and decomposition of waste material is avoided and enzymes can be reused again. Hence, microbial enzymes which are eco¬friendly are used in chemical industry.

Question 6.
Complete the following conceptual picture with respect to its uses. (Board’s Model Activity Sheet)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 3
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 4

Question 7.
Complete the following conceptual picture related to environmental management.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 5
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 6

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 8.
Answer the following questions.
a. What is the role of microbes in compost production?
Answer:

  • Microbes can bring about natural decomposition of the organic compounds.
  • During the biodegradation, some bacteria andmfungi bring about such decomposition and release the inorganic constituents back into the nature.
  • Compost is formed in such a way by recycling process.

b. What are the benefits of mixing ethanol with petrol and diesel?
Answer:
When only diesel or petrol is used as fuel, there is increased air pollution. Morevoer, since these are non-renewable and exhaustible fuels, they will be finished in next some years. When petrol and diesel is mixed with ethanol, the proportion of CO2, CO, and hydrocarbons which are emitted in the atmosphere becomes lesser.

The particulate pollutants which otherwise are emitted through combustion of petrol and diesel are not formed when fuels are mixed with ethanol. By adding ethanol to the fuels, the cost of expensive petrol or diesel also becomes less. The ethanol burns more efficiently hence ethanol is mixed with petrol and diesel.

c. Which plants are cultivated to obtain the fuel?
Answer:

  • The ethanol is obtained from wheat, maize, beet, sugarcane and molasses of sugarcane.
  • For biodiesel, the soybean, rapeseed, jatropa, mahua, flaxseed, mustard, sunflower, palm, jute and some types of algae are cultivated.

d. Which fuels are obtained from biomass?
Answer:
From biomass, the biogas and biodiesel are mainly obtained. The biogas is obtained from dung of cattle. The fermentation of cattle dung gives rise to methane. From methane, methanol is obtained. Ethanol is obtained from molasses of sugarcane and some other crops. In some countries, special crops are cultivated for the biodiesel.

e. How does the bread become spongy?
Answer:

  • When the dough for bread is prepared, the baker’s yeast – Saccharomyces cerevisiae is added to it.
  • This yeast carries out anaerobic fermentation.
  • This results in formation of CO2 and ethanol.
  • The CO2 formed tries to escape out of the flour and thus the dough rise. When such dough is baked, it produces spongy bread.

Project: (Do it your self)

Project 1.
Find the ways to implement the zero garbage system at domestic level.

Project 2.
Which are the microbes that destroy the chemical pesticides in soil?

Project 3.
Collect more information about reasons for avoiding the use of chemical pesticides.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Can you recall? (Text Book Page No. 77)

Question 1.
Which different microbes are useful to us?
Answer:
Many microbes are useful to us, such as bacteria which are used for making curds from milk, yeast used to ferment the batter of bread, bacteria used for making other milk products, bacteria and fungi used for making antibiotics. The bacteria are even used for pollution control.

Question 2.
Which different products can he produced with the help of Microbes?
Answer:
Milk products, cheese, cocoa, pickles made from vegetables, wine and other beverages, bread, probiotic substances and cattle feed are produced with the help of microbes.

Use your brain power. (Text Book Page No. 79)

Question 1.
In the earlier class, you had prepared the solution of dry yeast for observation of yeast. Which substance is prepared by its use on commercial basis?
Answer:
The commercial production of bread and other bakery products need yeast. In wine and beer making also solution of yeast is required.

(Use your brain power. (Text Book Page No. 81)

Question 1.
Food materials like cold drinks, ice creams, cakes, juices are available in various colours and flavours. Whether these colours and flavours are really derived from fruits?
Answer:
The eatables can be made directly from fruits or essence of fruits. But most of the food products purchased from markets use these colours and flavours which are derived from synthetic chemicals.

Let’s Think: (Text Book Page No. 83)

Question .1
Why is it asked to segregate wet and dry waste in each home?
Answer:
The wet waste decomposes on its own as most of the matter therein is biodegradable. This waste can be converted into manure by composting. The dry waste can be picked up by the bhangarwala or kabadiwala. This waste can be reused or recycled. Therefore, if dry and wet wastes are kept separately, the solid waste management becomes much easier.

On the contrary if everything is dumped indiscriminately, it adds to the total volume of the solid wastes. This becomes unmanageable. Therefore, to reduce the problems of solid waste management, the dry and wet waste segregation must be done at every point source. This also could fetch wealth from waste.

Question 2.
What is done with the segregated waste?
Answer:
In big cities, there is a mechanism to pick up the solid waste every day or even twice a day at some places. The segregated garbage is taken by the municipal garbage trucks at the land filling sites. Here it is buried deep in the ground. The dry waste that can be reused or recycled, is sold to the recycling units.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 3.
Which is most appropriate method of disposal of dry waste?
Answer:
Reuse and recycle is the most appropriate method of disposal of dry waste.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Enzyme ……….. obtained from fungi is used to produce vegetarian cheese.
(a) lipase
(b) protease
(c) amylase
(d) trypsin
Answer:
(b) protease

Question 2.
Milk is subjected to ………… at the beginning to destroy unwanted microbes.
(a) pasteurization
(b) fermentation
(c) coagulation
(d) decomposition
Answer:
(a) pasteurization

Question 3.
………….. like compounds are formed due to lactobacilli that gives characteristic taste to the yoghurt.
(a) Lactose
(b) Caesin
(c) Acetyldehyde
(d) All the above
Answer:
(c) Acetyldehyde

Question 4.
Methane can be obtained by …………. decomposition of urban agricultural and industrial waste.
(a) aerobic
(b) anaerobic
(c) microbial anaerobic
(d) chemical
Answer:
(c) microbial anaerobic

Question 5.
……….. gas is considered to be the fuel of future.
(a) Hydrogen
(b) Nitrogen
(c) Methane
(d) Butane
Answer:
(a) Hydrogen

Question 6.
………. are mixed with waste materials at land-filling sites for quicker decomposition.
(a) Microbes
(b) Bioreactors
(c) Fungi
(d) Worms
Answer:
(b) Bioreactors

Question 7.
…………. bacteria decompose the xenobiotic chemicals present in sewage.
(a) Hydrocarbonoclastic
(b) Decomposing
(c) E.coli
(d) Phenol oxidizing
Answer:
(d) Phenol oxidizing

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 8.
Microbes are used for ………… of environment polluted due to sewage.
(a) protection
(b) conservation.
(c) bioremediaiion
(d) decomposition
Answer:
(c) bioremediaiion

Question 9.
……….. is a powerful antibiotic against tuberculosis.
(a) Streptomycin
(b) Tetracycline
(c) Rifamycin
(d) Bacitracin
Answer:
(c) Rifamycin

Question 10.
Bacteria are used to clear the oil spills are called ………….. bacteria.
(a) phenol oxidizing
(b) electrolytic
(c) hydrocarbonoclastic
(d) decomposing
Answer:
(c) hydrocarbonoclastic

Question 11.
………… convert these salts of uranium into insoluble salts.
(a) Saccharomyces
(b) Thiobacillus
(c) Acidobacillus
(d) Geobacter
Answer:
(d) Geobacter

Question 12.
………….., a byproduct of fermentation is a biopesticide.
(a) Fluoroacetamide
(b) Vanillin
(c) Aspertame
(d) Spinosad
Answer:
(d) Spinosad

Question 13.
…………. beverage is obtained by fermentation of apple juice. (July ’19)
(a) Cider
(b) Wine
(c) Coffee
(d) Cocoa
Answer:
(a) Cider

Question 14.
Vinegar is the chemically ………… acid. (Board’s Model Activity Sheet)
(a) Citric
(b) Gluconic
(c) Glutamic
(d) Acetic
Answer:
(d) Acetic

Question 15.
In which of the following industries microbial enzymes are not used?
(a) Glass industry
(b) Cheese industry
(c) Tanning industry
(d) Paper industry
Answer:
(a) Glass industry

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 16.
Citric acid used in production of beverages, toffees, chocolates is obtained by fermentation of …….. by Aspergillus niger.
(a) grapes
(b) sugar molasses
(c) apple
(d) coffee nuts
Answer:
(b) sugar molasses

Match the pairs:

Question 1.

Column ‘A’Column ‘B’
(1) Vinegar(a) Polylactic acid
(2) Xanthan gum(b) Molasses
(c) Icecreams and puddings
(d) Acetic acid

Answer:
(1) Vinegar – Acetic acid
(2) Xanthan gum – Icecreams and puddings

Find the odd one out:

Question 1.
Lactobacillus acidophilus, Lactobacillus casei, Bifidobacterium bifidum, Streptococcus thermophilus
Answer:
Streptococcus thermophilus. (All others are bacteria producing probiotics.)

Question 2.
Lactobacillus lactis, Bifidobacterium bifidtim, Lactobacillus cremoris, Streptococcus thermophilus
Answer:
Bifidobacterium bifidum. (All others are bacteria used in cheese production.)

Question 3.
Dark chocolate, Miso soup, Wafers, Corn syrup
Answer:
Wafers. (All others are probiotic products.)

Question 4.
Vinegar, Soya sauce, Ketchup, Monosodium glutamate
Answer:
Ketchup. (All others are products prepared by microbial fermentation.)

Question 5.
Actinomycetes, Streptomyces, Nocardia, yeast
Answer:
Yeast. (All others have ability of decomposing rubber from garbage.)

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Find the correlation:

Question 1.
Bread Baker’s yeast : : Soya sauce : ……….
Answer:
Bread Baker’s yeast : : Soya sauce : Aspergillus oryzae

Question 2.
Coffee : Caffea arabica : : Cocoa : …………
Answer:
Coffee : Coffea arabica : : Cocoa : Theobroma cacao

Question 3.
Oil slick : Alcanovorax : Rubber from garbage : …………
Answer:
Oil slick : Alcanovorax : Rubber from garbage : Actinomycetes

Question 4.
Conversion of metals into comounds : Thiobacilli : : Conversion of uranium salts …………
Answer:
Conversion of metals into comounds : Thiobacilli : : Conversion of uranium salts Geobacter.

Name the following:

Question 1.
Microbial enzymes.
Answer:
Oxidoreductases, transferases, hydrolases, lyases, isomerases, ligases.

Question 2.
Emulsifiers.
Answer:
Polysaccharides and glycolipids.

Question 3.
Microbe used in preparation of wine and cider.
Answer:
Saccharomyces cerevisiae.

Question 4.
Effective antibiotic against tuberculosis.
Answer:
Rifamycin.

Question 5.
Antibiotics.
Answer:
Penicillin, cephalosporins, monobactam, erythromycin, gentamycin, neomycin, streptomycin, tetracyclins, vancomycin.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 6.
Bacteria that use sulphuric acid as source of energy.
Answer:
Acidobacillus ferroxidens, Acidophillium spp.

Question 7.
Substance that makes biodegradable plastic.
Answer:
Polylactic acid.

Question 8.
Curd like food product made from sheep milk.
Answer:
Kefir.

Question 9.
Enzyme used to make vegetarian cheese.
Answer:
Protease.

Question 10.
Fungus used for making soya sauce.
Answer:
Aspergillus oryzae.

Complete the charts:

Question 1.

FruitMicrobe usedName of beverage
______________________________________________________Coffee
Theobroma cacaoCandida, Hansenula, Pichia, Saccharomyces___________________________
Grapes______________________________________________________
AppleSaccharomyces cerevisiae___________________________

Answer:

FruitMicrobe usedName of beverage
Caffea arabicaLactobacillus  brevisCoffee
Theobroma cacaoCandida, Hansenula, Pichia, SaccharomycesCocoa
GrapesSaccharomyces  cerevisiaeWine
AppleSaccharomyces cerevisiaeCider

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 2.

SourceMicrobeAmino acidUse
Sugar and beet molasses, ammonia salt______________________________________Production of monosodium glutamate (Ajinomoto).
___________________Aspergillus niger___________________Drinks, toffees, chocolate production.
Glucose, corn steep liquor___________________Gluconic acid___________________
Molasses, corn steep liquorLactobacillus delbrueckii______________________________________
___________________Aspergillus itaconiusItaconic acid___________________

Answer:

SourceMicrobeAmino acidUse
Sugar and beet molasses, ammonia saltBrevibacterium, CorynobacteriumL-glutamic acidProduction of monosodium glutamate (Ajinomoto).
Sugar molasses, saltAspergillus nigerCitric acidDrinks, toffees, chocolate production.
Glucose, corn steep liquorAspergillus nigerGluconic acidProduction of minerals used as  supplement for calcium and iron. 
Molasses, corn steep liquorLactobacillus delbrueckiiLactic acidSource of nitrogen, production of vitamins.
Molasses, corn steep liquorAspergillus itaconiusItaconic acidPaper, textile, plastic industry, gum production

Question 3.

SourceMicrobeAmino acid
(1) Sugar molasses and salt___________________Citric acid
(2) ___________________Lactobacillus delbrueckii___________________
(3) Corn steep liquorAspergillus itaconius___________________

Answer:

SourceMicrobeAmino acid
(1) Sugar molasses and saltAspergillus nigerCitric acid
(2) Molasses, corn steep liquorLactobacillus delbrueckiiLactic acid
(3) Corn steep liquorAspergillus itaconiusItaconic acid

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Answer the following questions:

Question 1.
Which microbes are used in the baking industries? (Board’s Model Activity Sheet)
Answer:
Yeast i.e. Saccharomyces cerevisiae is used in the baking industries.

Question 2.
There is an oil layer on the water surface of river in your area. What will you do? (March 2019)
Answer:
If there is an oil layer on the water surface, we shall use hydrocarbonoclastic bacteria like Pseudomonas to clean up the oil spill.

Question 3.
(a) How are microbes used in sewage management?
(b) How is the sludge produced in this process utilized? (Board’s Model Activity Sheet)
Answer:
(a)

  • In cities, the sewage is sent to processing plant and is treated with microbes.
  • Microbes that carry out decomposition, are mixed with sewage. Such microbes are able to destroy, pathogens as well as decompose any compounds.
  • Some microbes bring about bioremediation of environment, that are used for treating sewage pollution.
  • Upon decomposition of the carbon compounds present in sewage, microbes release methane and CO2.

(b) The sludge formed in this process, is used as fertilizer.

Question 4.
Answer the following questions:
(a) What is clean technology?
Answer:
Clean technology is the method to use microbes for controlling air, soil and water pollution. These microbes can degrade the manmade chemicals.

(b) Why is it essential to ban plastic bags?
Answer:
Plastic is a non-biodegradable substance. It cannot be degraded back into its original constituents. It remains just like that for many hundreds of years. It causes solid waste pollution in any environment wherever it is thrown indiscriminately. If burnt, it releases very toxic gases. If dumped in landfills it obstructs the other decomposition processes.

If thrown in water bodies, it causes harm to aquatic life. Cattle graze on plastic unknowingly and are killed by it as it clogs inside their alimentary canal. The gutters and rain water drains get clogged due to plastic bags and this causes cities to submerge in water during heavy rains. Nowadays, the fishermen get more than half of plastic if they cast their net in the sea.

People use the plastic bags indiscriminately without any thought towards their environmental impact. There are better alternatives for plastic bags such as cloth bags which can be reused again and again. Therefore, it is absolutely necessary to ban the use of plastic bag.

Write short notes on the following:

Question 1.
Production of Yoghurt.
Answer:

  • Yoghurt is one of the milk product produced from milk with the help of lactobacilli (inoculant).
  • In the industrial production of yoghurt, the milk is added with condensed milk powder. This increases the protein content of the milk. Then this milk is subjected to fermentation.
  • Milk is boiled and then it is cooled till it becomes lukewarm.
  • Then the bacterial strains of Streptococcus thermophiles and Lactobacillus delbrueckii are added to this lukewarm milk in 1:1 proportion.
  • The Streptococcus bacteria convert the milk into solution containing lactic acid. This makes the proteins to gel out. It makes the yoghurt dense.
  • The lactobacilli help in the formation of acetaldehyde like compounds giving a characteristic taste to the yoghurt.
  • For commercial reasons, various fruit juices are mixed with yoghurt to impart different flavours forming strawberry yoghurt, banana yoghurt, etc.
  • The pasteurization is carried out to increase the shelf life of yoghurt and improve its probiotic properties.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Question 2.
Production of cheese.
Answer:
Cheese is made from cow’s milk throughout the world. The steps in the process of cheese manufacture are as follows:

  • Chemical and microbiological testing of milk is done.
  • Three types of bacteria, viz. Lactobacillus lactis, Lactobacillus cremoris and Streptococcus thermophilus along with some colour is added to the milk.
  • It imparts sourness to the milk and it is converted into yoghurt like substance.
  • The water from this yoghurt, i.e. whey is not removed to make the yoghurt denser.
  • Enzyme, rennet or protease is added to the mixture to make it more denser.
  • Later cutting the solid yoghurt into pieces, washing, rubbing, salting, land mixing of essential microbes, pigments and flavours is done in suitable steps.
  • The pressed cheese is then cut in to pieces and stored for ripening.

Question 3.
Land-filling sites.
Answer:

  • In the land-filling sites the degradable wastes are transferred. Usually such sites are in urban areas.
  • The land-filling sites are away from the residential areas for the hygienic reasons. Here large pits arb dug in open spaces.
  • These pits are lined with plastic sheets. Therefore, the leaching of toxic and harmful materials is avoided to reduce the chance of soil pollution due to leachates.
  • Compressed waste is put in the pit and is covered with layers of soil, saw dust, leafy waste.
  • Specific biochemical substances are added for speedy decomposition.
  • Bioreactors which are mixtures of bacteria are mixed at some places.
  • Soil microbes and other top layers decompose the waste.
  • Soil slurry is used to seal the pits completely.
  • After a certain period, best quality compost is formed. Such land filling sites can be reused after removal of compost.

Complete the paragraph by choosing the appropriate words given in the brackets:

Question 1.
(Nocardia, Geobacter, Ideonella sakaiensis, Pseudomonas, Alcanovorax borkumensis, hydrocarbonoclastic, Acidophillium, streptomyces)
Bacteria like ………… spp. and ………. have the ability to destroy the pyridines and other chemicals. Hence, these bacteria are used to clear the oil spills. These are called ………. bacteria. It has been observed that species like Vibrio, …………… can decompose the PET. Similarly, species of fungi like ………… have ability of decomposing rubber from garbage. Sulphuric acid is source of energy for some species of bacteria like ………… Hence, these bacteria can control the soil pollution occurring due to acid rain. …………..convert the salts of uranium into insoluble salts.
Answer:
Bacteria like Pseudomonas spp. and Alcanovorax borkumensis have the ability to destroy the pyridines and other chemicals. Hence, these bacteria are used to clear the oil spills. These are called hydrocarbonoclastic bacteria. It has been observed that species like Vibrio, Ideonella sakaiensis can decompose the PET. Similarly, species of fungi like Nocardia have ability of decomposing rubber from garbage. Sulphuric acid is source of energy for some species of bacteria like Acidophillium. Hence, these bacteria can control the soil pollution occurring due to acid rain. Geobacter convert the salts of uranium into insoluble salts.

Read the paragraph and answer the questions given below:

Remediation is the process of removing dangerous or poisonous substances from the environment, or limiting the effect that they have on it. When any biological organism is used for remediation, it is called bioremediation. When plant species are used for the purpose of remediation, it is called phytoremediation. When any microbes are used then it is named as microbial remediation. The methods of such remediation have helped to clean the environment from toxic effluents, especially sewage and crude oil. Dr. Anand Chakraborty, a scientist of Indian origin, has worked on Pseudomonas aeruginosa which have reduced the crude oil films into carbon dioxide and water.

Questions and Answers:

Question 1.
What is the meaning of remediation?
Answer:
Remediation is the process by which dangerous or toxic substances are removed from the environment.

Question 2.
What is the difference between phytoremediation and microbial remediation?
Answer:
When any plant species are used for remediation process, then it is called phytoremediation, whereas when any microbe species used for remediation then it is called microbial remediation.

Question 3.
Which environmental pollutant is mainly removed through bioremediation processes?
Answer:
Toxicants released through sewage and crude oil are removed by bioremediation processes.

Question 4.
What is the role of Pseudomonas aeruginosa?
Answer:
Pseudomonas aeruginosa helps in bioremediation by acting on film of crude oil and reduces it to carbon dioxide and water.

Question 5.
Why Dr. Anand Chakraborty’s work phenomenal?
Answer:
Dr. Anand Chakraborty discovered that Pseudomonas aeruginosa bacteria can act on oil film which is toxic and reduce it to nontoxic products. This helps in controlling the oil pollution of marine waters which otherwise is very difficult to control.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Diagram based questions:

Question 1.
Observe the diagram and answer the following questions:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 7
(a) Name the following method of solid waste management.
Answer:
The above diagram shows modern landfill site. This method is used for solid waste management.

(b) What type of waste is used in this method?
Answer:
In this method only degradable waste matter collected in cities can be used. Such solid waste can undergo biodegradation and hence can be managed in an eco-friendly way.

(c) What kind of useful substances can be obtained from such methods?
Answer:
From such decomposition, organic fertilizers and manure formed through composting are obtained. Methane gas is also obtained which is used as fuel.

Question 2.
Observe the Figure 7.1 and answer the following questions: (March 2019)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 8
(a) Identify the process shown in the figure.
Answer:
The figure shows modern land fill site where microbial biodegradation process is carried out.

(b) Explain the process in short.
Answer:
Land-filling sites:

  • In the land-filling sites the degradable wastes are transferred. Usually such sites are in urban areas.
  • The land-filling sites are away from the residential areas for the hygienic reasons. Here large pits arb dug in open spaces.
  • These pits are lined with plastic sheets. Therefore, the leaching of toxic and harmful materials is avoided to reduce the chance of soil pollution due to leachates.
  • Compressed waste is put in the pit and is covered with layers of soil, saw dust, leafy waste.
  • Specific biochemical substances are added for speedy decomposition.
  • Bioreactors which are mixtures of bacteria are mixed at some places.
  • Soil microbes and other top layers decompose the waste.
  • Soil slurry is used to seal the pits completely.
  • After a certain period, best quality compost is formed. Such land filling sites can be reused after removal of compost.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Activity based questions:

Question 1.
Collect Information Search : (Textbook page no. 84)
(i) Which materials should not be present in garbage for its proper microbial decomposition?
Answer:
If there are non-biodegradable materials in the garbage, they will not decompose. The plastic, glass, metals etc. will not undergo microbial decomposition, therefore, such items should not be- there in the garbage. The toxic matter, hazardous chemicals and e-waste should also be removed. If such materials are present in the garbage, the microbes will be killed and the entire process of decomposition will be suffered.

(ii) How the sewage generated in your house or apartment is disposed off ?
Answer:
The sewage generated in our house is carried by the drainage pipes to municipal sewage treatment plants. Here, primary, secondary and tertiary treatment is done on the sewage. The safe water is then released into the ocean.

Question 2.
Observe: (Textbook page no. 83)
Observe the garbage vans of gram panchayat and municipality. Nowadays, there is facility of decreasing the volume of garbage by compaction in those vans. Explain the advantages of this activity.
Answer:
When the garbage is compressed, its volume is reduced. The trips of the vans that pick up the garbage can be reduced due to such measures. The land filling sites can also accommodate more garbage if it is compacted.

Question 3.
Observe the figure and answer the following:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology 9
(i) Lack of management of which factor is shown in the picture?
Answer:
The above picture shows the lack of management of sewage resulting in waste water being dumped carelessly.

(ii) How can that factor be managed with the help of microbes?
Answer:
Microbes which can destroy the pathogens of cholera, typhoid, etc. are mixed with sewage. They release methane and CO2 by decomposition of the carbon compounds present in sewage. Other microbes that decompose chemical compounds are also released. Phenol oxidizing bacteria decompose the xenobiotic chemicals present in sewage.

(iii) How are the oil spills in oceans cleared?
Answer:
Hydrocarbonoclastic bacteria like
Alcanivorax borkumensis and Pseudomonas are used to clear the oil spillage from ocean water. These bacteria decompose the hydrocarbons. They bring about the reaction of released carbon with oxygen to produced CO2 and water.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 7 Introduction to Microbiology

Projects: (Do it your self)

Project 1.
Search: (Textbook page no. 81)
Read the ingredients and their proportion printed on bottles of cold drinks and juices and wrappers of ice creams. Find out the natural and artificial ingredients.

Project 2.
Internet is My Friend: (Textbook page no. 85)
Collect pictures of various useful microbes. Display chart of their information in the classroom.

Project 3.
Observe the figure given on Textbook page no. 82. Discuss about bio-fuel?

Maharashtra State Board Class 10 Science Solutions 

Social Health Class 10 Questions And Answers Maharashtra Board

Std 10 Science Part 2 Chapter 9 Social Health Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 9 Social Health Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 2 Chapter 9 Social Health Question Answer Maharashtra Board

Question 1.
Fill in the blanks with appropriate word.
a. Laughter club is a remedy to drive away …………..
(a) stress
(b) addictions
(c) lethargy
(d) epidemics
Answer:
(a) stress

b. Alcohol consumption mainly affects …………. system.
(a) digestive
(b) respiratory
(c) nervous
(d) excretory
Answer:
(c) nervous

c. IT Act 2000 is to control the ……….
(a) housebreaking
(b) cybercrimes
(c) cheating
(d) pickpocketing
Answer:
(b) cybercrimes

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 2.
Answer the following.
a. Which factors affect the social health?
Answer:
(1) In order to maintain the social health of any community there should be good amenities for the people. E.g. food, water, shelter, clothing, medicines and medical help, equal opportunities for education, cleanliness of the surroundings, transport facilities etc. should be properly provided.

(2) The social and political conditions of the surrounding should be such that there should not be any connections with world of criminals. The presence of such criminal ties can affect the social health to a great extent.

(3) The gardens, playgrounds, the empty plots for outdoor games, sports clubs, etc. are important criteria for overall development of the society. This results into personality development and make people happy and strong.

(4) Addictions, criminal tendencies, pervert behaviour and perverse thinking affects other people in the society and this reflects negatively on the social health.

(5) Having large number of friends and relatives, proper use of time when alone and when along the peer group, trust in others, respect and acceptance for others build stronger social health.

b. Which changes occur in persons continuously using the internet and mobile phones?
Answer:

  • When a person continuously remains in contact with mobile phones, many physical problems can arise.
  • Tiredness, headache, insomnia, forgetfulness, tinnitus, joint pains and problems in vision occur due to radiation emanating from the cell phones. For young children this is more disastrous as these radiations can penetrate through their bones.
  • By logging into the internet for a long time, persons become solitary. Such individuals are unable to establish harmonious relations with relatives and other people around.
  • They tend to become self-centred and selfish, They lose sensitivity towards others.
  • Such people never take any social responsibility and the social health is thus disturbed.

c. Which problems does the common man face due to incidences of cybercrime?
Answer:

  • The numbers of Aadhaar card, PAN card, credit or debit card are obtained by the cheaters. This is a cybercrime. The PIN number can be misused and the money can be withdrawn from the bank accounts. The looters withdraw cash from our accounts in this way.
  • People can be cheated during online shopping.
  • Fake account on Facebook is opened and false information is displayed on it. Through such accounts the girls are emotionally and financially exploited.
  • Electronic media are misused for sending derogatory and vulgar messages, obscene pictxfres and provocative statements.
  • Through the internet, hackers can send virus to crash someone’s computer or even mobile phones.
    In all such different ways, common people can be victimized by cybercrime.

d. Explain the importance of good communication with others.
Answer:

  • Nowadays, there is fierce competition, insecurity and criminal tendencies in the society.
  • This kind of atmosphere is increasing mental and emotional stress.
  • If the stress remains buried in the mind, persons are depressed or frustrated. This causes, mental disorders if not treated in time. Depression can lead to addictions. The suicidal thoughts hover in the mind. If at that phase we can open our heart by good communication, many problems can be solved.
  • Help from counsellors can be taken to relieve the stress.
  • By good communication with parents or family members harmonious relations can be re-established.

Question 3.
Solve the following crossword.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 1a
1. Continuous consumption of alcoholic and tobacco materials.
Answer:
Addiction

2. This app may cause the cybercrimes
Answer:
Facebook

3. A remedy to resolve stress.
Answer:
Singing

4. Requirement for stress free life.
Answer:
Goodfood

5. Various factors affect ……….. health.
Answer:
Social

6. Art of preparing food items.
Answer:
Cooking.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 4.
What are the various ways to minimize mental stress?
Answer:
The ways of stress-bursting are as follows:
(1) Laughter club: People gather together and laugh collectively to reduce stress.

(2) Good communication: One should establish good communication with friends, siblings, cousins, teachers, parents or anybody in whom we can confide and express our feelings.

(3) Writing: By writing and noting the thoughts we feel relieved. We can confess and analyse about our mistakes through writing to reduce our stress.

(4) Hobbies: Collecting curios, photography, reading good literature, music, cooking, gardening, bird watching, keeping a pet, sculpturing, drawing, rangoli, dancing, etc. are such hobbies which are necessary for utilizing our spare time by creativity. Persuading hobby is the best way to be stress-free. Music in particular is said to change the negative thoughts, therefore, listening to music, learning the music and singing helps to fight stress. By admiring nature too, stress is relieved.

(5) Outdoor games and physical exercise: By participating in the sports, there are various benefits such as physical exercise, improving discipline, interaction with others and creating the tendency of unity, becoming more social and reduce stress.

Question 5.
Give three examples of each.
a. Hobbies to reduce stress.
Answer:

  1. To listen to music
  2. Bird watching and nature trails
  3. Reading good books.

b. Diseases endangering the social health.
Answer:

  1. AIDS
  2. Tuberculosis
  3. Leprosy.

c. Physical problems arising due to excessive use of mobile phones.
Answer:

  1. Headache
  2. Vision problems
  3. Joint pains.

d. Activities under the jurisdiction of cybercrime laws.
Answer:

  1. To do bank transactions by procuring PIN number of somebody.
  2. Misuse of written material of someone or illegal sale of the same.
  3. Hacking the information of government institutes and companies.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 6.
What will you do? Why?
a. You are spending more time in internet/mobile games, phone, etc.
Answer:
In life, the time once spent never returns back. We therefore must use our time for studies, exercise or outdoor games and some entertainment. In the free time, we must also help our parents in house hold work. But if we are spending hours together on surfing the net without any perfect aim or playing the computer or cell phone games it is total waste of time.

There are many inappropriate sites on the internet, which should not be watched. This causes stress. Continuous use of mobile phone and being hooked to the social media slowly becomes an addiction. If these bad habits are creeping in us, we must try to leave the habits by conscious change.

b. Child of your neighbour is addicted to tobacco chewing. (July 2019)
Answer:
The hazards of tobacco chewing will be explained to this child. Different photographs and videos showing the conditions of oral cancer will be shown to this child to persuade him, so that he can stay away from tobacco. This addiction has to be removed, so help of his parents will be taken. They will be told about the child’s habit and asked to help ? him free from his addiction.

c. Your sister has become incommunicative. She prefers to remain alone. (July 2019)
Answer:
The individual who prefers to be incommunicative has lots of thoughts in his/her mind. If this is the case with sister, she will be taken into confidence and the reason behind this lack of communication will be found out. Most often such persons have depression. So she will not be left alone. Her friends will be invited at home, so that she can converse with them. She should be motivated to mix with her favourite people. She should be encouraged to pursue her hobbies. She should be helped in selecting such work, If nothing changes her, then the help of counsellor should be taken.

d. You have to use free space around your home for good purpose.
Answer:
The free space around our home can be used to make a small garden. The garden-soil can be bought and spread in this free space. Small saplings can be planted here and nurtured for further growth. Nursery of saplings can also be started in this free space.

The space can also be used for outdoor games. The net for Badminton can be fixed and evening times can be spent in playing the game. Also care will be taken to keep the space clean and without any garbage.

e. Your Mend has developed the hobby of snapping selfies. (July 2019)
Answer:
The habit of continuously taking selfie is bad. It shows that the friend is constantly thinking of himself only. His self-centredness is to be removed by counselling him. The reason behind this behaviour should also be understood. He should be diverted and motivated to take some other tasks so that his habit can be lessened. Taking selfies is not a hobby. It is a bad habit if someone is repeatedly engaged in it.

f. Your brother studying in XII has developed the stress.
Answer:
The syllabus for class XII is vast. If the studies are not taken seriously from the beginning of the academic year, then the stress develops due to the fear of examination and result. Therefore, instead of being stressed, he should practise time management and study schedule. He should think of only one subject at a time. The atmosphere in the house should be maintained happy and tension-free. Everybody in the house should interact with him so that he gets a feeling that he is not alone. He should be convinced, “study is for you and you are not for study”.

Question 7.
What type of changes occur in a home having chronically ill old person? How will you help to maintain good atmosphere?
Answer:
If there is a chronically ill old person in the house, the entire atmosphere of the house changes. There is tension and grief in the house. Doctor’s visits to the house become routine. The ill person’s diet and medicines are strictly followed.

In such times, everybody in the family should contribute to the work of taking care of the patient. We can help in bringing medicines. We can sit beside the patient during night time. We should maintain pleasant atmosphere in the house. We should help the person who is burdened by the duties towards the sick patient by helping in whatever little ways that we can.

Project: (Do it your self)

Project 1.
Enlist various factors affecting the social health in your residential area. Decide the necessary changes to correct the situation and implement those changes.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Let’s Think: (Text Book Page No. 101)

Question 1.
Elders always instruct you to get out of the home to interact with relatives and others and play outdoor games but not to spend time continuously with television, phone and internet. Why the children of your age are instructed same in each home?
Answer:
When we interact with the relative, it becomes easier to mix with other strangers later. The personality moulds when we talk and interact with different people. We can exchange the thoughts. We learn to converse in a rightful way. When we go to playground and take part in outdoor games, we get health benefits. Sitting at home and spending productive time in just mobile or computer games, does not benefit in any way.

Most of the serials on the television are of no use for any kind of personality development, instead they push us in a virtual world. Except for few channels like National Geographic and Animal Kingdom, we do not get any knowledge by television viewing.

The elders in the house are experienced people. They understand ‘what is good’ and ‘what is to be avoided’. They are also genuinely concerned about bright future of the youngsters in the house. By giving instructions to the youngsters, they never get benefitted but it is our generation that gets proper guidance. These instructions should be followed for a perfect personality and bright future.

Think: (Text Book Page No. 103)

Question 1.
Whether the incidence shown in the following picture is rational? Express your opinion.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 2
Answer:
In the picture is seen a woman asking the beggar to move away. The beggar looks dirty and sick. In one way, the picture looks proper as the beggar may be causing inconvenience to the people in that house. He is unhygienic and may spread the- infection. But from the humanitarian point of view, he may be needing help. He may be starving. He may be sick. In such a case, Jie should be given food and help.

However, if he is a drug addict the police should be called and person should be transferred immediately. From the picture, the exact condition of the man is not clearly understood and hence, the exact opinion about the rationality of the incidence cannot be made.

Observe: (Text Book Page No. 103)

Question 1.
Two caricatures presenting the situations of the year 1998 and 2017 about playing on playground are given below. Observe those caricatures. Express your opinion about arising of such different situations.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 3
Answer:
In 1998, the technology was not so much advanced as it is today. In every house, there were no computers or laptops. Mobile phones were not popular then. The children used to play games which were outdoor and physical. They used to spend quality time on the playground. They always wanted to rush to the playground after their school hours. Therefore, mothers of that time had the task to get back their children from the playground.

By 2017, the situations and the social and technological change became enormous. The constant growth of the cities also experienced the rising construction. This too resulted in loss of playgrounds. After school time, children started spending their time in mobile and computer games. The parents also became financially well-off and started providing all the amenities to the children.

Due to the internet and the computer at home, the children got hooked to these electronic media. They started spending all the available time in virtual games, Facebook, what’s app and other social media. Thus mothers, of recent times had to force their children out of the house, at least for some time, so that they can play physical games.

Two caricatures presenting the situations of the year 1998 and 2017 about playing on playground show the tremendous social change that has undergone in our society.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Observe: (Text Book Page No. 104)

Question 1.
Observe the images below. Is it rational? Why?
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 4
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 5
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 6
Answer:
In the above three pictures three incidences are shown. In the first picture (Fig. 9.3), the boy who is taking his lunch is shown. He is busy with his mobile while having his food. In second picture (Fig. 9.4), a young man is taking selfie while standing in the busy road. He is not aware of the approaching car too. In the third picture (Fig. 9.5), some men are taking pictures of the accident that has recently happened. The person is injured and bleeding. But these men-are busy in photographing him.

All the three pictures are showing irrational and improper behaviour. We should respect the food while eating. We should eat in a disciplined way. Standing in the middle of the road and taking selfie is like inviting the mishap. Selfie taken in such circumstances usually results in an accident. In the last picture, the sensitivity and the humanity to save the victim is lacking. If the victim is immediately rushed to hospital, his life can be saved. Instead of helping the victim if people are engrossed in taking pictures, then it is absolutely wrong.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Our ……… has been changed to some extent in the age of technology.
(a) lifestyle
(b) habit
(c) circumstance
(d) passion
Answer:
(a) lifestyle

Question 2.
………… influence is stronger in case of adolescents.
(a) Teacher’s
(b) Father’s
(c) Relative’s
(d) Peer group
Answer:
(d) Peer group

Question 3.
Tobacco containing substances has ……….. effect on mouth and lungs.
(a) acidic
(b) alkaline
(c) carcinogenic
(d) neutral
Answer:
(c) carcinogenic

Question 4.
Persons continuously using computers and the internet become …………..
(a) courageous
(b) timid
(c) solitary
(d) criminal
Answer:
(c) solitary

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 5.
…………. has been newly launched in Police Department.
(a) Cybercrime unit
(b) Women protection unit
(c) Senior citizen care unit
(d) Forensic unit
Answer:
(a) Cybercrime unit

Question 6.
…………. helps to improve concentration in the studies.
(a) Eatables
(b) Meditation
(c) Hobbies
(d) Sports
Answer:
(b) Meditation

Question 7.
Hobbies like rearing pet animal helps to create a …………..
(a) positive mindset
(b) negative attitude
(c) wealth
(d) concentration
Answer:
(a) positive mindset

Find the odd one out:

Question 1.
Transport facilities, Social security, Counselling, Toilets.
Answer:
Counselling. (All others are factors affecting social health. Counselling is the positive measure for mental health.)

Question 2.
Aadhaar card, PAN card, Greeting card, Credit card.
Answer:
Greeting card. (All others are important cards of personal use.)

Question 3.
What’s app, Instagram, Facebook, Textbook.
Answer:
Textbook. (All others are social media.)

Question 4.
Tobacco, Laughter club, Alcoholism, Drug abuse.
Answer:
Laughter club. (All others are addictions.)

Find out the correlation:

Question 1.
Movement against tobacco : Tata trust : : Education of slum children : …………..
Answer:
Movement against tobacco : Tata trust : : Education of slum children : Salaam Mumbai Foundation

Question 2.
Addictive substances : Drugs : : Carcinogenic (Cancer causing) substances : ………..
Answer:
Addictive substances : Drugs : : Carcinogenic (Cancer causing) substances : Tobacco

Question 3.
Radiations from cell phones : Headache : : ………….. : Hindrance to the brain development.
Answer:
Radiations from cell phones : Headache : : Alcoholism : Hindrance to the brain development.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Select the two options in the ‘B’ group related to ‘A’ group:

Question 1.

‘A’ Group‘B’ Group
(1) Salaam Mumbai Foundation(a) Work against alcoholism
(b) Freedom from tobacco
(c) Laughter club
(d) Help to improve student’s lifestyle

Answer:
(1) Salaam Mumbai Foundation – (b) Freedom from tobacco (d) Help to improve student’s lifestyle.

Give scientific reasons:

Question 1.
Nowadays school going children suffer from mental stress.
Answer:

  • These days children stay in nuclear families. Due to need for earning and also due to her career choices, mother of the house is also away for long period of time.
  • The grandparents or other elders are not in the home. This makes the children alone in the house.
  • At school and during studies, there is fierce competition. The modern technology like internet or mobile phones are luring the children away from their regular exercises or outdoor games.
  • The wrong kind of peer pressure introduces addictive substances at the young age.
  • There is insecurity in the outside world for the young children.

These facts create emotional burden on the young minds and thus they suffer from mental stress.

Question 2.
Girls are facing the problem of stress due to such gender inequality.
Answer:

  • In most of the households there are many bindings on girls and excessive freedom for boys.
  • Boys do not participate in the domestic duties whereas girls have compulsion for the same.
  • In society too, girls have to face the problems like teasing and molestation.
  • This creates insecurity among the minds of girls.
  • The social change has made women independent and equal but still the male dominated society and the gender inequality persists causing more stress for young girls.

Question 3.
Consuming liquor is always bad.
Answer:

  • When the liquor is produced from alcohol wrong processes can be carried out which makes the liquor highly toxic.
  • It may cost the life too. Due to alcohol in the liquor, there is directly effect on the nervous system and especially on the brain.
  • Other vital organs such as liver and kidneys are harmed due to alcohol.
  • The lifespan of person decreases due to alcoholism.
  • In students, the brain functioning is affected and the ability to memorize and think rationally is lost. The learning process becomes slow.
  • Due to all these effects, there is social, mental and familial problems in the society. Therefore, consuming liquor is always bad.

Question 4.
We need to keep the PIN number of the debit card secret.
Answer:

  • Debit card is used to withdraw our money from the bank account.
  • During withdrawal, we have to use our PIN number.
  • If this PIN number is known to anybody, he or she can withdraw all our money and loot us.
  • Therefore, to prevent such financial loss, we have to keep the PIN number of the debit card secret.

Question 5.
Importance of outdoor games is unparalleled.
Answer:

  • Outdoor games give good physical exercise. These games give many physical benefits.
  • It improves personal discipline, interaction with fellow players and created sense of unity.
  • Through play by driving away the loneliness, mental stress and depression is reduced.
  • The person becomes more social.
  • Therefore, it is said that the importance of outdoor games is unparalleled.

Answer the following questions:

Question 1.
What is alcoholism? What are its effects?
Answer:

  • Alcoholism is the addiction to have alcohol in the form of different types of liquor. Liquor is produced from alcohol. Alcohol is in turn obtained by fermentation of different substances.
  • Consuming liquor becomes an addiction for a long-term. Due to alcohol, the efficiency of nervous system and especially the brain is affected.
  • Other vital organs such as kidneys and liver are adversely affected.
  • Lifespan of an alcoholic decreases due to constant drinking and malnourishment.
  • Especially in adolescent age if alcohol is consumed the brain functioning does not take place properly. The mental ability of memorization and learning becomes slow. There is lack of concentration in studies.
  • The alcoholic person lacks the rational thinking and hence faces with social, mental and familial problems along with physical illness.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 2.
How the excessive use of social media and technology is proving harmful?
Answer:
Excessive use of social media and modem technology is disturbing the social health. It is also affecting physical and mental health. Increase in cybercrime take place. People waste their time by watching useless and obscene material. Violence develops by watching few weird cartoon serials. Dependency on machine rises and persons lose self-reliance.

Question 3.
Explain the importance of exercise, yoga and meditation.
Answer:

  • Exercise, yoga and meditation are the ways to reduce mental and physical stress.
  • In yoga various asanas and pranayama are performed. It also includes good food and discipline of the body and mind.
  • Deep breathing, yogic sleep can help in the building up health.
  • Meditation helps in concentration and brings positivity to the mind. Especially, the students increase the concentration in the studies.

Write short notes on the following:

Question 1.
Cybercrimes.
Answer:

  • No personal information should be shared on the phone, especially the details of bank account, Aadhaar card, PAN card, credit card or debit card number, etc. Cheating persons by using this information is a greatest cybercrime.
  • If PIN of any debit or credit card is known to a stranger, he or she can make fraudulent transactions.
  • The PIN number and CVV number should be kept total secret. Otherwise, the bank transactions, are done using PIN without the knowledge of consumers.
  • In on-line purchases, many a times consumers are cheated. In this, the consumers are shown superior items on websites but actually the inferior ones are sold to them.
  • ‘Hacking of information’ is done by some programmers in which the confidential information about government, institutes and companies is obtained from internet with the help of computer programs.
  • Fake Facebook accounts are opened and false information is displayed there. This is for harassing girls or financially exploiting others.
  • In internet piracy, written literature, software, photos, videos, music, etc. of other persons are misused or illegally used.

Misuse of electronic media sending derogatory messages, spreading vulgar pictures and provocative statements is also a cybercrime. Very rapid exchange of information through media like email, Facebook and Whatsapp takes place these days. But we have to take care about leaking of our own important information.

However, when our personal information and phone numbers are automatically spread and reached to fraudulent people, then they commit malpractices which can hinder the function or shut of the cell phones or computers. All these are cybercrimes which are also indicative of mental health.

Question 2.
Addiction.
Answer:
(1) In adolescent age group, there is tremendous pressure of peers. This peer-group influence can be at times wrong, if the friends are not good. Instead of following advice of parents, the adolescent girls and boys tend to listen to the wrong advices of their friends.
(2) Due to lack of parental supervision, children in their early age start using tobacco, cigarette, gutkha, alcoholic drinks, drugs, etc. This may be due to peer-pressure.
(3) The children fall into the trap of addictions either due to peer-group pressure or due to false
symbol of high standard living. Sometimes they try to imitate their elders.
(4) The addictive substances are hazardous, and they cause long term effects. Some are temporarily intoxicating substances obtained from the plants. While some of the chemical ingredients in them can permanently damage the human nervous system, muscular system, heart, etc. Some tobacco like substances Eire carcinogenic in action especially on the mouth and lungs.

Complete the paragraph by choosing the appropriate words given in the bracket:

Question 1.
(lungs, heart, carcinogenic, nervous, intoxicating, hazardous, addictions, peer-group)
The children fall into the trap of …………. either due to ……….. pressure or due to false symbol of high standard living. Sometimes they try to imitate their elders. The addictive substances are ………….., and they cause long term effects. Some are temporarily ………… substances obtained from the plants. While some of the chemical ingredients in them can permanently damage the human ……… system, muscular system, …………, etc. Some tobacco like substances are ………. in action especially on the mouth and ………….
Answer:
The children fall into the trap of addictions either due to peer-group pressure or due to false symbol of high standard living. Sometimes they try to imitate their elders. The addictive substances are hazardous, and they cause long term effects. Some are temporarily intoxicating substances obtained from the plants. While some of the chemical ingredients in them can permanently damage the human nervous system, muscular system, heart, etc. Some tobacco like substances are Carcinogenic in action especiailly on the mouth and lungs.

Paragraph based questions:

1. Read the paragraph and answer the questions given below:
Social health involves your ability to form satisfying interpersonal relationships with others.
It also relates to your ability to adapt comfortably to different social situations and act appropriately in a variety of settings. Spouses, co-workers and acquaintances can all have healthy relationships with one another. Each of these relationships should include strong communication skills, empathy for others and a sense of accountability. In contrast, traits like being withdrawn, vindictive or selfish can have a negative impact on your social health. Overall, stress can be one of the most significant threats to a healthy relationship. Stress should be managed through proven techniques such as regular physical activity, deep breathing and positive self-talk.

Questions and Answers:

Question 1.
How can you be socially healthy?
Answer:
If one has ability to form satisfying interpersonal relationships with others, he or she can be socially healthy. In all social situations and settings there should be appropriate behaviour.

Question 2.
Which qualities are needed for having good social contacts?
Answer:
Strong communication skills, empathy for others and sense of accountability are the qualities needed for having good social contacts.

Question 3.
Which traits have negative impacts on social health?
Answer:
Being withdrawn, vindictive or selfish, and stressed out personality has negative impacts on the social health.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 4.
What are the stress management techniques?
Answer:
Regular physical activity, deep breathing and positive self-talk can be the simple stress management techniques.

Question 5.
What is the significant threat to social health of an adolescent in your opinion?
Answer:
General stress, addictions, wrong peer pressure, too much screen time, lack of parental care are threats to social health of an adolescent.

Activity based questions:

Question 1.
Fill in the boxes with the help of the given clue: (March ’19)
Continuous consumption of alcohol and tobacco material …………
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 7
Answer:
ADDICTION

Question 2.
Observe the pictures and answer the questions. (March’19)
(a) Playing games on mobile while eating is right or wrong. Justify.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 8
Answer:
The boy taking his lunch is shown in the adjoining picture. He is busy with his mobile while having his food. His nutrition may affect due to such behavior.

(b) What do you conclude from the following picture?
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 9
Answer:
Cigarette contains carcinogenic nicotine. It should never be smoked. Similarly, always stay away from addictions such as drugs, alcohol, gutkha, etc. The pictures give message for control of addictions.

(c) Observe the following picture and state what can be the outcome?
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 10
Answer:
In picture, a young man is taking selfie while standing in the busy road. He is not aware of the approaching car too. This may cause an accident.

Question 3.
Complete the following:
Concept-diagram using factors harming the social health and based on it, answer the following questions :
(i) Tobacco products can be included in which of those factors?
(ii) How the tobacco products are harming the social health?
Answer:
(Answers are given in bold.)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 11
(i) Tobacco products are included under addiction.
(ii) Tobacco is carcinogenic product. By its consumption personal and social health is affected on a large scale. Spitting tobacco anywhere is also common practice among tobacco chewers. This too affects public hygiene and cleanliness.

Question 4.
(9) Observe the following figure and answer the given questions: (Board’s Model Activity Sheet)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 12
(a) What does this picture given in the textbook indicate?
(b) Explain any two causes for this problem.
(c) Describe any two measures to eradicate this problem.
Answer:
(a) Given picture indicates that person is suffering from mental problem. He is under sever depression and frustration. Person may be using the drugs.

(b) Causes of this problems are as under:

  1. Nuclear families and working parents.
  2. Poverty, divided family and unemployment. Addiction is major cause of this problem.

(c) Measures to eradicate this problem:

  • By good communication with parents or family members harmonious relation can be re-established.
  • Help from counsellors can be helpful to minimize the problem.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Question 5.
(i) Which mental illness is shown in the picture 9.5?
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 13
(ii) Which social message would you like to give through it.
Answer:
(i) The picture (figure 9.5) shows ‘insensitivity’, which is a type of human nature.
(ii)

  • Instead of shooting the accident the victim should be given first aid.
  • Call on 100 and 108 and seek immediate help from police and ambulance.
  • Disperse the crowd and try to save life of victim by giving CPR.

Question 6.
Write, which is an inappropriate action in the picture 9.5? (Board’s Model Activity Sheet)
Answer:
The picture shows lack of sensitivity and responsibility.

Question 7.
Observe the figure and answer the questions given below.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health 14
(a) What do these figures indicate?
(b) Which gadgets can be misused for these?
(c) Give two examples of such events.
(d) Name the act amended by Government of Maharashtra to control such events.
(e) What care should be taken by a person to avoid such events?
Answer:
(a) The above figures indicate different types of cybercrime.

(b) The gadgets that are usually used for cybercrime are internet connected computers, cell phones, ATM machines, debit and credit cards, etc. Also using aadhar and PAN cards of others.

(c) (1) Bank transactions are done without the knowledge of the account holder by stealing necessary numbers or pass codes. (2) By opening the fake accounts of social media and deceiving girls, harming them psychologically by teasing them. (3) Deceiving customers by showing superior options on the internet and providing inferior ones when bought. In online shopping many may be cheated in this way.

(d) IT Act-2000 is the act enacted since 17th Oct. 2000 and amended in 2008 that has been imposed by Government of Maharashtra to control cybercrimes.

(e) To prevent cybercrimes, one has to keep vigil over bank transactions. Never reveal any details on the phone. The ATM pin number and PAN or AADHAR details should not be revealed to anyone. While at ATM machines, the pin number should be covered. Always log out from the internet after the work is over.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 9 Social Health

Projects:

Project 1.
Observe and Discuss: Observe the chart given on textbook page 101. Discuss about the relationship of various factors shown therein with the social health. (Textbook page no. 101)

Project 2.
Try This! (Textbook page no. 101) Classify your classmates into following groups depending upon the observation for a week.
1. Highly interactive.
2. Occasionally interactive.
3. Non-interactive
Make a list of the friends of each of the above three group members and also mention the group to which you belong.

Project 3.
Compare: (Textbook page no. 103) Distribute the 24 hours of your daily routine as per various duties you have observed. Make two categories as time spent on your health and time spent on other responsibilities and compare both the categories.

Project 4.
Internet is my friend: Visit the website www.cyberswachhtakendra.gov.in

Maharashtra State Board Class 10 Science Solutions