Bhagya

Std 7 Science Chapter 20 In the World of Stars Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 20 In the World of Stars Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 20 In the World of Stars Question Answer Maharashtra Board

1. Write the proper words in the blanks:
(meridian, horizon, twelve, nine, apparent, celestial, ecliptic)

Question a.
When seen from a great distance, the sky seems to be touching the ground along a circle. This circle is called the …………… .
Answer:
horizon

Question b.
The ………….. is used while defining the zodiac sign.
Answer:
meridian

Question c.
Classified according to seasons, one season will have ………… nakshatras.
Answer:
nine

Question d.
The rising of the sun in the east and its setting in the west is the ………. motion of the sun.
Answer:
apparent

2. A star rises at 8 pm. tonight. At what time will it rise after a month? Why?

Question a.
A star rises at 8 pm. tonight. At what time will it rise after a month? Why?
Answer:

1. Str s rise and set 4 minutes earlier every day. If star rises at 8 pm tonight, it will rise at 7:56 pm tomorrow.
2. It will rise at 5:24 pm after a month.
3. The sun and the moon are seen to move from the west to the east against the background of stars.
4. The sun moves through one degree every day and the moon through 12 to 13 degrees.
5. This happens due to the motion of the earth around the sun and the moon around the earth which affects the duration of the stars and shortens its time period.

3. What is meant by “The sun enters a nakshatra?” It is said that in the rainy season the sun enters the mrug nakshatra. What does it mean?

Question a.
What is meant by “The sun enters a nakshatra?” It is said that in the rainy season the sun enters the mrug nakshatra. What does it mean?
Answer:

1. When we look at the sun we see not only the sun but also constellation behind the sun.
2. The constellation cannot be seen in bright sunlight but it is indeed present behind the sun.
3. As the earth changes its position, a different constellation or zodiac sign or raashi appears behind the sun.
4. This is what we express when we say that the sun enters a particular zodiac sign or raashi.
5. In rainy season due to the perceived motion of the sun, it enters mrug nakshatra and that is how it is expressed.

4. Answer the following questions. 

Question a.
What is a constellation?
Answer:
A group of stars occupying a small portion of the celestial sphere is called a constellation.

Question b.
What points should be considered before a skywatch?
Answer:

1. The place for sky watching should be away from the city and as far as possible it should be new moon night.
2. Binoculars or telescopes should be used for skywatch.
3. Identifying the pole star in the north makes the skywatch easier. Hence the pole star should be used as a reference point for skywatch.
4. As the stars in the west set early, sky watching should begin with stars in the west.
5. (a) On a sky map, the north and south are towards the bottom and top of the map respectively, (b) This is because the sky map is to be held overhead in such a way that the direction we face is at the bottom side.

Question c.
It is wrong to say that the planets, stars and nakshatras affect human life. Why?
Answer:

1. Science has proved that the constituents of the solar system e.g. planets, satellites and comets as also distant stars and constellations do not have any influence on human life.
2. Man has stepped on the moon and will conquer Mars in the 21st century.
3. Hence, in this age of science, holding on to beliefs which have been proved wrong by numerous scientific tests, is an unnecessary waste of time and energy.
4. It is important to consider all these issues with a scientific frame of mind.

5. Write a paragraph on the birth and life cycle of stars using following figure.

Question a.
Write a paragraph on the birth and life cycle of stars using following figure.
Answer:
Stars are born out of nebulae. Nebulae are clouds made up mainly of hydrogen gas and dust particles which are attracted towards one another by the force of gravity, (i) As a result of pressure, the internal temperature increases and the cloud becomes dense and spherical in shape, (ii) From the diagram, life cycle of two stars can be explained.

(a) Ordinary star: (i) The ordinary star forms a Red giant star at the later stage of its evolution when it runs out of hydrogen gas at its core. (ii) At the end stage of its life it forms a white dwarf. Stars like the sun become white dwarf when its nuclear fuel is totally exhausted. (iii) It is 1% in diameter of its original size.

(b) Massive Star: (i) Massive star forms Red super giant star at the end of its life cycle, (ii) They are also called super red giants with a relatively cool outer surface, (iii) Supernova is the explosive death of the star of the end of its life with the brightness of 100 million stars in a short amount of time, (iv) A neutron star is the dense core of the supernova. (v) It is the smallest and the densest star known to exist with a 10 km radius, (vi) Neutron stars sometimes end as a black hole, (vii) Black holes are not seen from telescopes and are identified by their intense gravitational pull where even light cannot escape.

Project:

Question a.
Visit a planetarium, collect information and present it in your school on Science day.

Class 7 Science Chapter 20 In the World of Stars Important Questions and Answers

Fill in the blanks:

Question 1.
The different group of stars is known as ………….. .
Answer:
constellation

Question 2.
The pole star is ………….. .
Answer:
North star

Question 3.
The moon moves around the earth in about ………….. days.
Answer:
27 days

Question 4.
The celestial objects are ………….. .
Answer:
The stars

Question 5.
The star that lies close to the aris of rotation of the earth ………….. .
Answer:
Pole star

Question 6.
The definite elliptical path in which a planet revolves around the sun is called ………….. .
Answer:
Orbit

Question 7.
Our earth as well as the sun belongs to the galaxy which has a spiral shape called ………….. .
Answer:
milky way

Question 8.
………….. is made up of five bright stars which are distributed along the figure of the letter M.
Answer:
Sharmishtha

Question 9.
The pole star has ……………. on one-side and ……………… on the other.
Answer:
Saptarashi, Sharmishtha

Question 10.
The continuous empty space between the planets and stars in the sky is called ………….. .
Answer:
space

Name the following:

Question 1.
The brightest star in the nakshatra.
Answer:
Yogatara

Question 2.
The stars forming a group that has a recognizable shape.
Answer:
Constellation

Question 3.
Millions of stars and planets present in the sky forming a group.
Answer:
Milky way

Question 4.
The clouds from which stars are bom.
Answer:
Nebulae

Question 5.
Saptarshi constellation in English.
Answer:
Great Bear

Question 6.
The festival celebrated when sun enters Makar raashi.
Answer:
Makara Sankranti

Question 7.
The brightest star in the Orion constellation.
Answer:
Sirius

State whether the following statements are True or False. Correct the false statements:

Question 1.
The hydrogen gas and dust particles in a nebulae are attracted towards each other by gravity.
Answer:
True

Question 2.
While standing on the ground, the celestial sphere exactly below our feet is called the Zenith.
Answer:
False. The point on the celestial sphere exactly above our head is called the Zenith

Question 3.
The circle describing the apparent motion of the earth around the sun is called the ecliptic.
Answer:
False. Ecliptic is the apparent motion of sun around the earth

Question 4.
Vrushchik or Scorpio is a constellation with 10 to 12 stars.
Answer:
True

Question 5.
Makar raashi is also known as Capricorn zodiac sign.
Answer:
True

Match the following:

Question a.

Answer:

Answer in one line:

Question 1.
In which direction stars move in the sky except polar star?
Answer:
When seen from earth, stars appear to move from east to west.

Question 2.
What activity does IUCAA carry out?
Answer:
IUCAA which is present in PUNE carries out fundamental research in astronomy.

Question 3.
Define constellation.
Answer: A group of stars occupying a small portion of the celestial sphere is called constellation.

Question 4.
What is Nakshatra?
Answer:
The moon completes one revolution around the earth in approximately 27 days. The portion celestial sphere traversed by the moon in one day is called a nakshatra.

Question 5.
What is yogatara?
Answer:
A nakshatra is known from the brightest star that it contains. The brightest star is called the yogatara.

Define the following:

Question 1.
Horizon
Answer:
Far away the sky seems to be touching the ground. The line at which they meet is caled horizon.

Question 2.
Zenith
Answer:
While standing on the ground the point on the celestial sphere exactly above our head is called the Zenith.

Question 3.
Nadir
Answer:
While standing on the ground the point on the celestial sphere exactly below our feet is called the nadir.

Question 4.
Meridian
Answer:
The great circle which passes through both the celestial poles and the observer’s zenith and nadir is called a meridian.

Question 5.
Celestial equator
Answer:
If we uniformly expand earth’s equator in all directions indefinitely, it will penetrate the celestial sphere along a circle. This circle is known as the celestial equator.

Question 6.
Ecliptic
Answer:
The earth moves around the sun, but seen from the earth, the sun appears to move along a circle on the celestial sphere. This circle describing the apparent motion of the sun around the ‘ earth is called the ecliptic.

Question 7.
Zodiac sign
Answer:
The ecliptic has been imagined to divided into 12 equal parts. Each part subtends 30 degrees at the centre of the celestial sphere. Each of these part is called a raashi or zodiac sign.

Question 8.
Mrug Nakshatra or Orion
Answer:
It has 7 – 8 starts of which four are at the comers of quadrangle. The line passing through the three middle stars of the constellation when extended meets a very bright star. This is Vyadh or Sirius.

Find out.

Answer the following questions

Question 1.
Using a Marathi calendar collect information about 27 nakshatras, and divide them into the following 3 categories.
(i) Monsoon Nakshatra (ii) Winter Nakshatra (iii) Summer Nakshatra
Answer:
27 nakshtras: Ashwini, Bharani, Krittika, Rohini, Mrigashirasha, Ardra, Punarvasu, Pushya, Ashlesha, Magha, PurvaPhalguni, Uttara Phalguni, Hasta, Chitra, Swati, Vishakha, Anuradha, Jyeshtha, Mula, Purva Ashadha, Uttara Ashadha, Abhijit, Shravana, Dhanishtha, Shatabhishta, Purva Bhadrapada, Uttara Bhadrapada, Revati

Question 2.
Write the difference between constellations Saptarshi and Mrug nakshatra
Answer:

Question 3.
Draw sketches to show the relative position of prominent stars in Ursa major and Orion.
Answer:
a. Ursa major (Saptarshi) appears like a big dipper, (or kite shape). There are 3 bright stars in the handle and 4 stars in the bowl of the dipper. (It can be seen during April in summer in northern skies).

(b) Orion appears like a hunter. Three bright stars appear in the belt, 5 bright stars are arranged in the form of a quadrilateral. (It is visible during winter in the northern skies)

Question 4.
Why is the pole star important for sky watch?
Answer:

1. Identifying the Pole Star in the north makes the sky watch easier. Hence the pole star should be used as a reference point for skywatch.
2. If we extend one side of the quadrangle of Saptarshi, it reaches the Pole Star.
   The pole star has Saptarshi on one side and Sharmishtha on the other.

Question 5.
What is the relation between the pole star and the constellations Saptarshi and Sharmishtha?
Answer:

1. Saptarshi is in the shape of a quandrangle with a tail made up 3 stars resembling a kite. If we extend one side of the quadrangle it reaches the Pole Star.
2. The constellations of Saptarshi and Sharmishtha are useful in locating the Pole Star.
3. The perpendicular bisector of the line joining the third and fourth stars in Sharmishtha goes towards the Pole star.
4. The Pole Star has Saptarshi on one side and Sharmishtha on the other.

Use your brain power!

Answer the following questions.

Question 1.
One Zodiac Sign = ………….. nakshatras
Answer:
27 nakshatras. Each nakshatra is divided in padas or charan. Every nakshatra has 4 padas. These 27 nakshatra complete the entire circle of 360° of zodiac.
The zodiac comprises of 360°.

Question 2.
Is sun the only star present in our Milky Way galaxy?
Answer:
No. Sun is not the only star present in the Milky Way. There are lakhs of stars in the Milky Way, some of them being many times bigger than our sun. Some of them have their own planetary systems with a great diversity in colour, brightness, as well as size.

Maharashtra State Board Class 7 Science Textbook Solutions

• Cell Structure and Micro-organisms Class 7 Science Textbook Solutions
• The Muscular System and Digestive System in Human Beings Class 7 Science Textbook Solutions
• Changes – Physical and Chemical Class 7 Science Textbook Solutions
• Elements, Compounds and Mixtures Class 7 Science Textbook Solutions
• Materials We Use Class 7 Science Textbook Solutions
• Natural Resources Class 7 Science Textbook Solutions
• Effects of Light Class 7 Science Textbook Solutions
• Sound: Production of Sound Class 7 Science Textbook Solutions
• Properties of a Magnetic Field Class 7 Science Textbook Solutions
• In the World of Stars Class 7 Science Textbook Solutions

Std 10 Science Part 1 Chapter 5 Heat Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 1 Chapter 5 Heat Question Answer Maharashtra Board

Question 1.
Fill in the blanks and rewrite the sentences:
a. The amount of water vapour in air is determined in terms of its………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

b. If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their………….
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

c. When a liquid is getting converted into solid, the latent heat is……….  (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Question 2.
Observe the following graph. Considering the change in volume of water as its temperature is raised from 0 °C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

Answer:
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C. It is minimum at 4 °C. The volume of water goes on increasing in the range 4 °C to 10 °C.

In general, when a substance is heated, its volume goes on increasing with temperature. Thus, in the range 0 °C to 4 °C, behaviour of water is different from other substances. It is called anomalous behaviour of water.

Question 3.
What is meant by specific heat capacity?
How will you prove experimentally that different substances have different specific heat capacities?
Answer:
The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object.

Question 4.
While deciding the unit for heat, which temperatures interval is chosen? why?
Answer:
While deciding the unit for heat, the temperature interval chosen is 14.5 °C to 15.5 °C. For the reason, see the information given in the following box.

Question 5.
Explain the following temperature vs time graph:

(Practice Activity Sheet – 1 and 4; March 2019)
Answer:
The graph shows what happens when a mixture of ice and water is heated continuously. The temperature of the mixture remains constant (0 °C) till all the ice melts as shown by the line AB. This temperature is the melting point of ice. On further heating, the temperature rises steadily from 0 °C to 100 °C as shown by the line BC, At 100 °C water starts converting into steam. This temperature is the boiling point of water. Further heating does not change the temperature and the conversion waters steam continues as shown by the line CD.

Question 6.
Explain the following:
a. the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contracting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature of the water at the surface continues to fall to 0 °C. Finally, the water at the surface is converted into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

b. How can you relate the formation of water droplets on the outer surface of a bottle taken out of a refrigerator with formation of dew?
Answer:
At a given temperature, there is a limit on how much water vapour the given volume of air can hold. The lower the temperature, the lower is the capacity of air to hold water vapour.

The temperature of a bottle kept in a refrigerator is lower than room temperature. Hence, when the bottle is taken out of the refrigerator, the temperature of the air surrounding the bottle is lowered. Therefore, the capacity of the air to hold water vapour becomes less. Hence, the excess water vapour condenses to form water droplets (like dew) on the outer surface of the bottle.

c. In cold regions in winter, the rocks crack due to anomalous expansion of water.
Answer:
Sometimes water enters into crevices of the rocks. When the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts a tremendous pressure on the rocks which crack and break up into small pieces.

Question 7.
a. What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

b. Which principle is used to measure the specific heat capacity of a substance?
Answer:
The principle of heat exchange is used to measure the specific heat capacity of a substance. This principle is as follows: If a system of two objects is isolated from the environment by keeping it inside a heat resistant box, then no energy can leave the box or enter the box. In this situation, heat energy lost by the hot object = heat energy gained by the cold object.

c. Explain the role of latent heat in the change of state of a substance.
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

d. what basis and how will you determine whether air is saturated with vapour or not?
Answer:
Whether the air is saturated with water vapour or not is determined on the basis of the extent of water vapour present in the air. If the relative humidity is 100%, the air is saturated with water vapour. In that case, we can see the formation of water droplets on the leaves of plants/grass.
If the relative humidity is less than 100%, the air is not saturated with water vapour.

Question 8.
Read the following paragraph and answer the questions:
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy.

The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses’ heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
(1) Heat is transferred from where to where?
(2) Which principle do we learn about from this process?
(3) How will you state the principle briefly?
(4) Which property of the substance is measured using this principle?
Answer:
(1) Heat is transferred from a hot object to a cold object.
(2) This process shows the principle of heat exchange.
(3) In this process, the cold object gains heat energy and the hot object loses energy. If a system of two objects is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.
(4) This principle is used to measure the specific heat capacity of a substance.

Question 9.
Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3 °C and B by 5°C. Which object has more specific heat? And by what factor?
Solution:
Data: m = 1 g, Δ T₁ = 3 °C, Δ T₂ = 5 °C,
Q same
Here, Q = mc₁ ΔT₁ = mc₂ ΔT₂

Thus, c₁ > c₂
The specific heat of A is more than that of B and

b. Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg, ice at 0 °C, how many grams of ammonia is to be evaporated?
(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)
Solution:
Data : m₁ = 2kg, ΔT₁=20 °C – 0 °C
= 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg,
L₂ (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m₂ =?
Q₁ (heat lost by water) = m₁c₁ ΔT₁ + m₁L₁
= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg
=40 kcal + 160 kcal = 200 kcal
Q₂ (heat absorbed by ammonia) = m₂L₂
= m₂ × 34l kcal/kg
According to the principle of heat exchange, Q₁ = Q₂
∴ 200 kcal = m₂ × 341 kcal/kg
∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g
586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to he mixed to it, so that water of temperature 50 °C will be obtained?
(Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Solution:
Data: m₁ = 150 g, ΔT₁ = 50 °C – 0 °C
= 50 °C, c_w = 1 cal/g.°C, L₁ = 80 cal/g, L₂ = 540 cal/g,
Δ T₂ = 100°C – 50 °C = 50 °C, m₂ = ?
Q₁ (heat absorbed by ice) = m₁L₁
= 150 g × 80 cal/g = 12000 cal
Q₂ (heat absorbed by water formed on melting of ice) =m₁ c_w ΔT₁
= 150 g × 1 cal/g·°C × 50 °C = 7500 cal
Q₃ (heat given out by steam) = m₂L₂
= m₂ × 540 cal/g
Q₄ (heat given out by water formed on condensation of steam)
= m₂ c_w ΔT₂ = m₂ × 1 cal/g·°C × 50 °C
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ 12000 cal + 7500 cal = m₂ × 540 cal/g + m₂ × 50 cal/g
∴ 19500 cal = m₂ (540 + 50) cal/g
∴ m₂ = \(\frac{19500}{590}\) g
33.5 g of steam is to be mixed.

d. A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ·°C. It contains 250 g of liquid at 30 °C having specific heat of 0.4 kcal/kg·°C. If we drop a piece of ice of mass 10 g at 0 °C into the liquid, what will be the temperature of the mixture?
Solution:
Data: m₁ = 100 g, c₁ = 0.1 kcal/kg·°C,
= 0.1 cal/g·°C, T₁ = 30 °C, m₂ = 250 g,
c₂ = 0.4 kcal/kg·°C = 0.4 cal/g·°C, T₂ = 30 °C,
m₃ = 10 g, T₃ = 0 °C, L = 80 cal/g,
c (water) = 1 cal/g·°C, T = ?
Q₁ (heat lost by calorimeter) = m₁c₁ (T- T₁),
Q₂ (heat lost by liquid) = m₂c₂ (T – T₂),
Q₃ (heat absorbed by ice) = m₃ L,
Q₄ (heat absorbed by water formed on melting of ice) = m₃c (T – 0 °C)
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ m₁c₁ (T₁ – T) + m₂c₂ (T₂ – T) = m₃L + m₃c (T – 0 °C)
∴ m₁c₁T₁ – m₁c₁T + m₂c₂T₂ – m₂c₂T = m₃L + m₃c (T – 0°C)
∴ m₁c₁T₁ + m₂c₂T₂ = m₃L + (m₁c₁ + m₂c₂ + m₃c)T
∴ 100g × 0.1 cal/g°C × 30 °C + 250g × 0.4 cal/g.°C × 30 °C J
= 10 g x× 80 cal/g + (100 g × 0.1 cal/g.°C + 250 g × 0.4 cal/g.°C + 10 g × 1 cal/g.°C) T
∴ (10 + 100 + 10) T = (300 + 3000 – 800)°C
∴ 120 T = 2500 °C
∴ T = \(\frac{2500}{120}\) °C = \(\frac{125}{6}\) °C = 20.83 °C
This is the temperature of the mixture.

Project:
Take help of your teachers to make a working model of Hope’s apparatus and perform the experiment. Verify the results you obtain. [Do it your self]

Can you recall? (Text Book Page No. 62)

Question 1.
What is the difference between heat and temperature?
Answer:
Heat is a form of energy. Particles of matter (atoms, molecules, etc.) possess potential energy and kinetic energy. Total energy (potential energy + kinetic energy) of all particles of matter in a given sample is called it’s thermal energy. When two bodies at different temperatures are in thermal contact with each other, there is transfer of thermal energy from a body at higher temperature to a body at lower temperature. This energy in transfer is called heat. It is expressed in joule, calorie and erg.

Temperature is a quantitative measure of degree of hotness or coldness of a body. It is expressed in °C, °F or K (kelvin). Temperature determines the direction of energy transfer.

Question 2.
What are the different ways of heat transfer?
Answer:
Ways of heat transfer: conduction, convection and radiation.
[Note: heat ≡ heat energy. In the textbook, both the terms are used.]

Use your brain power! (Text Book Page No. 63)

Question 1.
Is the concept of latent heat applicable during transformation of gaseous phase to liquid phase and from liquid phase to solid phase?
Answer:
Yes.

Question 2.
Where does the latent heat go during these transformations?
Answer:
During these transformations, the latent heat is given out by the substance to the surroundings.

Use your brain power! (Text Book Page No. 64)

Question 1.
In the above experiment, the wire moves through the ice slab. However, the ice slab does not break. Why?
Answer:
When the thin wire with two equal weights attached to its ends is hung over the block of ice, it exerts pressure on the ice below it. Due to this, the melting point of the ice below the wire is lowered and some ice melts. The wire passes through the water so formed.

The water above the wire is no longer under pressure and, therefore, refreezes. Once again the ice below the wire melts, and the wire passes through it, and the process continues. In this way, due to alternate melting of ice and refreezing of water, the wire cuts right through the block of ice leaving the block intact.

Question 2.
Is there any relationship of latent heat with regelation?
Answer:
Yes. when the ice melts, heat is absorbed, but the temperature does not change. Also, when water refreezes, heat is given out, but the temperature does not change. This heat absorbed or given out is the latent heat.

Question 3.
You know that as we go higher than the sea level, the boiling point of water decreases. What would be the effect on the melting point of a solid?
Answer:
As we go higher than the sea level, the melting point of solids (i) that expand on melting is lowered due to a decrease in pressure (ii) that contract on melting is raised due to a decrease in pressure.

[The wire used in the experiment is made of a metal (usually copper). Metals are good conductors of heat. Hence, exchange of heat between the portion of the ice above the wire and that below the wire takes place readily.]

Can you tell? (Text Book Page No. 64)

Question 1.
We feel that some objects are cold, and some are hot. Is this feeling related in some way to our body temperature?
Answer:
Yes. If the temperature of the object is lower than our body temperature, e.g., ice, we feel the object is cold. If the temperature of the object is higher than our body temperature, e.g., hot water, we feel the object is hot.

Use your brain power! (Text Book Page No. 66)

Question 1.
How will you explain the following statements with the help of the anomalous behaviour of water?
(1) In regions with cold climate, the aquatic plants and animals can survive even when the atmospheric temperature goes below 0 °C.
(2) In cold regions in winter the pipes for water supply break and even rocks crack.
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contraeting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature or the water at the surface continues to fall to 0 °c. Finally, the water at the surface is converted Into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

(2) Sometimes water enters into crevices of the rocks. when the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts tremendous pressure on the rocks which crack and break up Into small pieces.

In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. when the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is rormed, there is an increase in the volume.

As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Fill in the blanks and rewrite the sentences:

Question 1.
The amount of water vapour in air is determined in terms of its…………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

Question 2.
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their……………
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

Question 3.
When a liquid is getting converted into solid, the latent heat is…………. (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Rewrite the following statements by selecting the correct options:

Question 1.
……….is used to study the anomalous behaviour of water.
(a) Calorimeter
(b) Joule’s apparatus
(c) Hope’s apparatus
(d) Thermos flask
Answer:
(c) Hope’s apparatus

Question 2.
When water boils and is converted into steam, then………..
(a) heat is taken in and temperature remains constant
(b) heat is taken in and temperatures rises
(c) heat is given out and temperature lowers
(d) heat is given out and temperature remains constant
Answer:
(a) heat is taken in and temperature remains constant

Question 3.
When steam condenses to form water,………..
(a) heat is absorbed and temperature increases
(b) heat is absorbed and temperature remains the same
(c) heat is given out and temperature decreases
(d) heat is given out and temperature remains the same
Answer:
(d) heat is given out and temperature remains the same

Question 4.
The temperature of ice can be decreased below 0 °C by mixing………..in it. (Practice Activity Sheet – 3)
(a) saw dust
(b) sand
(c) salt
(d) coal
Answer:
(c) salt

Question 5.
Ice/water is a substance that………..
(a) expands on melting and contracts on freezing
(b) contracts on melting and does not undergo change in volume on freezing
(c) contracts on melting and expands on freezing
(d) does not undergo any change in volume on melting or freezing
Answer:
(c) contracts on melting and expands on freezing

Question 6.
Heat absorbed when 1 g of ice melts at 0 °C to form 1 g of water at the same temperature is………..cal.
(a) 80
(b) 800
(c) 540
(d) 54
Answer:
(a) 80

Question 7.
The latent heat of vaporization of water is………..
(a) 540 cal/g
(b) 800 cal/g
(c) 80 cal/g
(d) 54 cal/g
Answer:
(a) 540 cal/g

Question 8.
The latent heat of fusion of ice is………..
(a) 540 cal/g
(b) 80 cal/g
(c) 800 cal/g
(d) 4cal/g
Answer:
(b) 80 cal/g

Question 9.
If the temperature of water is decreased from 4 °C to 10 °C, then its………..
(a) volume decreases and density increases
(b) volume increases and density decreases
(c) volume decreases and density decreases
(d) volume increases and density increases
Answer:
(b) volume increases and density decreases

Question 10.
At 4 °C, the density of water is………..
(a) 10 g/cm³
(b) 4g/cm³
(c) 4 × 10³ kg/m³
(d) 1 × 10³ kg/m³
Answer:
(d) 1 × 10³ kg/m³

Question 11.
The density of water is maximum at………..
(a) 0 °C
(b) – 4 °C
(c) 100 °C
(d) 4 °C
Answer:
(d) 4 °C

Question 12.
………..heat is needed to raise the temperature of 1 kg of water from 14.5 °C to 15.5 °C.
(a) 4180 J
(b) 103 J
(c) 1 cal
(d) 4180 cal
Answer:
(a) 4180 J

Question 13.
………..heat is needed to convert 1 g of water at 0 °C and at a pressure of one atmosphere into 1 g of steam under the same conditions.
(a) 80 cal
(b) 540 cal
(c) 89 J
(d) 540 J
Answer:
(b) 540 cal

Question 14.
Water expands on reducing its temperature below………..°C. (March 2019)
(a) 0
(b) 4
(c) 8
(d) 12
Answer:
(b) 4

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

Question 1.
Specific latent heat of fusion is expressed in g/cal.
Answer:
False. (Specific latent heat of fusion is expressed in cal/g.)

Question 2.
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on increasing.
Answer:
False. (If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C and then goes on increasing in the range 4 °C to 10 °C.)

Question 3.
At dew point relative humidity is 100%.
Answer:
True.

Question 4.
1 kcal = 4.18 joules.
Answer:
False. (1 kcal = 4180 joules.)

Question 5.
Specific heat capacity is expressed in cal/g·°C
Answer:
True.

Question 6.
Latent heat of fusion, Q = mL.
Answer:
True.

Question 7.
If the relative humidity is more than 60%, we feel that the air is humid.
Answer:
True.

Question 8.
If the relative humidity is less than 60%, we feel that the air is dry.
Answer:
True.

Question 9.
Relative humidity has no unit.
Answer:
True.

Question 10.
Absolute humidity is expressed in kg/m3.
Answer:
True.

Identify the odd one and give the reason:

Question 1.
Temperature, conduction, convection, radiation.
Answer:
Temperature. It is a physical quantity. Others are modes of transfer of heat.

Question 2.
The joule, The erg, The calorie, The newton.
Answer:
The newton. It is a unit of force. Others are units of energy (as well as work.)

Question 3.
cal/g, cal/g·°C, k cal/kg·°C, erg/g·°C.
Answer:
cal/g. It is a unit of specific latent heat. Others are units of specific heat capacity.

Match the columns:

Answer:
(1) Latent heat – Q = mL
(2) Specific heat capacity – cal/g·°C
(3) Heat absorbed or given out by a body when its temperature changes – Q = mc ΔT.

Answer the following questions in one sentence each:

Question 1.
State units of temperature.
Answer:
Units of temperature: °C, °F and K (kelvin).

Question 2.
State units of energy.
Answer:
Units of energy: the erg, the joule, the calorie.

Question 3.
State the relation between the joule and the calorie.
Answer:
1 calorie = 4.18 joules.

Question 4.
State the relation between the erg and the joule.
Answer:
1 joule = 10⁷ ergs.

Question 5.
State the relation between the erg and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10¹⁰ ergs.

Question 6.
State the relation between the joule and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10³ joules.

Question 7.
When heat energy is absorbed by an object, ΔT represents the rise in temperature. What would ΔT represent if the object loses heat energy? (Practice Activity Sheet – 4)
Answer:
If the object loses heat energy, ΔT would represent the decrease in temperature.

Answer the following questions:

Question 1.
Define latent heat of fusion.
(OR)
What is latent heat of fusion? State its units.
Answer:
When a solid is converted into liquid at constant temperature (melting point of the substance) the amount of heat absorbed by it is called the latent heat of fusion.
Heat is a form of energy. Hence, latent heat is expressed in units joule, erg, calorie or kilocalorie.

Question 2.
Define specific latent heat of fusion.
(OR)
What is specific latent heat of fusion? State its units.
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.
It is expressed in units J/kg, erg/g, cal/g, kJ/ kg and kcal/kg.

[Note: Specific latent heat (L) = \(\frac{\text { latent heat }(Q)}{\text { mass of the substance }(m)}\)
:. SI unit of specific latent heat = SI unit of energy / SI unit of mass = J/kg]

Question 3.
Explain the term latent heat of vaporization.
Answer:
When a liquid is heated continuously, initially, its temperature increases. Later, at a certain stage, its temperature does not increase even when heat is supplied to it. At this temperature, heat absorbed by the liquid is used for breaking the bonds between its atoms or molecules, i.e., for doing work against the forces of attraction between the atoms or molecules and conversion into gaseous phase.

This heat is called the latent heat of vaporization and the constant temperature at which this change of state occurs is called the boiling point of the liquid.

Question 4.
Define boiling point of a liquid.
(OR)
What is boiling point of a liquid?
Answer:
The constant temperature at which a liquid transforms into gaseous state is called the boiling point of the liquid.
[Note: On application of pressure, the boiling point of a liquid is raised. On reducting the pressure, the boiling point is lowered.]

Question 5.
Define specific latent heat of vaporization.
OR
What is specific latent heat of vaporization?
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.

Question 6.
The specific latent heat of fusion of ice is 80 cal/g. Explain this statement.
Answer:
When 1 g of ice at a pressure of one atmosphere and at a temperature 0 °C is converted into 1 g of water, heat absorbed by the ice is 80 cal.

Question 7.
The specific latent heat of fusion of silver is 88.2 kJ/kg. Explain this statement.
Answer:
When 1 kg of silver at a pressure of one atmosphere and at a temperature of 962 °C (melting point of silver) is converted into 1 kg of silver in liquid phase, heat absorbed by the silver is 88.2 kJ.

Question 8.
The specific latent heat of vaporization of water is 540 cal/g. Explain this statement.
Answer:
When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal.

Question 9.
Define regelation.
(OR)
What is regelation?
Answer:
The phenomenon in which the ice converts to liquid due to applied pressure and then re-converts to ice once the pressure is removed is called regelation.

Question 10.
The terms hot and cold are used in relative context. Explain.
Answer:

(1) Take three large bowls, P, Q and R. Fill bowl P with cold water, bowl Q with lukewarm water, and bowl R with hot water.
(2) Immerse your right hand in bowl P, and left hand in bowl R for about five seconds.
(3) Now, immerse both the hands in bowl Q at the same time.
(4) You will find that the water in bowl Q appears warm to your right hand, and cold to your left hand. Thus, the hand immersed in cold water for some time finds the lukewarm water hot while the one immersed in hot water finds the same lukewarm water cold. This experiment shows that the terms hot and cold are relative.

Question 11.
Draw a neat labelled diagram of Hope’s apparatus. Explain how this apparatus can be used to demonstrate anomalous behaviour of water. Draw a graph of temperature of water against time.
Answer:

The figure shows Hope’s apparatus. Initially, the cylindrical container in Hope’s apparatus is filled with water at about 12 °C and the flat bowl is filled with a freezing mixture of ice and salt.

The temperature of water in the upper part of the container (T₂) is recorded by thermometer T₂ and that of water in the lower part of the container (T₁) is recorded by thermometer T₁. Figure shows variation of temperature of water with time.

Initially, both the thermometers show the same temperature (say, 12 °C). In a short time, the temperature shown by the lower thermometer starts decreasing, while the temperature shown by the upper thermometer does not change very much.

This process continues till the temperature shown by the lower thermometer falls to 4 °C and remains constant thereafter. This shows that in the temperature range 12 °C to 4 °C, the density of the water in the central part of the container goes on increasing and hence the water sinks to the bottom. It means that water contracts, i.e., its volume decreases as its temperature falls from 12 °C to 4 °C.

As the temperature of the water in the central part of the container becomes less than 4 °C, the temperature shown by the upper thermometer begins to fall rapidly to 0 °C. But the temperature shown by the lower thermometer remains constant (4 °C). Later, the heading shown by the lower thermometer decreases to 0 °C.

In the temperature range 4 °C to 0 °C, the water moves upward. This shows that the density of water goes on decreasing in this range. It means that water expands, i.e. its volume increases as its temperature falls from 4 °C to 0 °C.

Thus, the volume of a given mass of water is minimum at 4 °C, i.e., the density of water is maximum at 4 °C.

In the above figure, the point of intersection of the two curves shows the temperature at which the density of water is maximum. This temperature is 4 °C.

Question 12.
A mountaineer climbing on the Everest, experienced the following facts. Explain each fact with the scientific reason : (1) He found j fishes alive below the ice (2) Time required for cooking was more as he went higher (3) He saw many times cliffs falling suddenly (4) He saw tubes carrying water broken.
Answer:
Explanation:
(1) Water expands as its temperature decreases from 4 °C to 0 °C. Water is converted into ice at 0 °C. The density of water is more than that of ice. Fishes can remain alive in the water (at 4 °C) below the ice.
(2) At high altitudes, atmospheric pressure is low and hence water boils at a temperature lower than its normal boiling point. Therefore, the time required for cooking food is more at higher altitudes.
(3) Water expands while freezing. Hence, the water present in the crevices of the rocks exerts a tremendous pressure on the rocks, while freezing. Therefore, the cliffs fall.
(4) Water expands while freezing. Hence, the water in the tube exerts a large pressure on the tube, while freezing. Therefore, the tube carrying water breaks.

Question 13.
What is humidity?
Answer:
The moisture, i.e., the presence of water vapour, in the atmosphere is called humidity.

Question 14.
When is air said to be saturated with water vapour?
Answer:
When air contains maximum possible water vapour, it is said to be saturated with water vapour at that temperature.

Question 15.
What does the amount of water vapour needed to saturate air depend on?
Answer:
The extent of water vapour needed to saturate air depends on the temperature. The greater the temperature, the greater is the amount of water vapour needed to saturate air.

Question 16.
When is air said to be unsaturated with water vapour?
Answer:
When air contains water vapour less than its capacity to hold water vapour at that temperature, it is said to be unsaturated with water’vapour.

Question 17.
What is dew point temperature?
(OR)
Define dew point temperature.
Answer:
If the temperature of unsaturated air is decreased, a temperature is reached at which the air becomes saturated with water vapour. This temperature is called the dew point temperature.

Question 18.
Name the physical quantity used to express the amount of water vapour present in air.
Answer:
Absolute humidity.

Question 19.
Define absolute humidity.
(OR)
What is absolute humidity? State its unit.
Answer:
The mass of water vapour present in a unit volume of air is called absolute humidity. Generally it is expressed in kg/m³.

Question 20.
Define relative humidity.
(OR)
What is relative humidity? Write the formula for % relative humidity.
Answer:
The ratio of the actual mass of water vapour content in the air for a given volume and temperature to that required to make the same volume of air saturated with water vapour at the same temperature is called the relative humidity.

% Relative humidity = [the actual mass of water vapour content in the air for a given volume and temperature ÷ the mass of water vapour required to make the same volume of air saturated with water vapour at the same temperature] × 100%.

Question 21.
What is the value of relative humidity at the dew point temperature?
Answer:
At the dew point temperature, relative humidity is 100%.

Question 22.
The mass of water vapour in air enclosed in a certain space is 60 g and the mass of water vapour needed to saturate the same air with water vapour under the same conditions is 100 g. What is the corresponding % relative humidity?
Answer:
Here, % relative humidity = (\(\frac{60 \mathrm{g}}{100 \mathrm{g}}\)) × 100% = 60%

Question 23.
During winter, sometimes we see a white trail at the back of a flying aeroplane in a clear sky. Explain why.
Answer:
In winter, air temperature is low. Hence, when an aeroplane flies, the vapour released by its engine condenses and forms white clouds. If the relative humidity of the air surrounding the plane is high, we see this white trail at the back of the plane for a long time before it disappears. If.the relative humidity is low, the white trail is short and disappears quickly. If the relative humidity is very low, there is no formation of the white trail.

Question 24.
State two effects of humidity present in atmosphere.
Answer:
Effects of humidity present in atmosphere: When the temperature of air falls below the dew point, dew and fog are formed.

Question 25.
Explain how dew and fog are formed.
(OR)
Write a short note on formation of dew and fog.
Answer:
At a particular temperature, a given volume of air can contain a certain maximum amount of water vapour. Normally, the temperature of air during the day is such that air is not saturated with water vapour present in it.

As the temperature falls, the capacity of air to hold water vapour becomes less. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed. If the water vapour condenses on the fine dust particles present in the atmosphere, mist or fog is formed.

Question 26.
State the units of heat.
Answer:
Units of heat: joule, erg, calorie, kilocalorie.

Question 27.
Define the kilocalorie.
Answer:
The amount of heat necessary to raise the temperature of 1 kg of water by 1 °C from 14.5 °C to 15.5 °C is called one kilocalorie.

Question 28.
Define the calorie.
Answer:
The amount of heat necessary to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C is called one calorie.

Question 29.
State the relation between the kilocalorie and the calorie.
Answer:
1 kilocalorie = 10³ calories.

Question 30.
Study the following procedure and answer the questions below:  (Practice Activity Sheet – 2)
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note the depth that each sphere goes into the wax.
(i) Which property of a substance can be studied with this procedure?
(ii) Describe that property in minimum words.
(iii) Explain the rule of heat exchange with this property.
Answer:
(i) Specific heat.
(ii) Specific heat: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C.
(iii) According to the rule/principle of heat exchange, heat energy lost by the hot object = heat energy gained by the cold object.

In this activity, heat absorbed by the iron sphere is transmitted more in the wax, hence the sphere goes deepest into the wax, while the lead sphere absorbs less heat, resulting in less transmission of heat in the wax, hence, the sphere goes the least depth into the wax.

Question 31.
Write the symbol for specific heat capacity. State the units of specific heat capacity.
Answer:
Symbol for specific heat capacity: c.
Units of specific heat capacity: J/kg·°C,
erg/g·°C, cal/g·°C, kcal/kg·°C.
[ Notes: (1) Specific heat capacity

In SI, heat is expressed in joule (J), mass in kg and temperature in kelvin(K).
∴ SI unit of specific heat capacity = \(\frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\). (2) The specific heat capacity of a substance depends upon its constituent particles (atoms, molecules, etc.), interaction between them, structure of the substance (atomic/molecular arrangement), temperature of the substance, etc.]

Question 32.
Explain the principle of heat exchange. Ans. Suppose two objects A and B at different temperature T₁ and T₂ respectively are enclosed in a box of heat resistant material as shown in figure.

Let m₁ = mass of A, m₂ = mass of B, c₁ = specific heat capacity of A, c₂ = specific heat capacity of B and T = common temperature attained by A and B after the heat exchange between A and B. Here, no heat leaves the box or enters the box from outside. Hence, if T₁ > T₂, heat energy lost by A (Q₁) = heat energy gained by B (Q₂).
∴ m₁c₁ (T₁ – T) = m₂c₂ (T – T₂)
[Note: If m₁, c₁, T₁, T, m₂ and T₂ are known, c₂ can be determined.]

Question 33.
The specific heat capacity of silver is 0.056 kcal/kg·°C. Explain this statement.
Answer:
The amount of heat needed to raise the temperature of 1 kg of silver by 1 °C is 0.056 kcal.

Question 34.
Explain how the specific heat capacity of a solid can be determined (measured) by the method of mixture.
Answer:
A hot solid is put in water in a calorimeter. The mixture is stirred continuously and the maximum temperature of the mixture is measured with a thermometer. Heat exchange between the hot solid, water and calorimeter results in sill bodies attaining the same temperature after some time. Hence, according to the principle of heat exchange, heat lost by the solid = heat gained by the water in the calorimeter + heat gained by the calorimeter.

Now, heat lost by the solid (Q) = mass of the solid × its specific heat capacity × decrease in its temperature, heat gained by the water (Q₁) = mass of the water × its specific heat capacity × increase in its temperature and heat gained by the calorimeter (Q₂) = mass of the calorimeter × its specific heat capacity × increase in its temperature.

Heat lost by the hot object = heat gained by the calorimeter + heat gained by the water. Q = Q₂ + Q₁
Using this equation, the specific heat capacity of the solid can be determined (measured) when the other quantities are known.

Give scientific reasons:

Question 1.
Even though heat is supplied to boiling water, there is no increase in its temperature.
Answer:
Once water starts boiling, all the heat supplied to it is used in conversion of water into steam at the boiling point of water. Hence, there is no rise in its temperature.

Question 2.
Burns from steam are worse those from boiling water at the same temperature.
Answer:
1. A given quantity of steam contains more heat than the same quantity of boiling water at the same temperature.
2. When steam comes in contact with one’s body, it releases extra heat of 540 calories per gram and causes a more serious burn than that caused by boiling water.

Question 3.
In winter, the pipelines carrying water burst in cold countries.
Answer:
1. In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. When the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is formed, there is an increase in the volume.

2. As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Question 4.
If crushed ice is pressed and then the pressure is released, a lump of ice is formed.
Answer:
1. When crushed ice is pressed, its melting point is lowered and some ice melts to form water.
2. When pressure is released, the melting point becomes normal and the water freezes to form ice forming a lump.

Question 5.
In cold countries, in winter, even when the water of lakes freezes, aquatic animals and plants can survive.
Answer:
1. In cold countries, in winter, a layer of ice is formed on the surface of lakes when the atmospheric temperature falls below 0 °C. However, below this layer, there is water at 4 °C.
2. Ice, being a bad conductor of heat, does not allow transfer of heat from this water to the atmosphere. Hence, aquatic animals and plants can survive in this water.

Question 6.
Water droplets are seen on’ the outer surface of a cold drink bottle.
Answer:
1. The temperature of the outer surface of a cold drink bottle is less than that of the atmosphere.
2. Therefore, the excess of water vapour from the air condenses to form droplets on the outer surface of the cold drink bottle.

Question 7.
During cold nights, sometimes dew is formed.
Answer:
1. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. 2. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed.

Question 8.
When you enter a warm room after being outside on a frosty early morning, your spectacles ‘steam up’.
Answer:
1. On a frosty early morning, the temperature of air outside a warm room is lower than the dew point.
2. Hence, when you enter the room from outside, some water vapour in the room condenses on the glass of your spectacles, i.e., the spectacles ‘steam up’.

Question 9.
A plastic bottle, completely filled with water, when kept in a freezer, is likely to break.
Answer:
The temperature of air in the freezer (deep freeze) compartment of a refrigerator is less than 0 °C. 2. When a plastic bottle, completely filled with water, is kept in this compartment, the temperature of water falls below 4 °C and the water expands. Even when water freezes and ice is formed, there is an increase in the volume. It exerts a large pressure on the sides of the bottle and hence the bottle is likely to break.

Question 10.
The outer surface of a beaker containing ice cubes becomes wet in a short while.
Answer:
1. When ice cubes are placed in a beaker, ice starts melting. The heat required for melting is absorbed from the surrounding air and also from the beaker to some extent.
Hence, the temperature of the air and beaker falls.

2. The capacity of air to hold water vapour depends upon the temperature of the air, and this capacity decreases as the temperature decreases. At a certain low temperature, the surrounding air becomes saturated with water vapour present in it. As the temperature falls further, the air is unable to hold all the water vapour.

Hence, the extra water vapour starts condensing on the cold outer surface of the beaker in the form of minute drops. Therefore, the outer surface of the beaker containing ice cubes becomes wet in a short while.

Distinguish between the following:

Question 1.
Absolute humidity and Relative humidity.
Answer:
Absolute humidity:

1. Absolute humidity is the mass of water vapour present in a unit volume of air.
2. It is commonly expressed in kg/m³.

Relative humidity:

1. Relative humidity is the ratio of the mass of water vapour in a given volume of air at a given temperature to the mass of water vapour required to saturate the same volume of air at the same temperature.
2. It does not have unit.

Solve the following examples/Numerical problems:
[Use the data given in the Tables on pages 130 and 131.]

Problem 1.
Calculate the amount of heat required to convert 5 g of ice of 0 °C into water at 0 °C. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: Here, m = 5 g, L = 80 cal/g; Q = ?
Amount of heat required, Q = mL
= 5 g × 80 cal/g
= 400 calories.

Problem 2.
Find the amount of heat required to convert 10 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 10 g × 540 cal/g
= 5400 calories.

Problem 3.
Calculate the amount of heat required to convert 15 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
m = 15 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 15 g × 540 cal/g
= 8100 calories.

Problem 4.
How many calories of heat will be absorbed when 3 kg of ice at 0 °C melts?
Solution:
m = 3 kg = 3000 g; L = 80 cal/g; Q = ?
Quantity of heat absorbed, Q = mL
= 3000 g × 80 cal/g
∴ Q = 240000 calories.

Problem 5.
Calculate the amount of heat required to convert 10 g of water at 30 °C into steam at 100 °C. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
Here, m = 10 g; c = 1 cal/g·°C
T₂ – T₁ = 100 °C – 30 °C = 70 °C; L = 540 cal/g; Q = ?
Amount of heat required, Q = mc (T₂ – T₁) + mL
= 10 g × 1 cal/g·°C × 70 °C + 10 g × 540 cal/g
= 700 cal + 5400 cal
∴ Q = 6100 calories.

Problem 6.
If water of mass 80 g and temperature 45 °C is mixed with water of mass 20 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
Data : m₁ = 80 g, T₁ = 45 °C, m₂ = 20 g,
T₂ = 30 °C, T = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T1 – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₁T₁ + m₂T₂ = (m1 + m₂)T
∴ Maximum temperature of the mixture,

= (36 + 6) °C
= 42°C.

Problem 7.
When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.
Solution:
Data : m₁ = 70 g, T₁ = 50 °C, m₂ = 30 g, T = 41 °C, T₂ = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T₁ – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₂T₂ = (m₁ + m₂) T – m₁T₁

This is the required temperature.

Problem 8.
Find the heat needed to raise the temperature of a silver container of mass 100 g by 10 °C. (c = 0.056 cal/g.°C)
Solution:
Data: m = 100 g, ΔT = 10 °C, c = 0.056 cal/g·°C
Heat needed to raise the temperature of the container = mc ΔT
= 100 g × 0.056 cal/g·°C × 10 °C .
= 56 calories.

Problem 9.
If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Solution:
Data: m₁ = 100 g, L₁ = 540 cal/g,
T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c (water) = 1 cal/g·°C
According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:
Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal
Decrease in the temperature of this water to 0 °C:
Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal
Melting of ice: Q₃ = mL₂
= m × 80 cal/g
Now, Q₁ + Q₂ = Q₃
∴ (54000 + 10000) cal = m × 80 cal/g
∴ m = \(\frac{64000}{80}\) = 800 g
800 g of ice will melt.

Numerical problems for practice:

Problem 1.
Calculate the amount of heat required to convert 80 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
6400 cal

Problem 2.
Find the heat required to convert 20 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
1600 cal

Problem 3.
Calculate the quantity of heat released during the conversion of 10 g of ice cold water (temperature 0 °C) into ice at the same temperature. (Specific latent heat of freezing of water = 80 cal/g)
Solution:
800 cal

Problem 4.
How many calories of heat will be absorbed when 2 kg of ice at 0 °C melts? (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
160000 cal

Problem 5.
How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
10800 cal

Problem 6.
Find the heat absorbed by 25 g of water at 100 °C to convert into steam at the same temperature. (Specific latent heat of vaporization of water = 540 cal/g.)
Solution:
13500 cal

Problem 7.
If water of mass 60 g and temperature 50 °C is mixed with water of mass 40 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
42 °C

Problem 8.
If water of mass 60 g and temperature 60 °C is mixed with water of mass 60 g and temperature 40 °C, what will be the maximum temperature of the mixture?
Solution:
50 °C

Problem 9.
Find the heat needed to raise the temperature of a piece of iron of mass 500 g by 20 °C. (c = 0.110 cal/g·°C)
Solution:
1100 cal

Problem 10.
Water of mass 200 g and temperature 30 °C is taken in a copper calorimeter of mass 50 g and temperature 30 °C. A copper sphere of mass 100 g and temperature 100 °C is released into it. What will be the maximum temperature of the mixture? [c (water) = 1 cal/g·°C, c (copper) =0.1 cal/g·°C]
Solution:
33.26 °C

Problem 11.
A copper calorimeter of mass 100 g and temperature 30 °C contains water of mass 200 g and temperature 30 °C. If a piece of ice of mass 40 g and temperature 0 °C is added to it, what will be the maximum temperature of the mixture? [c (copper) = 0.1 cal/g·°C, c (water) = 1 cal/g·°C, L = 80 cal/g]
Solution:
12.4 °C

Maharashtra State Board Class 10 Science Solutions 

• Gravitation Class 10 Science Solutions
• Periodic Classification of Element Class 10 Science Solutions
• Chemical reactions and equations Class 10 Science Solutions
• Effects of electric current Class 10 Science Solutions
• Heat Class 10 Science Solutions
• Refraction of light Class 10 Science Solutions
• Lenses Class 10 Science Solutions
• Metallurgy Class 10 Science Solutions
• Carbon Compounds Class 10 Science Solutions
• Space Missions Class 10 Science Solutions

Std 10 Science Part 1 Chapter 9 Carbon Compounds Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 1 Chapter 9 Carbon Compounds Question Answer Maharashtra Board

Question 1.
Match the pairs.

Question 2.
Draw an electron dot structure of the following molecules. (Without showing the circles)
a. Methane.
Answer:
Molecular formula: CH₄

b. Ethene.
Answer:
Molecular formula: H₂C = CH₂

c. Methanol.
Answer:
Molecular formula: H₃C – OH

d. Water.
Answer:
Molecular formula: H₂O

Question 3.
Draw all possible structural formulae of compounds from their molecular formula given below.
a. C₃H₈
b. C₄H₁₀
c. C₃H₄
Answer:
a. C₃H₈ Propane:

b. C₄H₁₀ Butane:

c. C₃H₄ Propyne:

Question 4.
Explain the following terms with example.
a. Structural isomerism.
Answer:
The phenomenon in which compounds having different structural formulae have the same molecular formula is called structural isomerism. Butane is represented by two different compounds as their structural formulae are different. The first compound is a straight chain compound and the second compound is a branched chain compound. These two different structural formulae have the same molecular formula i.e. C₄H₁₀.

b. Covalent bond.
Answer:
The chemical bond formed by sharing of two valence electrons between the two atoms is called covalent bond.
Example:
1. Hydrogen molecule formation: The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

2. Formation of oxygen molecule:
(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

c. Hetero atom in a carbon compound.
Answer:
Carbon compounds are formed by formation of bonds of carbon with other elements such as halogens, oxygen, nitrogen, sulfur. The atoms of these elements substitute one or more hydrogen atoms in the hydrocarbon chain and thereby the tetravalency of carbon is satisfied. The atom of the element which is substitute for hydrogen is referred to as a heteroatom. Sometimes hetero atoms are not alone but exist in the form of certain groups of atoms.

d. Functional group.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.

e. Alkane.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: In methane, four hydrogen atoms are bonded to carbon atom by four single covalent bonds.

f. Unsaturated hydrocarbons.
Answer:
The carbon compounds having a double bond or triple bond between two carbon atoms are called unsaturated hydrocarbons. The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
e.g. Ethene (CH₂ = CH₂), Propene (CH₃ – CH = CH₂).
The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes e.g. Ethyne (CH ≡ CH).

g. Homopolymer.
Answer:
The polymers formed by repetition of single monomer are called homopolymer. e.g. polyethylene (CH₂ – CH₂)n.

h. Monomer.
Answer:
The small unit that repeats regularly to form a polymer is called monomer.
Example: Ethylene.

i. Reduction.
Answer:
In a chemical reaction, removal of oxygen from a compound or addition of hydrogen to a compound is called a reduction.

j. Oxidant.
Answer:
An oxidant is a reactant that oxidizes or removes electrons from other reactants during a redox reaction. An oxidant may also be called an oxidizer or oxidizing agent. When the oxidant includes oxygen, it may be called an oxygenation reagent or oxygen-atom transfer (OT) agent.
Examples of oxidants include:

1. Hydrogen peroxide
2. Ozone
3. Nitric acid
4. Sulfuric acid

Question 5.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH₂ – CH₃
Answer:
The number of carbon atopic in the longest chain: 4
Parent alkane: Butane IUPAC name: n-Butane

b. CH₃ – CHOH – CH₃ (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3

Parent alkane: Propane
Functional group: -OH (ol)
Assign the number: 2
The carbon atom to which the -OH group is attached is numbered as C₂. If the carbon chain of the compound contains a -OH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘ol’. (ol stands for alcohol)
Parent suffix: Propan-2-ol
IUPAC name: Propan-2-ol

c. CH₃ – CH₂ – COOH (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane: Propane
Functional group: -COOH (-oic acid)
If the carbon chain of the compound contains a -COOH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘oic acid’.
Parent suffix: Propanoic acid
IUPAC name: Propanoic acid

d. CH₃ – CH₂ – NH₂
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -NH₂ (amine)
If the carbon chain of the compound contains a -NH₂ group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘amine’.
Parent suffix: Ethanamine
IUPAC name: Ethanamine.

e. CH₃ – CHO
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -CHO (al)
If the carbon chain of the compound contains a -CHO group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘al’.
Parent suffix: Ethanal
IUPAC name: Ethanal

f. CH₃ – CO – CH₂ – CH₃
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -CO- (one)
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the function group.
If the carbon chain of the compound contains a (-CO-) group, then change the ending of the parent name, i.e., ‘e’ of butane is replaced by ‘one’.
Parent suffix: Butan-2-one
IUPAC name: Butan-2-one

Question 6.
Identify the type of the following reaction of carbon compounds.
1. CH₃ – CH₂ – CH₂ – OH + (O) → CH₃ – CH₂ – COOH
2. CH₃ – CH₂ – CH₃ + O₂ → 3CO₂ + 4H₂O
3. CH₃ – CH = CH – CH₃ + Br₂ → CH₃ – CHBr – CHBr – CH₃
4. CH₃ – CH₃ + Cl₂ → CH₃ – CH₂ – Cl + HCl
5. CH₃ – CH₂ – CH₂ – CH₂ – OH → CH₃ – CH₂ – CH = CH₂ + H₂O
6. CH₃ – CH₂ – COOH + NaOH → CH₃ – CH₂ – COONa⁺ + H₂O
7. CH₃ – COOH + CH₃ – OH → CH₃ – COO – CH₃ + H₂O
Answer:

Question 7.
Write the structural formulae for the following IUPAC names:
a. Pent-2-one
Answer:
Pent-2-one.
(1) Pent stands for 5 carbon atoms in a chain.
Number the carbon atoms in a chain as 1, 2, 3,…..

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

b. 2-Chlorobutane
Answer:
(1) In 2-chlorobutane, butane is parent alkane stands for 4 carbon atoms and number the carbon atoms in a chain as 1, 2, 3,….

(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C₂

(3) Now satisfy the valencies of each carbon atom

c. Propan-2-ol
Answer:
(1) Propan stands for 3 carbon atoms in a chain. Number the carbon atom in a chain as 1, 2, 3.

(2) ‘-ol’ stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C₂.

(3) Now satisfy the valencies of each carbon atom

d. Methanal
Answer:
(1) Meth – stands for one carbon atom and assigned the number ‘1’ to carbon in the functional group -CHO.
(2) ‘-al’ stands for functional group (-CHO) aldehyde.
(3) Now satisfy the valencies of carbon in -CHO.

e. Butanoic acid
Answer:
(1) But stands for 4 carbon atoms in a chain. Number the carbon atoms in a chain as 1, 2, 3,….

‘-oic acid’ stands for functional group -COOH. Assign the number 1 to carbon in the functional group -COOH.

Now satisfy the valencies of each carbon atom

f. 1-Bromopropane.
Answer:
(1) In 1-bromopropane, propane is parent alkane stands for 3 carbon atoms and number the carbon atoms in a chain as 1, 2, 3…….

(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C₁.

(3) Now satisfy the valencies of each carbon atom

g. Ethanamine
Answer:
(1) Eth stands for 2 carbon atoms in a chain and the parent alkane is ethane.
– C – C –
(2) ‘amine’ stands for (- NH₂) amino group. Show the amino (-NH₂) at any carbon atom.
– C – C – NH₂
(3) Now satisfy the valencies of each carbon atom

h. Butanone.
Answer:
(1) But stands for 4 carbon atoms in a chain and the parent alkane is butane. Number the carbon atoms in a chain 1, 2, 3,….

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

Question 8.
a. What causes the existance of very large number of carbon compound?
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(2) One, two or three covalent bonds can bond together two carbon atoms. These bonds are called single covalent bond, double covalent bond and triple covalent bond respectively. Due to the ability of carbon atoms to form multiple bonds as well as single bonds, the number of carbon compounds increases. For example, there are three compounds, namely, ethane (CH₃ – CH₃), ethene (CH₂ = CH₂) and ethyne (CH = CH) which contain two carbon atoms.

(3) Carbon being tetravalent, one carbon atom can form bonds with four other atoms (carbon or any other). This results in formation of many compounds. These compounds possess different properties as per the atoms to which carbon is bonded. For example, five different compounds are formed using one carbon atom and two monovalent elements hydrogen and chlorine: CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄. Similarly carbon atoms form covalent bonds with atoms of elements like O, N, S, halogen and P to form different types of carbon compounds in large number.

(4) Isomerism is one more characteristic of carbon compound which is responsible for large number of carbon compounds.

b. Saturated hydrocarbons are classified into three types. Write these names giving one example each.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called saturated hydrocarbons. Methane molecule contains only one carbon atom. In methane, four hydrogen atoms are bonded to carbon atom by four covalent bonds.

c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:

e. Give names of three natural polymers. write the place of their occurance and names of monomers from which they are formed.
Answer:

1. Poly saccharide is a natural polymer. It occurs in starch/carbohydrates. It is formed from monomer glucose.
2. Protein is a natural polymer. It occurs in muscles, hair, enzymes, skin, egg. It is formed from alpha amino acids.
3. Rubber is a natural polymer. It occurs in latex of rubber tree. It is formed from monomer isoprene.

f. What is meant by vinegar and gasohol? What are their uses?
Answer:
(1) Vinegar is a 5 – 8% aqueous solution of acetic acid. It is used as preservative in pickles. It is used to cook meat. 1t is used as a salad dressing.
(2) To increase the efficiency of petrol, it is mixed with 10% anhydrous ethanol, such a fuel is called gasohol. It is used as a fuel in cars and other vehicles.

g. what is a catalyst ? write any one reaction which is brought about by use of catalyst?
Answer:
Catalyst is a substance, which changes the rate of reaction, without causing any disturbance to it. Vegetable oil (unsaturated compound) undergoes addition reaction with hydrogen in the presence of nickel catalyst to form vanaspati ghee (saturated compound).

Project:
Prepare a chart giving detailed information of carbon compounds in everyday use. Display it in the cluss and discuss.

Can you recall? (Text Book Page No. 110)

Question 1.
What are the types of compounds?
Answer:
Organic and inorganic compounds are the two important types of compounds.

Question 2.
Objects in everyday use such as foodstuff, fibres, paper, medicines, wood, fuels are made of various compounds. Which constituent elements are common in these compounds?
Answer:
The constituent elements common in these compounds are carbon (C), hydrogen (H) and oxygen (O).

Question 3.
To which group in the periodic table does the element carbon belongs? Write down the electronic configuration of carbon and deduce the valency of carbon.
Answer:
The element carbon belongs to group 14 and its electronic configuration is 2, 4. The valency of carbon is 4.

Use your brain Power! (Text Book Page No. 115)

Question 1.
The molecular formula of ethyne is C₂H₂. From this draw its structural formula and electron-dot structure.
Answer:
Ethyne: Molecular formula: C₂H₂

Question 2.
How many bonds have to be there in between the carbon atoms in ethyne so as to satisfy their tetra valency?
Answer:
To satisfy their tetravalency, three double bonds have to be there in between two carbon atoms in ethyne.

Use your brain power! (Text Book Page No. 116)

Question 1.
Draw the electron-dot structure of cyclohexane.
Answer:
Cyclohexane: Molecular formula: C₆H₁₂

Use your brain power! (Text Book Page No. 112)

Question 1.
Atomic number of chlorine is 17. What is the number of electrons in the valence shell of the chlorine?
Answer:
There are seven electrons in the valence shell of the chlorine.

Question 2.
Molecular formula of chlorine is Cl₂. Draw electron-dot and line structure of a chlorine molecule.
Answer:

Question 3.
The molecular formula of water is H₂O. Draw electron-dot and line structures for triatomic molecule. (Use dots for electron of oxygen atom and crosses for electrons of hydrogen atoms.)
Answer:

Question 4.
The molecular formula of ammonia is NH₃. Draw electron-dot and line structures for ammonia molecule.
Answer:

Question 5.
The molecular formula of carbon dioxide is CO₂. Draw the electron-dot structure (without showing circle) and line structure for CO₂.
Answer:

Question 6.
With which bond C atom in CO₂ is bonded to each of the O atoms ?
Answer:
In CO₂, carbon atom is bonded to each of the O atoms by double bond.

Question 7.
The molecular formula of sulfur is S₈ in which eight sulphur atoms are bonded to each other to form one ring. Draw electron-dot structure for S₈ without showing the circles.
Answer:

The above S₈ molecule of sulphur has crown shaped structure. One molecule of sulpur is made up of eight atoms of sulphur.

Use your brain power! (Text Book Page No. 113)

Question 1.
Hydrogen peroxide decomposes of its own by the following reaction:
2H – O – O – H → 2H – O – H + O₂
From this, what will be your inference about the strength O – O covalent bond ?
Tell from the above example whether oxygen has catenation power or not.
Answer:
In hydrogen peroxide (H₂O₂), the O – O covalent bond is not strong as oxygen has no catenation power.

Use your brain power! (Text Book Page No. 120)

Question 1.
The above table shows the homologous series of alkenes. Inspect the molecular formulae of the members of this series. Do you find any relationship, in the number of carbon atoms and the number of hydrogen atoms in the molecular formulae?
Answer:
In the above homologous series, if we observe the molecular formulae of alkenes then the number of carbon atoms are half the number of hydrogen atoms.

Question 2.
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ what will be the number of hydrogen atoms?
Answer:
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ then the number of hydrogen atoms would be 2n.

Question 3.
What would be the general formula for the molecular formulae of the members of the homologous series of alkanes? What would be the value of ‘n’ for the first member of this series?
Answer:
The general formula for the homologous series of alkane is C_nH_{2n + 2}. The value of ‘n’ for the first member of homologous series is 1.
C_nH_2n+2 = C₁H_{2 × 1} + 2 = CH₄

Question 4.
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}. Write down the individual molecular formulae of the first, second and third members by substituting the values 2, 3 and 4 respectively for ‘n’ in this formula.
Answer:
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}
n = 2 C₂H_{2 × 2 – 2} = C₂H₂ Ethyne
n = 3 C₃H_{2 × 3 – 2} = C₃H₄ Propyne
n = 4 C₄H_{2 × 4 – 2} = C₄H₆ Butyne

Question 5.
Write down structural formulae of the first four members of the various homologous series formed by making use of the functional groups.
Answer:

Question 6.
General formula of the homologous series of alkanes is C_nH_{2n + 2}. Write down the molecular formula of the 8th and 12th member using this.
Answer:
General formula of alkanes is C_nH_{2n + 2}
n = 8 C₈H_{2 × 8 + 2} = C₆H₁₈ Octane
n = 12 C₁₂H_{2 × 12 + 2} = C₁₂H₂₆ Dodecane

Use your brain power! (Text Book Page No. 121)

Question 1.
Draw three structural formulae having molecular formula C₅H₁₂. Give the names n-pentane, i-pentane and neo-pentane to the above structural formulae.
Answer:

Question 2.
Draw all possible structural formulae having molecular formula C₆H₁₄. Give names to all the isomers.
Answer:

Try This! (Text Book Page No. 124)

Question 1.
Light a bunsen burner. Open and close the air hole at the bottom of the burner by means of the movable ring around it. When do you get yellow sooty flame? When do you get blue flame?
Answer:
When the air hole at the bottom of the burner is open, sufficient oxygen is mixed gaseous fuel for complete combustion and a clean blue flame is obtained. When the air hole is partially blocked by means of the movable ring around it, the air supply is limited which results in incomplete combustion. Hence, yellow sooty flame is produced.

(Text Book Page No. 126)

Question 1.
The names of four fatty acids separated from vegetable oils are given in the table. Identify the number of carbon – carbon double bonds from their structure and molecular formula from the below fatty acids which one when reacts with iodine will make the colour of iodine disappear.
Answer:

Use your brain power! (Text Book Page No. 128)

Question 1.
Explain by writing a reaction, what will happen when pieces of sodium metal are put in n-propyl alcohol.
Answer:
n-Propyl alcohol reacts with pieces of sodium metal, sodium propoxide and hydrogen gas are obtained.

Question 2.
Explain by writing a reaction, which product will be formed on heating n-butyl alcohol with concentrated sulphuric acid.
Answer:
When n-butyl alcohol is heated with concentrated sulphuric acid, one molecule of water is removed from its molecule to form 1-butene.

Use your brain power! (Text Book Page No. 129)

Question 1.
Which one of ethanoic acid and hydrochloric acid is stronger?
Answer:
Hydrochloric acid is stronger acid.

Question 2.
Which indicator paper out of blue litmus paper and pH paper is useful to distinguish between ethanoic acid and hydrochloric acid ?
Answer:
pH paper is useful to distinguish between ethanoic acid and hydrochloric acid.

Use your brain power! (Text Book Page No. 130)

Question 1.
Explain why does the lime water turns milky in the reaction of acetic acid with sodium carbonate.
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 2.
Explain the reaction that would take place when a piece of sodium metal is dropped in ethanoic acid.
Answer:
When a piece of sodium metal is dropped in ethanoic acid, sodium acetate and hydrogen gas is formed.

Question 3.
Two test tubes contain two colourless liquids ethanol and ethanoic acid. Explain by writing reaction which chemical test you would perform to tell which substance is present in which test tube.
Answer:
Ethanol does not react with sodium bicarbonate, while ethanoic acid reacts with sodium bicarbonate to form carbon dioxide gas.

Use your brain power! (Text Book Page No. 131)

Question 1.
When fat is heated with sodium hydroxide solution, soap and glycerin are formed. Which functional groups might be present in fat and glycerin? What do you think?
Answer:
The functional group carboxylic acid (-COOH) is present in fat whereas the functioned group hydroxyl group (-OH) is present in glycerin.

Can you tell? (Text Book Page No. 131)

Question 1.
What are the chemical names of the nutrients that we get from the foodstuff, namely, cereals, pulses and meat?
Answer:
The nutrients that we get from the foodstuff, namely cereals, pulses and meat are alpha amino acids.

Question 2.
What are the chemical substances that make cloth, furniture and elastic objects?
Answer:
The chemical substances that make cloth, furniture and elastic objects are cellulose and rubber.

Use your brain power! (Text Book Page No. 133)

Question 1.
Structural formulae of some monomers are given below. Write the structural formula of the homopolymer formed from them.

Answer:

Question 2.
From the given structural formula of polyvinyl acetate, that is used in paints and glues, deduce the name and structural formula of the corresponding monomer.

Answer:

(Text Book Page No. 133)

Question 1.
Complete the following table by writing their Structural formulae and Molecular formulae.
Answer:
(Answer is given directly in bold letters.)

Fill in the gaps in the table: (Text Book Page No. 119)
(Answer is given directly in bold letters.)
a. Homologous series of alkanes.
Answer:

b. Homologous series of alcohol.
Answer:

c. Homologous series of alkenes.
Answer:

(Text Book Page No. 123)

Question 1.
Complete the table by writing the IUPAC names in the third column.
(Answer is directly given with underline.)
Answer:

Use your brain power! (Text Book Page No. 119)

Question 1.
By how many -CH₂– (methylene) units do the formulae of the first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differ? Similarly, by how many -CH₂– units do the neighbouring members ethane (C₂H₆) and propane (C₃H₈) differ from each other?
Answer:
The first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differed by one -CH₂– unit. Similarly, ethane (C₂H₆) and propane (C₃H₈) differed by -CH₂– unit.

Question 2.
How many methylene units are extra in the formula of the fourth member than the third member of the homologous series of alcohols?
Answer:
There is only one, methylene unit extra in the formula of the fourth member and the third member of the homologous series of alcohols.

Question 3.
How many methylene units are less in the formula of the second member than the third member of the homologous series of alkenes?
Answer:
There is only one methylene unit less in the formula of the second member of and the third member of the homologous series of alkenes.

Fill in the blanks and rewrite the completed statements:

Question 1.
The organic compounds having double or triple bonds in them are termed as …………
Answer:
The organic compounds having double or triple bonds in them are termed as unsaturated hydrocarbons.

Question 2.
The general formula of alkane is ……………
Answer:
The general formula of alkane is C_nH_{2n + 2}.

Question 3.
The compounds of homologous series have the same ………….. group.
Answer:
The compounds of homologous series have the same functional group.

Question 4.
A double bond is formed between carbon atoms by ………… pairs of electrons.
Answer:
A double bond is formed between carbon atoms by two pairs of electrons.

Question 5.
The compounds having different structural formulae having the same molecular formula is called ……….
Answer:
The compounds having different structural formulae having the same molecular formula is called structural isomerism.

Question 6.
The functional group of ether is …………..
Answer:
The functional group of ether is -O-.

Question 7.
The general formula of alkene is …………
Answer:
The general formula of alkene is C_nH_2n.

Question 8.
The bond between two atoms of nitrogen is a ………… bond.
Answer:
The bond between two atoms of nitrogen is a triple bond.

Question 9.
Benzene ring is made up of ………….. carbon atoms.
Answer:
Benzene ring is made up of six carbon atoms.

Question 10.
Due to …………., vegetable oil is converted into vanaspati ghee.
Answer:
Due to hydrogenation, vegetable oil is converted into vanatspati ghee.

Question 11.
………….. control the heredity at molecular level.
Answer:
Nucleic acids control the heredity at molecular level.

Question 12.
The regular repetition of a small unit is called …………..
Answer:
The regular repetition of a small unit is called polymer.

Question 13.
The structural formula of polypropylene is ……………….
Answer:
The structural formula of polypropylene is

Question 14.
The monomers of proteins are ……………..
Answer:
The monomers of proteins are alpha amino acids.

Question 15.
The monomer of cellulose is …………
Answer:
The monomer of cellulose is glucose.

Question 16.
…………. have sweet odour.
Answer:
Esters have sweet odour.

Choose the correct alternative and rewrite the statement:

Question 1.
The property of direct bonding between atoms of the same element to form a chain is called ………..
(a) catenation
(b) isomerism
(c) dehydration
(d) polymerization
Answer:
The property of direct bonding between atoms of the same element to form a chain is called catenation.

Question 2.
The molecular weight of two adjacent members in homologous series of an alkane differ by ………. units.
(a) 16
(b) 20
(c) 14
(d) 12
Answer:
The molecular weight of two adjacent members in homologous series of an alkane differ by 14 units.

Question 3.
Consecutive members of a homologous series differ by ………. group.
(a) -CH
(b) -CH2
(C) -CH3
(d) -CH4
Answer:
Consecutive members of a homologous series differ by CH₂ group.

Question 4.
……….. is used to prepare carbon black.
(a) Methane
(b) Ethene
(c) Propane
(d) Butane
Answer:
Methane is used to prepare carbon black.

Question 5.
……….. is the general formula of alkene.
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{n – 2}
Answer:
C_nH_2n is the general formula of alkene.

Question 6.
The reaction of methane with chlorine in the presence of sunlight is called ………..
(a) pyrolysis
(b) an elimination reaction
(c) a substitution reaction
(d) an addition reaction
Answer:
The reaction of methane with chlorine in the presence of sunlight is called a substitution reaction.

Question 7.
The general formula for alkynes is ………….
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{2n – 1}
Answer:
The general formula for alkynes is C_nH_{2n – 2}

Question 8.
The reaction of ………… with ethanol is a fast reaction.
(a) calcium
(b) magnesium
(c) sodium
(d) aluminum
Answer:
The reaction of sodium with ethanol is a fast reaction.

Question 9.
Ethylene has …………. bond between two carbon atoms.
(a) a single
(b) a double
(c) a triple
(d) an ionic
Answer:
Ethylene has a double bond between two carbon atoms.

Question 10.
The saturated hydrocarbons are those in which carbon atom are linked by ………….
(a) a single bond
(b) a double bond
(c) a triple bond
(d) an ionic bond
Answer:
The saturated hydrocarbons are those in which carbon atom are linked by a single bond.

Question 11.
C₇H₁₆ is ………….
(a) hexane
(b) octane
(c) methane
(d) heptane
Answer:
C₇H₁₆ is heptane.

Question 12.

(a) carboxylic acid group
(b) aldehyde group
(c) ketonic group
(d) alcohol group
Answer:
(a) carboxylic acid group

Question 13.
The possible isomers for C₅H₁₂ are ……………
(a) 2
(b) 4
(c) 1
(d) 3
Answer:
The possible isomers for C₅H₁₂ are 3.

Question 14.
………. contains alcoholic functional group.

Answer:
(d) all of these

Question 15.
Oxygen molecule has ………… bond between two oxygen atoms.
(a) a double
(b) a single
(c) a triple
(d) an ionic
Answer:
Oxygen molecule has a double bond between two oxygen atoms.

Question 16.
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be ………..
(a) colourless
(b) blue
(c) green
(d) yellow
Answer:
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be colourless.

Question 17.
Ethanoic acid has a ……… odour.
(a) rotten eggs
(b) pungent
(c) mild
(d) vinegar-like
Answer:
Ethanoic acid has a vinegar-like odour.

Question 18.
Acetic acid ………..
(a) turns red litmus blue
(b) has pungent odour
(c) is red in colour
(d) is odourless
Answer:
Acetic acid has pungent odour.

Question 19.
When acetic acid reacts with sodium metal ……….. gas is formed.
(a) oxygen
(b) hydrogen
(c) chlorine
(d) nitrogen
Answer:
When acetic acid reacts with sodium metal hydrogen gas is formed.

Question 20.
The molecular formula of acetic acid (ethanoic acid) is …………
(a) HCOOH
(b) CH₃COOH
(c) C₂H₅COOH
(d) C₃H₇COOH
Answer:
The molecular formula of acetic acid (ethanoic acid) is CH₃COOH.

Question 21.
When sodium bicarbonate solution is added to dilute acetic acid …………
(a) a gas is evolved
(b) a solid settles at the bottom
(c) the mixture becomes warm
(d) the colour of the mixture becomes yellow
Answer:
When sodium bicarbonate solution is added to dilute acetic acid a gas-is evolved.

Question 22.
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in ………..
(a) test tube A
(b) test tube B
(c) test tube C
(d) all the test tubes
Answer:
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in all the test tubes.

Question 23.
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ……….. ester is produced.
(a) ethanol
(b) ethanoic
(c) ethyl ethanoate
(d) ethyl ethanol (Practice Activity Sheet – 1)
Answer:
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ethyl ethanoate ester is produced.

Question 24.
The following structural formula belongs to which carbon compound?

(a) Camphor
(b) Benzene
(c) Starch
(d) Glucose (Practical activity sheet- 2)
Answer:
(b) Benzene

Question 25.
What type of reaction is shown below?
\(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text { Sunlight }}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{Cl}+\mathrm{HCl}\)
(a) Addition
(b) Substitution
(c) Decomposition
(d) Reduction (Practice Activity Sheet – 3)
Answer:
(b) substitution

Question 26.
The carbon compound is used in daily life is ………..
(a) edible oil
(b) salt
(c) carbon dioxide
(d) baking soda (March 2019)
Answer:
The carbon compound is used in daily life is edible oil

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

Question 1.
Generally the melting and boiling points of carbon compounds are high.
Answer:
False. (Generally the melting and boiling points of carbon compounds are low.)

Question 2.
Till now the number of known carbon compounds is about 10 million.
Answer:
True.

Question 3.
Unsaturated hydrocarbons are less reactive than saturated hydrocarbons.
Answer:
False. (Unsatured hydrocarbons are more reactive than saturated hydrocarbons.)

Question 4.
Benzene is an aromatic compound.
Answer:
True.

Question 5.
The carbon-carbon double and triple bonds are also recognised as functional groups.
Answer:
True.

Question 6.
The general formula of alkyne is C_nH_2n.
Answer:
False. (The general formula of alkyne is C_nH_{2n – 2})

Question 7.
Naphthalene burns with a yellow flame.
Answer:
True.

Question 8.
When vegetable oil and tincture iodine react, the color of iodine does not change.
Answer:
False. (When vegetable oil and tincture iodine react, the colour of iodine changes.)

Question 9.
Saturated fats are healthy.
Answer:
False. (Saturated fats are harmful to health.)

Question 10.
Aqueous solution of ethanol is found to be neutral.
Answer:
True.

Question 11.
Denatured ethanol is used as industrial solvent.
Answer:
True.

Question 12.
Vinegar is a 12-15 % aqueous solution of acetic acid.
Answer:
False. (Vinegar is a 5-8 % aqueous solution of acetic acid.)

Question 13.
The functional group of ethanoic acid is a carboxylic group.
Answer:
True.

Question 14.
Sodium hydroxide is used in the preparation of soap from fats and oils.
Answer:
True.

Question 15.
Rubber is a manmade macromolecule.
Answer:
False. (Rubber is a natural macromolecule.)

Question 16.
Polyvinyl chloride is used in the manufacture of P.V.C. pipes and bags.
Answer:
True.

Question 17.
Polyethylene is a homopolymer.
Answer:
True.

Question 18.
The chemical bonds in carbon compounds do not produce ions.
Answer:
True.

Find the odd one out:

Question 1.
Propane, methane, ethene, pentane
Answer:
Ethene. (Others are saturated hydrocarbons.)

Question 2.
Methane, butane, benzene, sodium chloride
Answer:
Sodium chloride. (Others are organic compounds.)

Question 3.
CH₄, C₂H₆, C₃H₈, CaCO₃
Answer:
CaCO₃. (Others are organic compounds.)

Question 4.
C₂H₂, C₃H₈, C₂H₆, CH₄
Answer:
C₂H₂. (Others are saturated hydrocarbons.)

Question 5.
C₂H₄, C₄H₁₀, C₃H₈, CH₄
Answer:
C₂H₄. (Others are saturated hydrocarbons.)

Question 6.
Polyethylene, Polysaccharide, Polystyrene, Polypropylene
Answer:
Polysaccharide (Others are manmade polymers.)

Question 7.
-NH₂, -COOH,-SO₄, -Br
Answer:
-SO₄ (Others are functional groups.)

Question 8.
Methane, Ethane, Propene, Propane, Butane
Answer:
Propene (Others are members of homologous series of alkanes.)

Match the columns:

Question 1.

Answer:
(1) CH₄ – Methane
(2) Ethane – C₂H₆
(3) Alkene – CH₂ = CH₂
(4) Alkyne – C_nH_{2n – 2}.

Question 2.

Answer:
(1) Aromatic hydrocarbon – Benzene
(2) Alkane – Saturated hydrocarbon
(3) Alkyne – Propyne
(4) Alkene – C_nH_2n.

Question 3.

Answer:
(1) Cyclohexane – C₆H₁₂
(2) Methanol – CH₃OH
(3) Acetaldehyde – CH₃CHO
(4) Ethanoic acid – CH₃COOH.

Question 4.

Answer:
(1) (-OH) – Alcohol
(2) (-COOH) – Carboxylic acid
(3) (-CHO) – Aldehyde

Question 5.

Answer:
(1) Ethyne – C₂H₂
(2) Ethene – C₂H₄
(3) Ethane – C₂H₆
(4) Propyne – C₃H₄.

Question 6.

Answer:
(1) Cellulose – Wood
(2) R.N.A. – Chromosomes of plants
(3) Polyacrylonitrile – Blankets
(4) Polyvinyl chloride – P.V.C. pipes, bags.

Consider the relation between Column I and II. Fill in Column IV to match Column III.

Answer:
(1) Teflon
(2) Styrene
(3) Alpha aminoacid
(4) Nucleotide
(5) R.N.A.

Define the following:

Question 1.
Define Alkane
Answer:
Alkane: In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: Methane (CH₄), Ethane (C₂H₆)

Question 2.
Define Alkene.
Answer:
Alkene: The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
Example : Ethene (CH₂ = CH₂)

Question 3.
Define Alkyne.
Answer:
Alkyne: The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes.
Example: Ethyne C₂H₂ (CH ≡ CH).

Question 4.
Define Addition reaction.
Answer:
Addition reaction: When a carbon compound combines with another compound to form a product that contains all the atoms in both the reactants; it is called an addition reaction.

Question 5.
Define Substitution reaction.
Answer:
Substitution reaction: The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms, is called substitution reaction.

Question 6.
Define Esterification.
Answer:
Esterification: A carboxylic acid reacts with an alcohol in presence of an acid catalyst, an ester is formed. The reaction is known as esterification.

Question 7.
Define Saponification.
Answer:
Saponification: When an ester reacts with the alkali, i.e. sodium hydroxide, the corresponding alcohol and sodium salt of carboxylic acid are obtained. This reaction is called saponification reaction. It is used in the preparation of soap.
Ester + Sodium hydroxide → Sodium salt of carboxylic acid + Alcohol.

Question 8.
Define Polymerization.
Answer:
Polymerization: The reaction by which monomer molecules are converted into a polymer is called polymerization.

Name the following:

Question 1.
The higher homologue of hexane.
Answer:
Keptane.

Question 2.
The number of double bonds in benzene.
Answer:
Three.

Question 3.
The functional group in ether and halogen.
Answer:
Functional groups:
Ether: – O –
Halogen: – X (-Cl, -Br, -I).

Question 4.
Polymer of tetrafluoroethylene.
Answer:
Teflon.

Question 5.
The monomer of polysaccharide.
Answer:
Glucose.

Question 6.
The Polymer of nucleotide.
Answer:
D.N.A./R.N.A.

Question 7.
The Monomer of rubber.
Answer:
Isoprene.

Question 8.
Two oxidising compounds.
Answer:
Potassium permanganate, Potassium dichromate.

Question 9.
IUPAC name of sodium acetate.
Answer:
Sodium ethanoate.

Question 10.
The main component of natural gas.
Answer:
Methane.

Question 11.
Two isomers of butane.
Answer:
n-butane and i-butane.

Question 12.
A nomenclature system based on the structure of the compounds and it was accepted all over the world.
Answer:
International Union of Pure and Applied Chemistry (IUPAC).

Answer the following questions in one sentence each:

Question 1.
State the atomic number and electronic configuration of carbon.
Answer:
The atomic number of carbon is 6 and the electronic configuration of carbon is (2, 4).

Question 2.
State number of electrons in the outermost orbit of carbon and valency of carbon.
Answer:
Four electrons are present in the outermost orbit of carbon and the valency of carbon is 4.

Question 3.
What are hydrocarbons? Give one example.
Answer:
The compounds containing only carbon and hydrogen are called hydrocarbons. These compounds are known as organic compounds. E.g. Methane, Ethane.

Question 4.
What is the molecular formula and structural of methane?
Answer:
The molecular formula of methane is CH₄.

Question 5.
How many atoms of carbon and hydrogen are present in methane?
Answer:
The molecule of methane has one carbon atom and four hydrogen atoms.

Question 6.
State the general formula of alkane.
Answer:
The general formula of an alkane is C_nH_{2n + 2}.

Question 7.
Give two examples of alkanes.
Answer:
Methane (CH₄) and ethane (C₂H₆) are alkanes.

Question 8.
Give two examples of alkenes.
Answer:
Ethene (CH₂ = CH₂) and propene (CH₃ – CH = CH₂) are alkenes.

Question 9.
Give two examples of alkynes.
Answer:
Ethyne (HC ≡ CH) and propyne (CH₃ – C ≡ CH) are alkynes.

Question 10.
Write the name and molecular formula of a higher homologue of propane.
Answer:
Butane (C₄H₁₀) is a higher homologue of propane.

Question 11.
Write the structure and molecular formula of ethane.
Answer:
Structure of ethane:

Molecular formula of ethane: C₂H₆

Question 12.
What is meant by catenation power?
Answer:
Carbon has a unique ability to form strong covalent bonds with other carbon atoms, this result in formation of big molecules. This property of carbon is called catenation power.

Question 13.
State the structural and molecular formula of benzene.
Answer:

Question 14.
Which functional groups are present in aldehyde and ketone?
Answer:
The functional group -CHO is present in aldehyde and the functional group

is present in ketone.

Question 15.
Which functional group is present in

Answer:

Question 16.
Compare: The proportions of carbon atoms in ethanol (C₂H₅OH) and naphthalene (C₁₀H₈).
Answer:
Ethanol contains two carbon atoms while naphthalene contains 10 carbon atoms. Ethanol is a saturated hydrocarbon and naphthalene is an unsaturated hydrocarbon.

Question 17.
What are the products of combustion of methane?
Answer:
Carbon dioxide (CO₂) and water (H₂O) are the products of combustion of methane.

Question 18.
Which gas is evolved when ethanol reacts with sodium?
Answer:
Hydrogen gas (H₂) is evolved when ethanol reacts with sodium.

Question 19.
Compare: How is the transformation of ethanol into ethanoic acid on oxidation reaction?
Answer:
The transformation of ethanol into ethanoic acid is an oxidation process, in which ethanol accepts oxygen.

Question 20.
Complete the following:

Answer:

Question 21.
Which of the following hydrocarbons undergo addition reactions:
C₂H₆, C₃H₈, C₂H₄, C₃H₈?
Answer:
C₂H₄ (ethene) and C₃H₆ (propene) undergo addition reactions.

Question 22.
How many covalent bonds are there in a molecule of cyclohexane?
Answer:
A molecule of cyclohexane contains 18 covalent bonds.

Question 23.
Give the IUPAC name for CH₃COOH.
Answer:
The IUPAC name for CH₃COOH is ethanoic acid.

Question 24.
Write the IUPAC name of CH₃COONa.
Answer:
IUPAC name of CH₃COONa is sodium ethanoate.

Question 25.
What is meant by denatured alcohol?
Answer:
Ethanol is the important commercial solvent. To prevent the misuse of this solvent, it is mixed with the poisonous methanol. Such ethanol is called denatured spirit.

Question 26.
What is meant by glacial acetic acid ?
Answer:
The melting point of pure acetic acid is 17 °C. Therefore, during winter in old countries acetic acid freezes at room temperature itself and looks like ice. Therefore it is named glacial acetic acid.

Question 26.
Which useful components of hydro¬carbon are obtained by fractional distillation of crude oil?
Answer:
Various useful components of hydrocarbon such as CNG, LPG, petrol (gasoline), rockel, diesel, engine oil, lubricant, etc. are obtained by separation of crude oil using fractional distillation.

Question 28.
Which functional groups are present in ester and amine?
Answer:
Ester: -COO-
Amine: -NH₂–

Question 29.
Give two examples of natural macromolecules.
Answer:
Examples: Polysaccharide, protein and nucleic acid.

Question 30.
Write the structure of polystyrene and give its uses.
Answer:
Structure:

Polystyrene is used to make thermocoal articles.

Question 31.
Write the name and the structure of monomer of polyacrylonitrile.
Answer:
The name and structure of monomer: Acrylonitrile CH₂ = CH – CN

Question 32.
Write the name and the structure of monomer of teflon and its uses.
Answer:
The name and structure of monomer: Tetrafluro ethylene CF₂ = CF₂
Teflon is used to make nonstick cookware.

Question 33.
What is meant by copolymers?
Answer:
The polymers formed from two or more monomers are called copolymers.
Examples: Poly ethylene terephthalate.

Answer the following questions:

Question 1.
How is hydrogen molecule formed?
Answer:

The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

Question 2.
Describe the formation of oxygen molecule (O₂).
Answer:

(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

Question 3.
Describe the formation of nitrogen molecule.
Answer:

(1) The atomic number of nitrogen is 7. The electronic configuration of nitrogen is (2, 5). Nitrogen has 5 electrons in the outermost shell.
(2) It requires three more electrons to complete the L shell and attain the configuration of neon (Ne).

(3) Two nitrogen atoms come close together and share three pairs of electrons with each other, resulting in the formation of a triple bond.
(4) Thus, two nitrogen atoms are bound with a triple bond (=) to form a nitrogen molecule.

Question 4.
How is the methane molecule formed?
Answer:
(1) The electronic configuration of carbon is (2, 4). Carbon has four electrons in the outermost shell, hence it is tetravalent.
(2) The electronic configuration of hydrogen is 1, hence it is monovalent.
(3) Carbon needs four electrons to complete the L shell and attain the configuration of neon (Ne).
(4) Four atoms of hydrogen share 1 electron each with 4 electrons of carbon.
(5) A single covalent bond is formed by sharing of two electrons.

Thus, the methane molecule contains four single bonds between the carbon and hydrogen atoms.

Question 5.
State the various compounds and its formulae formed by a single atom of carbon with monovalent hydrogen and chlorine.
Answer:

Question 6.
Observe the straight chain hydrocarbons given below and answer the following questions:

(i) Which of the straight chain compounds from A and B is saturated and unsaturated straight chains?
(ii) Name these straight chains.
(iii) Write their chemical formulae and number of -CH₂ units. (Practice Activity Sheet – 2)
Answer:
(i) A is a saturated hydrocarbon, B is an unsaturated hydrocarbon.
(ii) A = Propane, B = Propene
(iii) The chemical formula of A = C₃H₈ and number of -CH₂ units are 3.
The chemical formula of B = C₃H₆ and number of -CH₂ unit is 1.

Question 7.
Draw electron-dot and line structure of an ethane molecule.
Answer:
The molecular formula of ethane is C₂H₆.

Question 8.
The molecular formula of propane is C₃H₈. From this draw its structural formula. (Practice Activity Sheet – 3)
Answer:
The structural formula of propane:

Question 9.
Draw the structure and carbon skeleton for cyclohexane.
Answer:

Question 10.
Classify into saturated and unsaturated hydrocarbons: (1) Methane (2) Ethene (3) Ethane (4) Ethyne (5) Propene (6) Propyne (7) Butane (8) Cyclohexene (9) Cyclopentane (10) Heptane.
Answer:
(i) Saturated hydrocarbons: (1) Methane (2) Ethane (3) Butane (4) Cyclopentane (5) Heptane.
(ii) Unsaturated hydrocarbons: (1) Ethene (2) Ethyne (3) Propene (4) Propyne (5) Cyclohexene.

Question 11.
Classify into alkanes, alkenes and alkynes: (1) Ethane (2) Ethene (3) Methane (4) Propene (5) Ethyne (6) Propyne (7) Butane (8) Pentane.
Answer:
Alkanes: (1) Ethane (2) Methane (3) Butane. (4) Pentane
Alkenes: (1) Ethene (2) Propene
Alkynes: (1) Ethyne (2) Propyne

Question 12.
Classify into straight chain carbon compounds, branched chain carbon compounds and ring carbon compounds:
(1) Propene (2) Butane (3) Iso-butane (4) Cyclopentane (5) Benzene (6) Isobutylene.
Answer:
Straight chain carbon compounds:

1. Propene
2. Butane.

Branched chain carbon compounds:

1. Iso-butane
2. Isobutylene.

Ring carbon compounds:

1. Cyclopentane
2. Benzene.

Question 13.
Draw chain and ring structures of organic compound having six carbon atoms in it.
Answer:
Chain structures of an organic compound having six carbon atoms:

Ring structures of an organic compound having six carbon atoms:

Question 14.
Explain the structure of benzene.
Answer:
The molecular formula of benzene is C₆H₆. It is a cyclic unsaturated hydrocarbon. Benzene ring is made of six carbon atoms. In benzene, each carbon atom is linked to two other carbon atoms, on one side by a single bond and on the other side by a double bond, i.e. three alternate single bonds and double bonds in the six membered ring structure of benzene. The compound having this characteristic unit in their structure are called aromatic compounds.

Question 15.
Draw the structures of isomers of pentane (C₅H₁₂).
Answer:

Question 16.
Recognize the carbon chain type for each of the following:

(Practice Activity Sheet – 1)
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 17.
What is meant by functional group? Give examples.
(OR)
Explain the term functional group with example.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.
Example: Methyl alcohol, acetic acid.

In methane (CH₄), when one hydrogen atom is replaced by an -OH group, methyl alcohol (CH₃OH), is formed. The -OH is known as the alcoholic functional group.
Similarly, from methane (CH₄) when one hydrogen atom is replaced by -COOH group, acetic acid (CH₃COOH) is formed. The -COOH group is known as the carboxylic acid functional group.

Question 18.
Define functional group and complete the following table:

Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the
functional groups.

Question 19.
What is meant by homologous series?
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula CnH_{2n + 2}. The members of this series are as follows:

Question 20.
State the four characteristics of homologous series.
Answer:
Characteristics of Homologous series:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit ( -CH₂– ) gets added
(b) molecular mass increases by 14 u (c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) While going in an increasing order of the length there is gradation in the physical properties i.e. the boiling and melting points.

Question 21.
Write names of first four homologous series of alcohols: (March 2019)

Answer:
First four homologous series of alcohols are

1. Methanol CH₃ – OH
2. Ethanol C₂H₅ – OH
3. Propanol C₃H₇ – OH
4. Butanol C₄H₉ – OH

Question 22.
Describe the IUPAC rules of naming organic compounds.
Answer:
IUPAC nomenclature system: International Union for Pure and Applied Chemistry (IUPAC) put forth a nomenclature system based on the structure of the compounds and it was accepted all over the world. There are three units in the IUPAC name of any carbon compound: parent, suffix and prefix. These are arranged in the name as follows:

Prefix-parent-suffix:
An IUPAC name is given to a compound on the basis of the name of its parent alkane. The name of the compound is constructed by attaching appropriate suffix and prefix to the name of the parent-alkane. The steps in the IUPAC nomenclature of straight chain compounds are as follows:

Step 1: Draw the structural formula of the straight chain compound and count the number of carbon atoms in it. The alkane with the same number of carbon atoms is the parent alkane of the concerned compound. Write the name of this alkane.

In case the carbon chain of concerned compound contains a double bond, change the ending of the parent name from ‘ane’ to ‘ene’. If the carbon chain in the concerned compound contains a triple bond, change the ending of the parent name from ‘ane’ to ‘yne’.

Step 2: If the structural formula contains a functional group, replace the last letter ‘e’ from the parent name by the condensed name of the functional group as the suffix. (Exception: The condensed name of the functional group ‘halogen’ is always attached as the prefix.)

Step 3: Number the carbon atoms in the carbon chain from one end to the other. Assign the number T to carbon in the functional group -CHO or -COOH, if present. Otherwise, the chain can be numbered in two directions. Accept that numbering which gives smaller number to the carbon carrying the functional group. In the final name, a digit (number) and a character (letter) should be separated by a small horizontal line.

Question 23.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH = CH₂
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butene
Functional group: double bond
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the double bond and the other c-atoms are numbered accordingly.
Parent suffix: But-1-ene
IUPAC name: But-1-ene

b. CH₃ – C ≡ C – H
Answer:
Number of carbon atoms in the longest chain: 3
Parent alkane: Propyne
Functional group: triple bond
Parent suffix: Propyne
IUPAC name: Propyne

c.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Prefix functional group: Chloro
Assign the number: 2

The carbon atom to which the -Cl atom is attached is numbered as C₂ and the other C atoms are numbered accordingly.
Prefix parent: 2-Chloropentane
IUPAC name: 2-Chloropentane

d. CH₃ – CH₂ – CH₂ – Br
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane : Propane Prefix functional group: Bromo
Assign the number:

The carbon atom to which the -Br atom is attached is numbered as C₁ and the other C atoms are numbered accordingly.
Prefix parent: 1-Bromopropane
IUPAC name: 1-Bromopropane

e.

Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -OH (ol)
Assign the number:

The carbon atom to which the -OH group is attached is numbered as C₂.
If the carbon chain of the compound contains a -OH group, then change the ending ‘e’ of the parent name, i.e. ,‘e’ of butane is replaced by ‘ol’ (ol for alcohol).
Parent suffix: Butan-2-ol
IUPAC name: Butan-2-ol

f.

The number of carbon atoms: 3
Parent alkane: Propane
Functional group: -NH₂ (amine)
Assign the number:

If the carbon chain of the compound contains a -NH2 group then change the ending of the parent name, i.e. ‘e’ of propane is replaced by ‘amine’.
Parent suffix: 2-Propanamine
IUPAC name: 2-Propanamine

g. HCOOH
Answer:
The number of carbon atoms: 1
Parent alkane: Methane
Functional group: -COOH (-oic cid)
If the carbon chain of the compound contains a -COOH group, then change the ending of the parent name, i.e. ‘e’ of methane is replaced by ‘-oic acid’.
Parent suffix: Methanoic acid
IUPAC name: Methanoic acid

h. CH₃ – CH₂ – CH₂ – CHO
Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane : Butane Functional group: -CHO (al)
Assign the number: 1
Assign the number ‘1’ to carbon in the functional group

If the carbon chain of the compound contains a -CHO group then change the ending of the parent name, i.e. ‘e’ of the butane is replaced by ‘al’.
Parent suffix: Butanal
IUPAC name: Butanal

i.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Functional group:

Assign the numbering:

In the longest chain, the numbering of carbon atom starts from the carbon nearest to the functional group (both the numbering equivalent).
If the carbon chain of compound contains a

group, then change the ending of the parent name i.e. ‘e’ of pentane is replaced by ‘one’.
Parent suffix: Pentan-3-one
IUPAC name: Pentan-3-one.

Question 24.
What happens when methane is burnt in air? Write the balanced chemical equation for the same.
Answer:
When methane burns in air, carbon dioxide and water are formed. The reaction is exothermic with release of large amount of heat and light.

Question 25.
What happens when ethanol is burnt in air?
Answer:
When ethanol is burnt in air, it burns with a clean blue flame, carbon dioxide and water are formed. In this reaction, release of large amount of heat and light takes place.

Question 26.
What happens when ethanol is treated with alkaline potassium permanganate? Write the balanced chemical equation for the same.
Answer:
When ethanol is treated with alkaline potassium permanganate, ethanol gets oxidised by alkaline potassium permanganate to’ form ethanoic acid.

Question 27.
What happens when vegetable oil is hydrogenated? Write the balanced chemical equation.
Answer:
When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, vanaspati ghee (saturated) compound is formed.

Question 28.
What happens when chlorine is treated with methane?
(OR)
Describe the action of chlorine on methane.
(OR)
Write a note on chlorination of methane.
Answer:
Methane reacts rapidly with chlorine in the presence of sunlight to form four products. In this reaction, chlorine atoms replace, one by one, all the hydrogen atoms in the methane.

The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms is called substitution reaction. Chlorination of methane is a substitution reaction.

Question 29.
What happens when ethanol is reacted with sodium?
Answer:
When ethanol is reacted with sodium at room temperature, sodium ethoxide is formed and hydrogen gas is liberated.

Question 30.
What happens when ethanol is heated at 170 °C with excess of conc. sulphuric acid?
Answer:
When ethanol is heated at 170 °C with excess of conc. sulphuric acid, one molecule of water is removed from its molecule to form ethene (unsaturated compound).

Question 31.
What happens when ethanoic acid is treated with sodium hydroxide? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with sodium hydroxide, neutralization takes place to form sodium acetate (sodium ethanoate) and water.

Question 32.
What happens when ethanoic acid is treated with sodium carbonate?
Answer:
When ethanoic acid is treated with sodium carbonate, sodium ethanoate, carbon dioxide and water is formed.

Question 33.
What happens when ethanoic acid is treated with sodium bicarbonate?
Answer:
When ethanoic acid is treated with sodium bicarbonate, sodium ethanoate, water and carbon dioxide is formed.

Question 34.
What happens when ethanoic acid is treated with ethanol? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with ethanol in the presence of an acid catalyst, an ester, i.e., ethyl ethanoate is formed.

Question 35.
What happens when ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst?
Ans. When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

Question 36.
State the physical properties of ethyl alcohol ethanol.
Answer:

1. Ethanol is a colourless liquid and it is soluble in water in all proportions and has pleasant odour.
2. The boiling point of ethanol is 78 °C and the freezing point is -114 °C.
3. It is combustible and burns with a blue flame.
4. An aqueous solution of ethanol is neutral to litmus paper.

Question 37.
State the properties of ethanoic acid.
Answer:

1. Ethanoic acid is a colourless liquid with boiling point 118 °C and melting point 17 °C. It has a pungent odour.
2. Its aqueous solution is acidic and turns blue litmus red.
3. A 5-8% aqueous solution of acetic acid is used as vinegar.
4. It is a weak acid.

Write short notes:

Question 1.
Catenation power.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power.

(2) Carbon shows catenation. Two or more carbon atoms can share their valence electrons and bond with each other. Thus, carbon chains can be straight or branched or closed chain ring structure forming large molecules. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(3) Hence, carbon atoms can form an unlimited number of compounds.

Question 2.
Characteristics of Carbon.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms: this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

Question 3.
Functional group.
Answer:
(1) The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of the length and nature of the carbon chain in that compound. Therefore these hetero atoms or the groups of atoms containing hetero atoms are called functional groups.

All organic compounds are derivatives of hydrocarbons. The derivatives are formed by replacing one or more H-atom/atoms of hydrocarbon by some other hetero atom or groups of atoms containing hetero atoms. After replacement, a new compound is formed which has properties different from the parent hydrocarbon.

Examples: For methane, if one hydrogen atom is replaced by an – OH group, then a compound is methyl alcohol (CH₃OH). The -OH group is known as the alcoholic functional group.
Functional group is organic compound:
1. Alcohol: – OH (hydroxy group)

4. Carboxylic acid : -COOH

Question 4.
Homologous series.
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula C_nH_{2n + 2}
The members of this series are as follows:

Characteristics:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit (-CH₂-) gets added
(b) molecular mass increases by 14 u
(c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) while going in an increasing order of the length there is gradation in the physical properties i.e., the boiling and melting points.

Question 5.
Polymerization.
Answer:
(1) The reaction by which monomer molecules are converted into a polymer is called polymerization. A macromolecule formed by regular repetition of a small unit is called polymer. The small unit that repeats regularly to form a polymer is called monomer. The important method of polymerization is to make a polymer by joining alkene type of monomers.

(2) When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

(3) The polymer polystyrene is used to make thermocoal articles. The polymer polyvinyl chloride is used to make P.V.C. pipes, doormats, etc. The polymer teflon is used to make nonstick cookware. The polymer polypropylene is used to make injection syringe, furniture, etc.

Give scientific reasons:

Question 1.
Carbon atoms are capable of forming an unlimited number of compounds.
Answer:

1. Carbon has the property of catenation. Two or more carbon atoms can share some of their valence electrons to form (single, double and triple) bonds.
2. The straight chains or branched chains or rings may have different shapes and sizes. This results in formation of many compounds. Hence, carbon atoms are capable of forming an unlimited number of compounds.

Question 2.
Ethylene is an unsaturated hydrocarbon.
Answer:
(1) Ethylene (CH₂ = CH₂) contains a double bond between carbon atoms.
(2) Thus, the valencies of the two carbon atoms are not fully satisfied by single covalent bonds. Hence, ethylene is an unsaturated hydrocarbon.

Question 3.
Naphthalene burns with a yellow flame.
Answer:
(1) Naphthalene is an unsaturated compound. In unsaturated hydrocarbon the proportion of carbon is larger than that of saturated hydrocarbon. As a result, some unburnt carbon particles are also formed during combustion of unsaturated compounds.

(2) In the flame. these unburnt hot carbon particles emit yellow light and therefore the flame appears yellow. Hence, naphthalene burns with a yellow flame.

Question 4.
The colour of iodine disappears in the reaction between vegetable oil and iodine.
Answer:
(1) Vegetable oils (unsaturated compound) contains a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.
(2) The addition reaction of vegetable oil with iodine takes place instantaneously at room temperature. The colour of iodine disappears in this reaction. This iodine test indicates the presence of a multiple bond in vegetable oil.

Question 5.
The hydrogenation of vegetable oil in the presence of nickel catalyst forms vanaspati ghee.
Answer:
(1) The molecules of vegetable oil contain long and unsaturated carbon chains. These unsaturated hydrocarbons contain a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.

(2) When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, the addition reaction takes place, vanaspati ghee (saturated compound) is formed.

Distinguish between the following:

Question 1.
Saturated hydrocarbons and Unsaturated hydrocarbons.
Answer:
Saturated hydrocarbons:

1. In saturated hydrocarbons, the carbon atoms are linked to each other only by single covalent bonds.
2. They contain only a single bond.
3. They are chemically less reactive.
4. Substitution reaction is a characteristic property of these hydrocarbons.
5. Their general formula is C_nH_{2n + 2}.

Unsaturated hydrocarbons:

1. In unsaturated hydrocarbons, the valencies of carbon atoms are not fully satisfied by single covalent bonds.
2. They contain carbon to carbon double or triple bonds.
3. They are chemically more reactive.
4. Addition reaction is a characteristic property of these hydrocarbons.
5. Their general formula is C_nH_2n or C_nH_{2n – 2}

Question 2.
Open chain hydrocarbons and closed chain hydrocarbons.
Answer:
Open chain hydrocarbons:

1. A hydrocarbon in which the chain of carbon atoms is not cyclic is called an open chain hydrocarbon.
2. All aliphatic hydrocarbons contain open chains.

Closed chain hydrocarbons:

1. A hydrocarbon in which the chain of carbon atoms is present in a cyclic form or ring form is called a closed chain hydrocarbon.
2. All aromatic hydrocarbons contain closed chains.

Question 3.
Alkane and Alkene.
Answer:
Alkane

1. Alkanes in which the carbon atoms are linked to each other only by single bonds.
2. The general formula of an alkane is C_nH_{2n + 2}
   
3. They are chemically less reactive.

Alkene:

1. Alkenes in which carbon atoms are linked to each other by double bonds.
2. The general formula of an alkene is C_nH_2n.
3. They are chemically more reactive.

Project:

Question 1.
Prepare a list of carbon compounds which occur in nature and discuss their uses in daily life.

Maharashtra State Board Class 10 Science Solutions 

• Gravitation Class 10 Science Solutions
• Periodic Classification of Element Class 10 Science Solutions
• Chemical reactions and equations Class 10 Science Solutions
• Effects of electric current Class 10 Science Solutions
• Heat Class 10 Science Solutions
• Refraction of light Class 10 Science Solutions
• Lenses Class 10 Science Solutions
• Metallurgy Class 10 Science Solutions
• Carbon Compounds Class 10 Science Solutions
• Space Missions Class 10 Science Solutions

Std 7 Science Chapter 15 Materials We Use Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 15 Materials We Use Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 15 Materials We Use Question Answer Maharashtra Board

1. Fill appropriate terms in the blanks:
(white cement, soap, detergent, wearing of bones, tooth decay, hard, soft, portiand, fatty acid)

Question a.
The substance that helps water to remove dirt from the surface of material is called …………… .
Answer:
soap

Question b.
Fluoride is used in toothpaste to prevent …………… .
Answer:
tooth decay

Question c.
Soap is a salt of …………… and sodium hydroxide.
Answer:
fatty acid

Question d.
Synthetic detergents can be used in …………… water as well.
Answer:
hard

Question e.
For construction purposes …………… Cement is the most commonly used cement.
Answer:
white cement

2. Write answers to the following questions. 

Question a.
How does the use of a detergent help to clean soiled clothes?
Answer:

1. A molecule of a detergent holds on to a water molecule at one end and an oil molecule at the other.
2. As a result the molecules of oil mix with the water.
3. This is how detergent acts on our soiled clothes. So detergent removes any oil or dirt sticking on to our clothes.
4. Due to the property of holding on to both oil and water, soap water spreads easily on many types of surfaces.
5. This property of spreading on a surface is called surface activity.
6. Detergents are surface active.
7. One effect of surface activity is lather formation.

Question b.
How will you check with the help of soap powder whether water is hard?
Answer:

1. In the hard water, (of a well or a tube-well), soap does not give lather but forms a scum.
2. As a result, soap loses its cleansing property. So with help of soap we will come to know that it is a hard water.

Question c.
What are the important ingredients of a tooth paste and what is the function of each?
Answer:

1. The important ingredients of a toothpaste are calcium carbonate and calcium hydrogen phosphate.
2. They remove the dirt on teeth. These ingredients also polish the teeth.
3. A certain proportion of fluoride in the tooth paste helps prevent tooth decay.
4. It is essential for the enamel covering of teeth.

Question d.
What are the ingredients of cement?
Answer:

1. Cement is a dry, greenish grey powder with fine particles.
2. It is made from silica (sand), alumina (aluminium oxide), lime, iron oxide and magnesia (magnesium oxide).

Question e.
What will happen if cement is not used in making concrete?
Answer:

1. In making concrete strong, the ingredients cement, water, sand and gravel should usually be mixed in the ratio of 1 : 2 : 3 : 0.5 to achieve maximum strength.
2. If enough cement is not used then the concrete will eventually fall apart, as cement is used as a binding agent.

Question f.
Make a list of detergents, that you use. Answer: There are two types of detergents that we use, (a) Natural (b) Man-made detergents.

1. Natural detergents are soap nut (ritha) soap pod (shikekai).
2. Man-made detergents are soap, hard soaps, soft soaps.
3. Synthetic detergents have taken the place of soap.
   • Detergents are commonly available as powders or concentrated solution.
   • Detergents are for laundry, washing clothes and cleaning dishes.
4. Alkaline detergents used for hard surface cleaning.

Question g.
What should be expected from a detergent for delicate garments?
Answer:

1. Detergents should not be strong, they may spoil the texture, colour of delicate garments.
2. Now a days many detergents are available especially for woollen, delicate clothes.
3. They should not contain bleach, phosphorous which will harm delicate clothes.

Question h.
What is meant by surface activity? Name three chemicals responsible for the surface activity of various detergents.
Answer:

1. Due to the property of holding on to both oil and water, soap water spreads easily on many types of surfaces.
2. The property of a substance of spreading on a surface is called surface activity and the substance is called surfactant.
3. Detergents are surface active.
4. They increase the spreading and wetting ability of water by reducing its surface tension.
5. Chemicals responsible for the surface activity of various detergents is phenol, Aprotinin.

3. What are the similarities and differences between:

Question a.
Natural detergents and Man-made detergents
Answer:

Question b.
Soap and Synthetic detergents
Answer:

Question c.
Bath soap and Soap for washing clothes
Answer:

Question d.
Modern cement and Ancient cement
Answer:

4. Explain why –

Question a.
Soap cannot be used in hard water.
Answer:

1. In the hard water of a well or a tube well, soap does not give lather but forms a scum.
2. As a result soap loses its cleansing property. So soap cannot be used in hard water.

Question b.
Oil does not mix in water. However, oil and water become homogenous if a sufficient quantity of detergent is added.
Answer:

1. A molecule of a detergent holds on to a water molecule at one end and an oil molecule at the other.
2. As a result the molecules of oil mix with the water, and we will see that the water and oil have become homogenous and the colour of the mixture appears milky.

Question c.
Synthetic detergents are superior to soap.
Answer:

1. Soap is a man-made detergent which has been in use since ancient times.
2. It was prepared by using animal fat and wood ash. In hard water, soap does not give lather, it loses its cleansing property.
3. So we can not use soap in hard water. Now synthetic detergents have taken the place of soap.
4. They can be used in hard water as well
5. Synthetic detergents have many more advanced properties like they are added with perfumes, conditioner for fabric, etc.

Question d.
Often coloured spots are formed on clothes during washing.
Answer:

1. Curry stains stick fast to the criss-crossing threads of the material of our clothes.
2. Curry contains turmeric a natural indicator which turns red in a basic solution.
3. Soaps are sodium salts of fatty acids and are basic in nature.
4. While washing the clothes, turmeric reacts with soap solution and turns red forming red spots on the clothes.
5. But the spots disappear after washing the clothes with plenty of water.

Question e.
Tobacco masheri should not be used for cleaning teeth.
Answer:

1. Masheri is the smokeless form of tobacco. It is tobacco, containing teeth cleaning powder
2. It contains tobacco leaves.
3. Tobacco contains toxic and Carcinogenic chemicals which can cause cancer, oral cancer, mouth and throat cancer, gum disease, tooth decay.
4. Its use can cause abnormal delivery in pregnant women.

Project:

Question 1.
Visit a cement factory. See how cement is prepared and discuss the process.

Question 2.
Write a conversation based on cement houses, mud-houses and wattle-and-daub houses.

Class 7 Science Chapter 15 Materials We Use Important Questions and Answers

Fill in the blanks:

Question 1.
The principal ingredients of a toothpaste are ………………. and ………………. which remove the dirt on the teeth.
Answer:
Calcium carbonate, calcium hydrogen phosphate

Question 2.
………………. in the toothpaste helps prevent tooth decay.
Answer:
Fluoride

Question 3.
The property of a substance of spreading on a surface is called ………………. .
Answer:
surface activity

Question 4.
………………. and ………………. are the natural detergent in common use.
Answer:
Soap nut, Soap pod

Question 5.
Soap nut and soap pod contain a chemical named ………………. .
Answer:
Saponin

Question 6.
………………. is a man-made detergent.
Answer:
Soap

Question 7.
………………. is used for washing clothes.
Answer:
Hard soap

Question 8.
………………. is used for bathing.
Answer:
Soft soap

Question 9.
Hard soap is ………………. salt of fatty acids.
Answer:
Sodium

Question 10.
Soft soap is ………………. salt of fatty acids.
Answer:
Potassium

Name the following:

Question 1.
A substance which is spread on a given surface and used for cleaning.
Answer:
Surfactant

Question 2.
A chemical contained in soap nut and pod.
Answer:
Saponin

Question 3.
An element which helps prevent tooth decay which is used in toothpastes.
Answer:
Fluoride

Question 4.
The latin word which means detergent.
Answer:
Detergere

Question 5.
A cement used for construction purpose.
Answer:
Portland cement

Question 6.
A mixture of cement, water, sand and gravel.
Answer:
Concrete

Question 7.
A natural detergent.
Answer:
Soap nut

Choose the correct option:

Question 1.
Which one of these material grows on an animal
(cotton, rubber, wood, wool)
Answer:
Wool

Question 2.
Which of these is man-made?
(oranges, apples, plastic bags, tomatoes)
Answer:
Plastic bags

Question 3.
The statue is made from marble, marble is a ………….. material.
(weak, elastic, man-made, natural)
Answer:
Natural

Question 4.
Toy Duck is made from plastic. Plastic is a ………….. material.
(natural, man-made, raw, precious)
Answer:
Man-made

Question 5.
Which of these materials is natural?
(Nylon, Rock, Plastic, Polythene)
Answer:
Rock

Question 6.
Which one of these is a natural material?
(Lemonade, Cola, Water, Ice cream)
Answer:
Water

Question 7.
The toy plane made from wood. Wood is a ………….. material.
(man-made, natural, weak)
Answer:
Natural

Question 8.
Which one of these materials is natural?
(Nylon, cotton, polythene, polyesters)
Answer:
Cotton

Match the following:

Question 1.

Answer:

Find out:

Answer the following questions:

Question 1.
What is the source of the fluoride in a tooth paste or tooth powder?
Answer:
A naturally occurring mineral found in tooth paste and drinking water. Sodium Fluoride (NaF) is the source of fluoride in toothpaste.

Question 2.
Note down all the information given on a tooth powder/toothpaste container or carton and discuss.
Answer:
1. The carton shows the name of the company and name of toothpaste and its contains all the ingredients present in toothpaste.

2. Licence No. of company, Regd. Trade Mark of Colgate Palmolive Co-manufactured by Colgate-Palmolive (India) Ltd. Licenced user of Trade Mark made in India. Tooth paste contains 1000 ppm max of available fluoride when packed.

3. Ingredients: Calcium carbonate, sorbitol, sodium lauryl sulphate, silica, sodium silicate flavour, sodium monoflouro phosphate, sodium bicarbonate, benzyl alcohol.

4. Direction for use: Brush thoroughly atleast twice a day.
5. Children under 6 years of age should have adult supervision and use only appropriate amount.

Question 3.
Now a days why are the roads made of concrete?
Answer:

1. Concrete is prepared by mixing cement, limestone, sand, gravel, and water. It is solid, more durable and strong.
2. There is no erosion for many years and roads are smoother. So the roads are made of concrete.

Question 4.
What causes the hardness of water?
Answer:

1. Hardness is a measure of amount of dissolved salts in water.
2. It is caused by dissolved salts like carbonates, chlorides mostly of calcium and magnesium
3. Presence of these makes washing of clothes by soap difficult.
4. Hard water is water that has high mineral content.
5. Hard water is formed when water percolates through deposits of limestone and chalk.

Can you tell?

Answer the following questions:

Question 1.
Which substances were used earlier for cleaning teeth?
Answer:
In olden times neem twigs, coal powder ash, tooth powder, salt, pomegranate rind, were used for cleaning teeth.

Question 2.
What do we use today to clean our teeth?
Answer:
Now a days we use variety of toothpastes and tooth powders to clean our teeth.

Question 3.
What do we use for cleansing our body?
Answer:
Soap, many liquid body wash are also used to clean our body.

Question 4.
What are the materials used for construction?
Answer:
Metals, wood, stone, cement, concrete, timber, bricks, metal sheet, soil, Earth, marble, aluminium, iron, steel, bamboo, glass, plastic.
Concrete: is a wet mixture of sand, gravel, cement, and water used to create building foundations, footpaths or roads.

Question 5.
Which of the houses seen in the pictures here have a strong structure? Why?

Answer:
The houses made from stone, bricks, and cement and concrete have a strong structure.

Write short notes on or Explain:

Question 1.
Natural detergent
Answer:

1. It is naturally available soap nut (ritha) and soap pod (shikekai) are the natural detergents in common use.
2. They contain a chemical named saponin.
3. Soap nut and soap pod do not have any harmful effect on human skin or on silk, woollen threads, and cloth.

Question 2.
Man-made detergents
Answer:

1. Detergent which is made by processing naturally available material, soap is a man-made detergent which has been in use since ancient times.
2. In those days soap was prepared by using animal fat and wood ash.
3. Therefore two types of soap (a) Hard Soap is used for washing clothes. It is a sodium salt of fatty acids, (b) Softsoap is used for bathing. It is a potassium salt of fatty acids.
4. It does not cause irritation of the skin.
5. But we can’t use soap in hard water because soap does not give lather but forms a scum. So, soap loses its cleansing property so now synthetic detergent have taken the place of soap.
6. Synthetic detergents are obtained by subjecting these raw materials (fats and kerosene) to a variety of chemical processes. These can be used in hard water as well.

Question 3.
Soap
Answer:
Two types of soap are:
1. Hard soap is used for washing clothes. It is a sodium salt of fatty acids.
2. (a) Softsoap is used for bathing. It is a potassium salt of fatty acids, (b) It does not cause irritation of the skin, (c) But we can’t use soap in hard water because soap does not give lather but forms a scum, (d) So, soap loses its cleansing property so now synthetic detergents have taken place of soap, (e) Synthetic detergents are obtained by subjecting these raw materials (fats and kerosene) to a variety of chemical processes, (f) These can be used in hard water as well.

Question 4.
Concrete
Answer:

1. Concrete is prepared by mixing cement, water, sand and gravel.
2. For making a strong and leak proof slab certain substances are mixed in concrete.
3. Now a days, roads are made of concrete because they are very durable, strong and smooth.

Question 5.
Surface activity
Answer:

1. Soap molecule has the property of holding on to both oil and water soap water spreads easily on many types of surfaces.
2. The property of a substance of spreading on a surface is called surface activity and the substance is said to be a surfactant.
3. Detergents are surface-active. One effect of surface activity is lather formation.

Question 6.
Explain the method of preparation of soap.
Answer:
Material required for preparation soap are 15 g sodium hydroxide, 60 ml coconut oil, 15 g salt, perfume, a glass rod, beaker, tripod, wire gauze, burner, water mould etc.
Procedure:

1. Take 60 ml of coconut oil in a beaker.
2. Dissolve 15 g sodium hydroxide in 50 ml water. Mix the sodium hydroxide solution in the oil slowly, while stirring with a glass rod.
3. Heat the mixture, and boil it for 10 – 12 minutes, stirring it all the while.
4. Take care that the mixture does not boil over while heating.
5. Dissolve 15 g salt in 200 ml water, pour this solution into the above mixture and stir.
6. The soap formed by the chemical reaction now floats on the water. After some time, it becomes thick.
7. Now separate the thick soap and add the perfume to it, shape the bar of soap using the mould.
8. In this process, fat and alkali combine to form salts of fatty acids.
9. Chemically, soap is a sodium or potassium salt 4 of fatty acids.

Maharashtra State Board Class 7 Science Textbook Solutions

• Cell Structure and Micro-organisms Class 7 Science Textbook Solutions
• The Muscular System and Digestive System in Human Beings Class 7 Science Textbook Solutions
• Changes – Physical and Chemical Class 7 Science Textbook Solutions
• Elements, Compounds and Mixtures Class 7 Science Textbook Solutions
• Materials We Use Class 7 Science Textbook Solutions
• Natural Resources Class 7 Science Textbook Solutions
• Effects of Light Class 7 Science Textbook Solutions
• Sound: Production of Sound Class 7 Science Textbook Solutions
• Properties of a Magnetic Field Class 7 Science Textbook Solutions
• In the World of Stars Class 7 Science Textbook Solutions

Std 10 Science Part 2 Chapter 8 Cell Biology and Biotechnology Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 8 Cell Biology and Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 2 Chapter 8 Cell Biology and Biotechnology Question Answer Maharashtra Board

Question 1.
Fill in the blanks and complete the statements.
a. Methods like artificial insemination and embryo transplant are mainly used for ………..
(a) animal husbandry
(b) wild life
(c) pet animals
(d) for infertile women
Answer:
(a) animal husbandry

b. ……….. is the revolutionary event in biotechnology after cloning.
(a) Human genome project
(b) DNA discovery
(c) Stem cell research
(d) All the above
Answer:
(c) Stem cell research

c. The disease related with the synthesis of insulin is …………..
(a) cancer
(b) arthritis
(c) cardiac problems
(d) diabetes
Answer:
(d) diabetes

d. Government of India has encouraged the ……….. for improving the productivity by launching NKM-16.
(a) aquaculture
(b) poultry
(c) piggery
(d) apiculture
Answer:
(a) aquaculture

Question 2.
Match the pairs.

[Note: In examination match the column question will have 2 components in Column A’ with 4 alternatives in Column B’.]
Answer:
(1) Interferon – Viral infection
(2) Factor VIII – Haemophilia
(3) Somatostatin – Dwarfness
(4) Interleukin – Cancer
[Note: Factor VIII* is an important protein factor and it should not be just factor as given in the textbook.]

Question 3.
Rewrite the following wrong statements after corrections:
a. Changes in genes of the cells are brought about in non-genetic technique.
Answer:
Non-genetic biotechnology involves use of either cell or tissue.

b. Gene from Bacillus thuringiensis is introduced into soyabean.
Answer:
Gene from Bacillus thuringiensis is introduced with gene of cotton.

Question 4.
Write short notes.
a. Biotechnology: Professional uses. (Commercial uses)
Answer:
(1) Biotechnology can be used in the following professional fields, viz. crop biotechnology, animal husbandry, human health, etc.

(2) In crop biotechnology, improvement in the yield and variety of agricultural field is done. The hybrid seeds, genetically modified crops, herbicide tolerant plants are some of the areas in which lot of biotechnological research is being done. By such research, high yielding and disease resistant varieties and varieties which can tolerate stresses such as alkalinity, weeds, cold and drought etc. are produced. BT cotton, BT Brinjal and golden rice are some GMO plants which have become popular in India.

Due to herbicide tolerant plants, the weeds are now selectively destroyed. By using biofertilizers, the use of chemical fertilizers is reduced. Use of bacteria such as Rhizobium, Azotobacter, Nostoc, Anaixiena and plants like Azolla the nitrogen fixation and phosphate solubilization abilities of the plants are improved.

(3) Animal husbandry is now using the methods of artificial insemination and embryo transfer by which the breeds of cattle are improved.

(4) To improve and to manage the human health, diagnosis ahd treatment of diseases have to be focussed. Diagnosis of diabetes, heart diseases and infectious diseases such as AIDS and dengue can be done rapidly due to biotechnology.

(5) The treatment and prevention of diseases need hormones, interferons, antibiotics and different vaccine which are now manufactured through biotechnology. Gene therapy is also used to treat hereditary disorders.

(6) Industrial products and clean technology to combat environmental pollution uses biotechnology practices.

(7) DNA fingerprinting has revolutionized the profession of forensic science.

b. Importance of medicinal plants.
Answer:

• In Ayurveda practices, the natural remedies were used. Since India had great biodiversity and traditional knowledge of herbal medicinal uses, therefore, people depended on such medicinal plants.
• In olden days, such herbs were collected by roaming in the jungles.
• Such important medicinal herbs are now cultivated with care.
• In entire world people have understood the importance of holy basil (tulsi), Adulsa, Jyesthmadh, etc.
• In some of the allopathy medicines too, the plant extracts are used.
• Medicines made from harmful chemicals have side effects and are not safe to be used unless there is medical supervision. Therefore, world-wide herbal remedies are gaining more popularity.

Question 5.
Answer the following questions in your own words.
a. Which products produced through biotechnology do you use in your daily life?
Answer:

• The simplest use of biotechnology that we practice at home is making curd and buttermilk.
• The primary type of biotechnology is used in the process of fermentation while making food stuffs, like bread, idli-dosa, dhokla, etc.
• Nowadays, different types of cheese, paneer, yoghurt, energy drinks, etc. are produced with the help of biotechnology. We are consuming these in our daily life.
• Seedless grapes, papaya, and watermelons are available in the market these days.
• Violet cabbage, yellow capsicum and exotic vegetables used for salad are also biotechnology products.
• The vaccines, antibiotics and the injections of human insulin are in regular use in many house-holds.

b. Which precautions will you take during spraying of pesticides?
Answer:

• Pesticides are toxic chemicals. By using them indiscriminately, they contaminate the water, soil and also crops.
• The D.D.T., chloropyriphos and malathion are very dangerous. They spread through the food chain causing biomagnification.
• Therefore, we shall not use such insecticides and pesticides. We shall use organic pesticides. Excessive use will be avoided.
• At the time of spraying, nose, eyes and skin will be covered and protected.
• Care will be taken not to allow children or domestic animals to come in, contact with a pesticide.

c. Why some of the organs in human body are most valuable?
Answer:

• The body can be in best health,if all the vital organs of the body are also in the best condition.
• Brain, kidney, heart, liver, etc. are some such vital organs which are most essential for proper metabolism and functioning of the body. The sense organs of the body are also of utmost importance, especially eyes.
• One cannot survive if any of these vital organs are not functioning properly. Some of the organs like brain will never regenerate too.
• Some of the organs can be brought back to functionality by performing surgeries. However, any problem with these vital organs make life miserable, therefore, they are said to be valuable.

d. Explain the importance of fruit processing in human life?
Answer:
(1) Fruits are perishable food stuff. They are spoilt soon if not consumed immediately. Hence for storage and usage for a long term, their preservation is absolutely essential.
(2) For year-long use of the fruits they are dried, salted, packed in air tight containers, used for preparing jams and jellies or condensed into pulps or syrups. Beverages, pickles, sauce, and various other products made from the fruits are largely used by us.
(3) The preserved products also fetch financial benefits.
(4) In national and international markets, Indian fruits like mangoes are in great demand. We can get foreign currency through exports of fruits and fruit products. The local horticulturists get good benefit from their orchards.
(5) Processed fruit products also give vitamins and minerals that help in maintaining good health. Thus fruit processing is important for human life.

e. Explain the meaning of vaccination.
Answer:

• Vaccination is the administering of vaccine. Vaccine is the ‘antigen’, given to a person or even to animals for acquiring immunity against particular pathogens or diseases.
• In olden days, vaccipes were prepared with the help of completely or partially killed pathogens. But this method causes some inconvenience. Some persons were allergic to such raw vaccines or they contracted the same disease through such vaccines.
• Hence in recent times the vaccines are produced by using biotechnology. These vaccines are artificial which are synthesised in the laboratories.
• The antigen is produced with the help of gene of the pathogen. Such vaccine becomes safe for administering.
• These antigenic proteins are injected to people to make their immune systems strong. This process of vaccination is absolutely safe. The vaccines are more thermostable and active for a long period of time.

Question 6.
Complete the following chart.

Answer:

Question 7.
Write the correct answer in blank boxes.

Answer:

Question 8.
Identify and complete the following correlations:
a. Insulin : Diabetes : : Interleukin : …………….
Answer:
Insulin : Diabetes : : Interleukin : Cancer

b. Interferon : ………. : : Erythropoietin : Anaemia.
Answer:
Interferon : Viral infection : : Erythropoietin : Anaemia.

c. ……….. : Dwarfness : : Factor VIII : Haemophilia.
Answer:
Somatostatin : Dwarfness : : Factor VIII : Haemophilia.

d. White revolution : Dairy : : Blue revolution : ………
Answer:
White revolution : Dairy : : Blue revolution : Fishery

Question 9.
Write a comparative note on usefulness and harmfulness of biotechnology.
(OR)
“Biotechnology is not only beneficial but it has some harmful effects too”. Express your opinion about this statement.
Answer:
(1) Biotechnology has proved to be useful in the field of agriculture, medicine, clean technology and industrial products.
(2) Due to various biotechnological experiments, the food production is increased substantially. The milk and milk products are now freely available. People no longer die of hunger due to abundant food supply.
(3) The sophisticated vaccines have stopped the spread of epidemics.
(4) The diseases like diabetes can be controlled due to human insulin injections that can be manufactured by biotechnology.
(5) The problems of pollution control, solid waste management and fuels are partially tackled by biotechnological alternatives.
(6) Though all such positive aspects are there, the biotechnology also poses some problems. The genetic changes are breaking the principles of nature. By inserting human genes in bacteria or virus, the products that are needed only for humans are produced.
(7) Human cloning is also a debatable issue. It will cause social and ethical problems. The new generations formed by cloning will have mothers but no fathers. If man tries to manipulate the genomes of other living organisms, it will cause disturbances in the natural balance. The long ternT effects of all such genetic manipulations can be disastrous. Thus, according to some views, biotechnology can be dangerous too.

Projects: (Do it your self)

Project 1.
Visit the organic manuring projects nearby your place and collect more information.

Project 2.
What will you do to increase public awareness about organ donation in your area?

Project 3.
Collect information about ‘green corridor’. Make a news-collection about it.

Can you recall? (Text Book Page No. 88)

Question 1.
What is cell?
Answer:
The structural and functional unit of the body is called a cell.

Question 2.
What is tissue? What are the functions of tissue?
Answer:
Tissue is a group of cells that performs a similar and definite function. E.g. The muscular tissues in the body perform contraction and extensions thereby helping in locomotion. The conducting tissues of the plants like xylem and phloem transport the water and food respectively.

Question 3.
Which technique in relation to tissues have you studied in earlier classes?
Answer:
The technique of tissue culture and genetic engineering has been studied last year. Tissue culture is ‘Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’. Genetic engineering and its use has also been studied under, ‘Introduction to biotechnology’.

Question 4.
Which are the various processes in tissue culture?
Answer:
Various step-wise processes are done while performing the-tissue culture. These processes are primary treatment, reproduction/cell division/multiplication, shooting or rooting, primary hardening, secondary hardening, etc.

Observe: (Text Book Page No. 88)

Question 1.
Assign names in the figure given below. Explain the various stages those are kept blank:

Answer:

Tissue Culture: Tissue culture is the technique in which ‘ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is done. While performing experiments of tissue culture, a liquid, solid or gel-like ” medium prepared from agar, is used. Such medium supplies nutrients and energy necessary for tissue culture technique. Different processes are to be done while performing tissue culture, viz. primary treatment, reproduction or multiplication, shooting and rooting, primary hardening, secondary hardening, etc. From the source plant, required tissues are taken out and all the processes are carried in an aseptic medium in laboratory.

(Use your brain power. (Text Book Page No. 89)

Question 1.
Just like the grafting in plants, is the organ transplantation possible in humans?
Answer:
The grafting as done in case of plants, cannot be done in human beings. But the transplantation of certain organs can be done. Liver, kidney, heart, eyes, etc. can be transplanted. But for these transplantations the donor and the recipient should match with each other in respect of their bloodr groups, age, disease condition, etc. In future, the stem cell research can bring about certain changes in the field of transplantations.

(Text Book Page No. 94)

Question 1.
What will happen if the transgenic potatoes are cooked before consumption?
Answer:
Some types of transgenic potatotes that contain edible vaccine against Hepatitis can be cooked. The cooking does not destroy the antigen incorporated into these transgenic potatoes. But according to some scientists, transgenic potatoes with enterotoxin vaccine, if cooked shows denaturation of vaccine.

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

Question 1.
The property of stem cells is called ………….
(a) diversity
(b) equality
(c) differentiation
(d) pluripotency
Answer:
(d) pluripotency

Question 2.
Cell ……….. starts from 14^th day of conception.
(a) development
(b) specialization
(c) growth
(d) differentiation
Answer:
(d) differentiation

Question 3.
Availability of ………… is an important requirement in organ transplantation.
(a) doctor
(b) clinic
(c) donor
(d) ambulance
Answer:
(c) donor

Question 4.
The toxin which is lethal for ……….. was produced in leaves and bolls of BT cotton.
(a) bollworm
(b) locust
(c) birds
(d) frogs
Answer:
(a) bollworm

Question 5.
Transgenic raw potatoes generate the immunity against ………… disease.
(a) plague
(b) cholera
(c) leprosy
(d) TB
Answer:
(b) cholera

Rewrite the following wrong statements after corrections:

Question 1.
High-class varieties of crops have been developed through the technique of transplantation.
Answer:
High-class varieties of crops have been developed through the technique of tissue-culture.

Question 2.
Earlier, insulin was being collected from, the pancreas of pigs.
Answer:
Earlier, insulin was being collected from the- pancreas of horses.

Question 3.
Malaria arises due to genetic changes in hepatocytes.
Answer:
Phenylketonuria (PKT) arises due to genetic changes in hepatocytes.

Question 4.
The E.coli bacteria are useful for cleaning the hydrocarbon and oil pollutants from soil and water.
Answer:
The Pseudomonas bacteria are useful for cleaning the hydrocarbon and oil pollutants from soil and water.

Question 5.
Various essential elements like N, P, K are removed and hence become unavailable to the crops due to earthworms and fungi.
Answer:
Various essential elements like N, P, K become available to crops due to earthworms and fungi.

Question 6.
We do not have any tradition that cures the diseases with the help of natural resources.
Answer:
We have a great tradition of ayurveda that cures the diseases with the help of natural resources.

Match the pairs:

Question 1.

Answer:
(1) Dr. Anand Mohan Chakravarti – Cleaning the oil spill
(2) Dr. M.S. Swaminathan – Green revolution in India

Question 2.

Answer:
(1) Pseudomonas – Hydrocarbons
(2) Pteris vitata – Arsenic

Find the odd man out:

Question 1.
Green revolution, Industrial revolution, White revolution, Blue revolution
Answer:
Industrial revolution. (All others are concerned with food.)

Question 2.
DDT, malathion, chloropyriphos, Humus
Answer:
Humus. (All others are insecticides.)

Question 3.
Sodium, Aluminium, Potassium, Phosphorus
Answer:
Aluminium. (All others are essential elements for plant growth.)

Question 4.
Diabetes, Anaemia, Leukaemia, Thalassemia
Answer:
Diabetes. (All other diseases involve reduction in the number of blood cells.)

Question 5.
Drying, Salting, Cooking, Soaking with sugar
Answer:
Cooking. (All others are food preservative methods.)

Identify and complete the following correlations:

Question 1.
White revolution : Increase in dairy production : : Green revolution : ………. (March 2019)
Answer:
White revolution : Increase in dairy production : : Green revolution : Increase in agricultural production or crop yield

Question 2.
Nostoc, Anabaena : Biofertilizers : : Alfalfa : ………..
Answer:
Nostoc, Anabaena : Biofertilizers : : Alfalfa : Phytoremediation.

Give definition/Give meanings:

Question 1.
Stem cell or what are stem cells?
Answer:
The special cells having pluripotency and ability to divide and differentiate into new cells are called stem cells. They are present in multicellular living beings.

Question 2.
Biotechnology.
Answer:
Technology that brings about artificial genetic changes and hybridization in organisms for human welfare is called biotechnology.

Question 3.
Genetically modified crops.
Answer:
Crops having desired characters are developed by integrating foreign gene with their genome, such crops have modified genome and are known as genetically modified crops.

Question 4.
Golden rice.
Answer:
Biotechnologically developed variety of rice in which gene synthesizing the vitamin A (Beta carotene) has been incorporated and which contains 23 times more amount of beta carotene than that of the normal variety is called golden rice. It was developed in 2005.

Question 5.
Vaccine.
Answer:
The ‘antigen’ containing material given to a person or animal to acquire either permanent or temporary immunity against a specific pathogen or disease is called a vaccine.

Question 6.
Cloning.
Answer:
Production of replica of any cell or organ or entire organism through biotechnological process is called cloning.

Question 7.
DNA fingerprint.
Answer:
The nucleotide sequence present on the DNA of each person is unique just like the fingerprint, thus for establishing the identity of any person DNA can be analysed, this technique is known as DNA fingerprinting.

Question 8.
Green revolution.
Answer:
All the methods applied for harvesting maximum yield from minimum land are collectively referred to as green revolution.

Question 9.
White revolution.
Answer:
Achieving the self-sufficiency in dairy business, by performing various experiments for quality control, bringing about newer dairy products and their preservation and thus raising economic standards is called white revolution.

Question 10.
Blue revolution.
Answer:
The aquaculture practices to increase the yield of edible aquatic organisms is called blue revolution.

Name the following:

Question 1.
Research institutes involved with cell science.
Answer:

• National Centre of Cell Science, Pune
• Instem, Bengaluru.

Question 2.
Sources of stem cells.
Answer:

• Umbilical cord
• Embryonic cells
• Redbone marrow
• Adipose connective tissue and blood of adult human being.

Question 3.
Types of Stem cells.
Answer:

• Embryonic stem cells
• Adult stem cells.

Question 4.
Organs that can be donated.
Answer:
Eyes, heart, pancreas, liver, kidneys, skin, J bones, lungs.

Question 5.
Organisms used as biofertilizers.
Answer:
Rhizobium, Azotobacter, Nostoc, Anabaena, Azolla.

Question 6.
Two main methods used in animal husbandry.
Answer:

1. Artificial insemination
2. Embryo transfer.

Question 7.
Two important aspects of human health management.
Answer:

1. Diagnosis
2. Treatment of diseases.

Question 8.
Place where DNA fingerprinting research is done in India.
Answer:
Centre of DNA fingerprinting and Diagnostics, Hyderabad.

Question 9.
One benefit of biotechnology to the agriculture.
Answer:
Expenses on the pesticides are reduced.

scientific reasons:

Question 1.
Nowadays, safer vaccines are being produced.
Answer:

• Before the advent of biotechnology, the vaccines were made from inactive or dead pathogens of that disease.
• But now the vaccine is made artificially using biotechnological processes.
• Such vaccines produced some disease symptoms in some cases.
• The antigen of the disease is researched upon and its genetic code is found out.
• A similar antigen is made in the laboratories which is used as a vaccine.
• Such vaccines are more thermostable and remain active for longer duration. Therefore, the vaccines are now safer.

Question 2.
Awareness about organ donation after death is increasing.
Answer:

• Due to accidents or illness, some of the vital organs may get damaged and may not work to fullest capacity.
• In such cases, if organ transplantation is done, it will be very helpful for that needy patient.
• The dead person’s organs can be used for organ transplantation and a life can be saved.
• Many government and social organizations are spreading awareness about such donations. Therefore, gradually the awareness about organ transplantation is increasing.

Answer the following questions:

Question 1.
Write two uses of biotechnology related to human health. (Board’s Model Activity Sheet)
Answer:

1. Biotechnology is used to manufacture vaccines for controlling diseases.
2. Different hormones such as insulin, somatotropin and somatostatin can be prepared in laboratories by using new biotechnological processes. The clotting factors are also manufactured through such techniques.

Question 2.
Answer the following questions:
(a) What is biotechnology?
(b) Explain any two commercial applications of it. (March 2019)
Answer:
(a) Biotechnology: Technology that brings about artificial genetic changes and hybridization in organisms for human welfare is called biotechnology.

(b)

• The treatment and prevention of diseases need hormones, interferons, antibiotics and different vaccine which are now manufactured through biotechnology. Gene therapy is also used to treat hereditary disorders.
• Industrial products and clean technology to combat environmental pollution uses biotechnology practices.
• DNA fingerprinting has revolutionized the profession of forensic science.

Question 3.
What is mainly included under biotechnology?
Answer:
Biotechnology includes the following main areas:

• Abilities of microbes are used in producing yoghurt from milk and making alcohol from molasses.
• Production of antibiotics and vaccines, etc. is carried out by with the help of specific cells using their productivity.
• Bio-molecules like DNA and proteins are used for human welfare.
• By performing gene manipulation, plants, animals and products of desired quality are produced. Genetically modified bacteria are used to produce human hormones such as Human Growth Hormone and insulin.
• Tissue culture is a non-genetic technique which is used for production of new cells or tissues. Hybrid seeds are also produced in a similar way.

Question 4.
What are edible vaccines?
Answer:

• Edible vaccines are those which are given as a food by incorporating them into the food-stuff.
• Such edible vaccines are produced through biotechnology.
• Transgenic potatoes are produced with the help of biotechnology which contain vaccine that act against bacteria like Vibrio cholera, Escherichiatoli.
• If raw potatoes are consumed, then the immunity is generated in the body of a person. However, eating only raw potatoes generates the immunity against cholera and the disease caused due to E. coli.

Question 5.
What is DNA fingerprinting? Explain it in brief. Where is this technique used? Give any two examples. (Board’s Model Activity Sheet)
Answer:

• As the fingerprints are unique for every individual, similarly the nucleotide sequence in the DNA molecule is also unique.
• By knowing this sequence, one can find out the identity of any person. Such technique to establish the identity of a person by taking into consideration the nucleotide sequence is called DNA fingerprinting.
• Its main use is in forensic sciences to confirm the identity of the criminal.
• Similarly, identity of parents in case of disputed parentage for any child can be understood by taking DNA fingerprints of both the parents and a child.

Write short notes on:

Question 1.
Uses of stem cells.
Answer:
Stem cells are used for following purposes:

• In regenerative therapy stem cells are used.
• In case of diseased conditions like diabetes, myocardial infarction, Alzheimer’s disease, Parkinson’s disease, etc., stem cells can be used to replace the damaged or functionless cells.
• In conditions such as anaemia, thalassaemia, leukaemia, etc. there is always the need of newer blood cells. Here, stem cells can be used to restore the number of blood cells.
• In techniques of organ transplantation stem cells can be used and they can help in the transplantation of new organs such as kidney and liver The defective organs can be replaced by those that are produced with the help of stem cells and transplanted.

Question 2.
Cloning.
Answer:

• Cloning is the modern technique in which there is production of replica of any cell or organ or entire organism is done.
• There are two types of cloning, viz. (i) Reproductive cloning and (ii) Therapeutic cloning.
• Reproductive cloning: In reproductive cloning, a clone is produced by fusion of a nucleus of diploid somatic cell with the enucleated ovum of anybody. In the process, the sperm or male gamete is not needed.
• Therapeutic cloning: This technique is largely used for treatment purpose. Stem cells are derived from the cell formed in laboratory by the union of somatic cell nucleus with the enucleated egg cell.
• This technique is used for therapy of various diseases.
• Gene cloning can also be done to form millions of copies of same gene. Such genes are used for gene therapy and other purposes.
• Due to cloning technique, the inheritance of hereditary diseases can be controlled, continuation of generations can be achieved and certain characteristic genes can be enhanced.
• However, for human cloning, there is world-wide opposition due to ethical reasons.

Question 3.
Dolly.
Answer:

• Dolly was the first mammalian cloned sheep.
• Dolly was born on 5th July 1996 in Scotland by the process of cloning.
• The Finn Dorset sheep was chosen and her diploid nucleus from the udder cell was introduced into the ovum whose haploid nucleus was removed. This enucleated ovum was of Scottish sheep.
• The egg was then introduced into uterus of another Scottish sheep and it grew into Dolly.
• Dolly resembled exactly like Finn Dorset sheep whose diploid nucleus was used. None of the characters of Scottish sheep were seen in Dolly.
• In this way, Dolly had three mothers but no father.
• Dolly gave birth to many young ones. She died on 14th February 2003 due to cancer of the lungs.

Question 4.
Green revolution.
Answer:

• In agriculture, different methods used to harvest maximum yield from minimum land, these methods are collectively called green revolution.
• Dr. M.S. Swaminathan is called father of Green Revolution in India while Dr. Norman Borlaug has done the similar efforts in the U.S.
• Before the Green Revolution in India, there was always the dearth of the food grains. The overflowing Indian population was badly affected due to poor quality and quantity of food.
• But due to the Green Revolution in India, attention was focussed on the agricultural research.
• Improvised dwarf varieties of wheat and rice, proper use of fertilizers and pesticides and water management were the proper methods that increased production of food grains.
• This created abundance of the grains for Indian population.

Question 5.
White revolution.
Answer:

• Few years back, there was scarcity of milk in various parts of India. At some places, milk and milk products were abundant but they did not reach all the consumers.
• Dr. Verghese Kurien ^ho was then the founder director of Anand Milk Union Limited (AMUL) started thecooperative movement in the direction to produce “operation flood”, i.e. abundance of milk everywhere.
• The use of biotechnology was also done to increase the milk production.
• Dr. Kurien’s efforts have reached all-time high status as India is now self-sufficient in dairy business.
• This is popularly known as White Revolution. Different experiments were performed for quality control, newer dairy products were thought off and preservation methods were improved.
• This created White Revolution. AMUL from Anand has now reached international standards.

Question 6.
Blue Revolution.
Answer:

• Utilization of aquaculture practices for obtaining edible and commercial aquatic organisms is called blue revolution.
• In East Asian countries where water bodies and fish population is abundant, the aquaculture was started.
• On similar lines, in India, the aquaculture of different fresh water and marine organisms is being done with the help of fishery scientists.
• Government of India has vowed to increase the aquaculture production by encouraging the people for aquaculture by launching the program ‘Nil- Kranti Mission-2016’ (NKM-16).
• Pisciculutre is culturing of fish, mariculture is culture of marine organisms such as prawns/shrimps and lobsters. Sea weeds, oysters, clams are also cultured.
• For carrying out aquaculture, 50% to 100% subsidies are offered by the Government.
• Fresh water fishes like rohu, catla and other edible varieties like shrimp and lobsters are being cultured on a large scale which can bring about Blue Revolution.

Complete the paragraph by choosing the appropriate words given in the bracket:

Question 1.
(degenerated, red bone marrow, adipose connective tissue, blastocyst, umbilical cord, Differentiation)
………… of stem cells form can form various tissues, in the body. Stem cells are present in the ………….. by which the foetus is joined to the uterus of the mother. Stem cells are also present in the ……….. stage of embryonic development. Stem cells are present in ……….. and ………… of adult human beings. It has become possible to produce different types of tissues and the ……… part of any organ with the help of these stem cells.
Answer:
Differentiation of stem cells form can form various tissues in the body. Stem cells are present in the umbilical cord by which the foetus is joined to the uterus of the mother. Stem cells are also present in the blastocyst stage of embryonic development. Stem cells are present in red bone marrow and adipose connective tissue of adult human beings. It has become possible to produce different types of tissues and the degenerated part of any organ with the help of these stem cells.

Paragraph-based questions :

1. Green corridor refers to a special road route that enables harvested organs meant for transplants to reach the destined hospital. A 45-year-old woman, a victim of a railway accident, was declared brain dead, her husband and children agreed to donate her kidneys, liver and heart. One of her kidneys was transplanted to a patient in MGM Hospital and the second kidney helped a patient in Jaslok hospital. Her liver helped the transplant of a patient in Wockhardt Hospital. And her heart was sent to Fortis to the patient on a super urgent priority list, transported via a green corridor covering 18km in less than 16 minutes. This was possible due to Green corridor.
Questions and Answers :

Question 1.
What is Green corridor?
Answer:
Green corridor is a special road route that enables harvested organs meant for transplants to reach the destined hospital

Question 2.
Which organs of brain-dead lady were transplanted?
Answer:
Two kidneys, liver and heart of the brain- dead lady were transplanted.

Question 3.
How many lives were saved from organs of one lady?
Answer:
Four patients lives were saved due to organ donation of one lady.

Question 4.
How was distance of 18km covered in 16 minutes? Why?
Answer:
The distance was covered because the concept of Green corridor was applied. The heart was sent from one hospital to another, where the recipient was kept ready. The quick transportation is necessary to keep heart in living condition.

Question 5.
Who takes the decision to donate the organs?
Answer:
The close relatives of deceased person take the decision to donate the organs.

2. Read the following extract and answer the questions that follow: (March 2019)
A liberal view behind the concept of organ and body donation is that after death our body should be useful to other needful persons so that their miserable life would become comfortable. Awareness about these concepts is. increasing in our country and people are voluntarily donating their bodies.

Life of many people can be saved by organ and body donation. Blinds can regain their vision. Life of many people can be rendered comfortable by donation of organs like liver, kidneys, heart, heart valves, skin, etc. Similarly, body can be made available for research in medical studies. Many government and social organizations are working towards increasing the awareness about body donation.
Questions and Answers :

Question 1.
What is the liberal view behind the organ and body donation?
Answer:
By body donation, research in medical studies is possible. The needy persons can get vital organs which can save their lives.

Question 2.
Name any four organs that can be donated.
Answer:
Liver, Kidneys, heart, eyes, skin, etc. can be donated.

Complete the following table:

Question 1.

Answer:

Diagram/chart based questions:

Question 1.

(A) Which process is shown in the above figure? *
Answer:
The figure shows process to make transgenic

(B) Describe in brief the steps I, II, III and IV.
Answer:

Question 2.
Draw well labelled diagram of Stem cell therapy.
Answer:

Question 3.
Label the following diagram :
(i) Stem cells and organ transplantation,
Answer:

(ii) Organs that can be donated:
Answer:

Question 4.
(i) Which therapy is shown in the Fig. 8.5?

(ii) Which will be possible benefits of this therapy in organ transplantation ?
Answer:
(i) The figure 8.5 shows the ‘regenerative therapy’ using stem cells. Also called stem cell therapy.
(ii) With the help of above therapy organs like liver, kidney from stem cells can be redeveloped to replace the failed ones.

Activity based questions:

Question 1.
Bring a packet of ‘Balghuti’ from ayurveda shop. Learn the information about each component in it. Collect information about various other medicines and prepare the chart as shown below. (Try this: Textbook page no. 99)
Answer:

Question 2.
Give five examples of each of the fruiting and flowering plants developed through tissue culture and mention their benefits. (Make a list and discuss: Textbook page no. 93)
Answer:
I. Fruiting trees: Banana, Chikoo (Sapota), Tomato, Fig, Pineapple.

II. Flowering trees: Orchids, Roses, Chrysanthemum, Gerbera, Begonia, Carnation, Lili. Benefits of such plants may be varied. Mostly fruits developed are made seedless and tastier.

III. Benefits of plants produced through technique of tissue culture:

• Techniques of tissue culture can produce more copies of same plant with better characters. ’ The plant grower likes to have bigger and more fruits from fruit trees. On the flowering trees, colourful flowers with good fragrance are favoured.
• Plants which do not depend on particular climate and local seasonal changes are produced by tissue culture methods. This helps to rise the yield in an area which otherwise may not produce a specific crop.
• For tissue culture, saplings and seedlings are made available throughout the year through laboratory. The limitations of getting natural seeds are not there thus planting can be done throughout the year.
• Tissue culture techniques create the plants of uniform size, shape and yield. Since they are exactly alike, it becomes beneficial.
• In lesser time period, the crops reach maturity.
• The crops are pest and disease resistant.
• Tissue culture techniques are cost effective and easy to carry out.

Question 3.
Which new species of the rice have been developed in India? (Collect Information: Textbook page no. 97)
Answer:

1. Species in 2015-16: High zinc species (DRR Dhan 45), Pusa 1592, Punjab basmati 3, Pusa 1609, Telangana Sona.
2. Species in 2014: CR Dhan 205, CR Dhan 306, CRR, 451.

Question 4.
Discuss about stem cells and organ transplantation in the class with the help of figures given on textbook page no. 90. (Observe: Textbook page no. 90)
Answer:


Organ transplantation:
Various organs in the human body either become less efficient or completely functionless due to various reasons like aging, accidents, infections, disorders, etc. Life of such person becomes difficult or even fatality may occur under such conditions. However, if a person gets the necessary organ under such conditions, its life can be saved.

Availability of donor is an important requirement in organ transplantation. Each person has a pair of kidneys. As the process of excretion can occur with the help of single kidney, person can donate another one. Similarly, skin from certain parts of the body can also be donated.

Various factors like blood group, diseases, disorders, age, etc. of the donor and recipient need to be paid attention during transplantation.

However, other organs cannot be donated during life time. Organs like liver, heart, eyes can be donated after death only. This has lead to the emergence of concepts like posthumous (after death) donation of body and organs.

Organ and Body Donation: human bodies are disposed off after death as per traditional customs. However due to progress in science, it has been realized that many organs remain functional for certain period even after death occurs under specific conditions. Concepts like organ donation and body donation have emerged recently after realization that such organs can be used to save the life of other needful persons. A liberal view behind the concept of organ and body donation is that after death, our body should be useful to other needful persons so that their miserable life would become comfortable. Awareness about these concepts is increasing in our country and people are voluntarily donating their bodies.

Life of many people can be saved by organ and body donation. Blinds can regain the vision. Life of many people can be rendered comfortable by donation of organs like liver, kidneys, heart, heart valves, skin. etc. Similarly, body can be made available for research in medical studies. Many government and social organizations are working towards increasing the awareness about body donation.

Question 5.
Which fruits processing industries you observe in your surrounding? What is their effect? (Make a list and discuss: Textbook page no. 99)
Answer:

Fruit Processing:
we are daily using various products prepared from fruits. All are consuming the products like chocolates, juices, jams and jellies. All these products can be produced by processing on fruits. Fruits are perishable agro-produce. It needs the processing in such a way that it can be used throughout the year. Fruit processing includes various methods ranging from storage in cold storage to drying, salting, air tight pucking, preparing murabba, evaporating, etc.

Projects: (Do it your self)

Project 1.
Collect information about various hybrid varieties of animals. What are their benefits? Make a presentation of various pictures and videos. (Use of ICT: Textbook page no. 93)

Project 2.
Visit the websites: http://www.who.int/transplantation/organ/en/ and www.organindia.org / approaching-the- transplant/and collect more information about ‘brain dead’, organ donation and body donation (Internet is my friend: Textbook page no. 90)

Project 3.
Collect more information about the Human Genome Project, one of the important projects in the world.
(Internet is my friend: Textbook page no. 95)

Project 4.
Collect the information and make the chart about the work of various state and national-level institutes related with biotechnology. (Internet is my friend: Textbook page no. 97)

Maharashtra State Board Class 10 Science Solutions 

• Heredity and Evolution Class 10 Science Solutions
• Life Processes in Living Organisms Part – 1 Class 10 Science Solutions
• Life Processes in Living Organisms Part – 2 Class 10 Science Solutions
• Environmental management Class 10 Science Solutions
• Towards Green Energy Class 10 Science Solutions
• Animal Classification Class 10 Science Solutions
• Introduction to Microbiology Class 10 Science Solutions
• Cell Biology and Biotechnology Class 10 Science Solutions
• Social health Class 10 Science Solutions
• Disaster Management Class 10 Science Solutions

Semiconductors Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 14 Semiconductors Textbook Exercise Questions and Answers.

Class 11 Physics Chapter 14 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 14 Exercise Solutions

1. Choose the correct option.

Question 1.
Electric conduction through a semiconductor is due to:
(A) Electrons
(B) holes
(C) none of these
(D) both electrons and holes
Answer:
(D) both electrons and holes

Question 2.
The energy levels of holes are:
(A) in the valence band
(B) in the conduction band
(C) in the band gap but close to valence band
(D) in the band gap but close to conduction band
Answer:
(C) in the band gap but close to valence band

Question 3.
Current through a reverse biased p-n junction, increases abruptly at:
(A) Breakdown voltage
(B) 0.0 V
(C) 0.3V
(D) 0.7V
Answer:
(A) Breakdown voltage

Question 4.
A reverse biased diode, is equivalent to:
(A) an off switch
(B) an on switch
(C) a low resistance
(D) none of the above
Answer:
(A) an off switch

Question 5.
The potential barrier in p-n diode is due to:
(A) depletion of positive charges near the junction
(B) accumulation of positive charges near the junction
(C) depletion of negative charges near the junction,
(D) accumulation of positive and negative charges near the junction
Answer:
(D) accumulation of positive and negative charges near the junction

2. Answer the following questions.

Question 1.
What is the importance of energy gap in a semiconductor?
Answer:

1. The gap between the bottom of the conduction band and the top of the valence band is called the energy gap or the band gap.
2. This band gap is present only in semiconductors and insulators.
3. Magnitude of the band gap plays a very important role in the electronic properties of a solid.
4. Band gap in semiconductors is of the order of 1 eV.
5. If electrons in valence band of a semiconductor are provided with energy more than band gap energy (in the form of thermal energy or electrical energy), then the electrons get excited and occupy energy levels in conduction band. These electrons can easily take part in conduction.

Question 2.
Which element would you use as an impurity to make germanium an n-type semiconductor?
Answer:
Germanium can be made an n-type semiconductor by doping it with pentavalent impurity, like phosphorus (P), arsenic (As) or antimony (Sb).

Question 3.
What causes a larger current through a p-n junction diode when forward biased?
Answer:
In case of forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction.

Question 4.
On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?
Answer:
For pure semiconductor, the number density of free electrons and number density of holes is equal. Thus, at a given temperature, the conductivity of pure semiconductor depends on the number density of charge carriers in the semiconductor.

Question 5.
Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?
Answer:

1. In a p-type semiconductor, holes are majority charge carriers.
2. When a p-type semiconductor is connected to terminals of a battery, holes, which are not actual charges, behave like a positive charge and get attracted towards the negative terminal of the battery.
3. During transportation of hole, there is an indirect movement of electrons.
4. The drift speed of these electrons is less than that in the n-type semiconductors. Mobility of the holes is also less than that of the electrons.
5. As, electrical conductivity depends on the mobility of charge carriers, the conductivity of a n-type semiconductor is greater than that of p-type semiconductor even when both of these have same level of doping.

3. Answer in detail.

Question 1.
Explain how solids are classified on the basis of band theory of solids.
Answer:
i. The solids can be classified into conductors, insulators and semiconductors depending on the distribution of electron energies in each atom.

ii. As an outcome of the small distances between atoms, the resulting interaction amongst electrons and the Pauli’s exclusion principle, energy bands are formed in the solids.

iii. In metals, conduction band and valence band overlap. However, in a semiconductor or an insulator, there is gap between the bottom of the conduction band and the top of the valence band. This is called the energy gap or the band gap.

iv. For metals, the valence band and the conduction band overlap and there is no band gap as shown in figure (b). Therefore, electrons can easily gain electrical energy when an external electric field is applied and are easily available for conduction.

v. In case of semiconductors, the band gap is fairly small, of the order of 1 eV or less as shown in figure (c). Hence, with application of external electric field, electrons get excited and occupy energy levels in conduction band. These can take part in conduction easily.

vi. Insulators, on the contrary, have a wide gap between valence band and conduction band of the order of 5 eV (for diamond) as shown in figure (d). Therefore, electrons find it very difficult to gain sufficient energy to occupy energy levels in conduction band.

vii. Thus, an energy band gap plays an important role in classifying solids into conductors, insulators and semiconductors based on band theory of solids.

Question 2.
Distinguish between intrinsic semiconductors and extrinsic semiconductors
Answer:

Question 3.
Explain the importance of the depletion region in a p-n junction diode.
Answer:
i. The region across the p-n junction where there are no charges is called the depletion layer or the depletion region.

ii. During diffusion of charge carriers across the junction, electrons migrate from the n-side to the p-side of the junction. At the same time, holes are transported from p-side to n-side of the junction.

iii. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

iv. The potential barrier thus developed, prevents continuous flow of charges across the junction. A state of electrostatic equilibrium is thus reached across the junction.

v. Free charge carriers cannot be present in a region where there is a potential barrier. This creates the depletion region.

vi. In absence of depletion region, all the majority charge carriers from n-region (i.e., electron) will get transferred to the p-region and will get combined with the holes present in that region. This will result in the decreased efficiency of p-n junction.

vii. Hence, formation of depletion layer across the junction is important to limit the number of majority carriers crossing the junction.

Question 4.
Explain the I-V characteristic of a forward biased junction diode.
Answer:

1. Figure given below shows the I-V characteristic of a forward biased diode.
2. When connected in forward bias mode, initially, the current through diode is very low and then there is a sudden rise in the current.
3. The point at which current rises sharply is shown as the ‘knee’ point on the I-V characteristic curve.
4. The corresponding voltage is called the knee voltage. It is about 0.7 V for silicon and 0.3 V for germanium.
5. A diode effectively becomes a short circuit above this knee point and can conduct a very large current.
6. To limit current flowing through the diode, resistors are used in series with the diode.
7. If the current through a diode exceeds the specified value, the diode can heat up due to the Joule’s heating and this may result in its physical damage.

Question 5.
Discuss the effect of external voltage on the width of depletion region of a p-n junction.
Answer:

1. A p-n junction can be connected to an external voltage supply in two possible ways.
2. A p-n junction is said to be connected in a forward bias when the p-region connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
3. In forward bias connection, the external voltage effectively opposes the built-in potential of the junction. The width of depletion region is thus reduced.
4. The second possibility of connecting p-n junction is in reverse biased electric circuit.
5. In reverse bias connection, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased

11th Physics Digest Chapter 14 Semiconductors Intext Questions and Answers

Internet my friend (Textbookpage no. 256)

i. https://www.electronics-tutorials.ws/diode
ii. https://www.hitachi-hightech.com
iii. https://nptel.ac.in/courses
iv. https://physics.info/semiconductors
v. http://hyperphysics.phy- astr.gsu.edu/hbase/Solids/semcn.html

[Students are expected to visit above mentioned links and collect more information regarding semiconductors.]

Maharashtra State Board Class 11 Physics Textbook Solutions 

• Units and Measurements Class 11 Physics Textbook Solutions
• Mathematical Methods Class 11 Physics Textbook Solutions
• Motion in a Plane Class 11 Physics Textbook Solutions
• Laws of Motion Class 11 Physics Textbook Solutions
• Gravitation Class 11 Physics Textbook Solutions
• Mechanical Properties of Solids Class 11 Physics Textbook Solutions
• Thermal Properties of Matter Class 11 Physics Textbook Solutions
• Sound Class 11 Physics Textbook Solutions
• Optics Class 11 Physics Textbook Solutions
• Electrostatics Class 11 Physics Textbook Solutions
• Electric Current Through Conductors Class 11 Physics Textbook Solutions
• Magnetism Class 11 Physics Textbook Solutions
• Electromagnetic Waves and Communication System Class 11 Physics Textbook Solutions
• Semiconductors Class 11 Physics Textbook Solutions

Electromagnetic Waves and Communication System Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 13 Electromagnetic Waves and Communication System Textbook Exercise Questions and Answers.

Class 11 Physics Chapter 13 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 13 Exercise Solutions

1. Choose the correct option.

Question 1.
The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to the greenhouse effect is
(A) Infra-red radiation
(B) X-ray
(C) Microwave
(D) Visible light
Answer:
(A) Infra-red radiation

Question 2.
Earth’s atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer:
(B) IR

Question 3.
How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(A) depends on the values of p and e
(B) increases
(C) decreases
(D) remains same
Answer:
(D) remains same

Question 4.
The direction of EM wave is given by
(A) \(\bar{E}\) × \(\bar{B}\)
(B) \(\bar{E}\).\(\bar{B}\)
(C) along \(\bar{E}\)
(D) along \(\bar{B}\)
Answer:
(A) \(\bar{E}\) × \(\bar{B}\)

Question 5.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) h½
(B) h
(C) h^3/2
(D) h²
Answer:
(A) h½

Question 6.
The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer:
(A) Microwave

Question 7.
If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer:
(A) 32000 m

2. Answer briefly.

Question 1.
State two characteristics of an EM wave.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

Question 2.
Why are microwaves used in radar?
Answer:
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Question 3.
What are EM waves?
Answer:
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4.
How are EM waves produced?
Answer:

1. According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
2. It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
3. EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
4. An electric charge at rest has an electric field in the region around it but has no magnetic field.
5. When the charge moves, it produces both electric and magnetic fields.
6. If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
7. But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
8. Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

Question 5.
Can we produce a pure electric or magnetic wave in space? Why?
Answer:
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6.
Does an ordinary electric lamp emit EM waves?
Answer:
Yes, ordinary electric lamp emits EM waves.

Question 7.
Why light waves travel in vacuum whereas sound wave cannot?
Answer:
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8.
What are ultraviolet rays? Give two uses.
Answer:
Production:

1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
2. Used in burglar alarms and security systems.
3. Used to distinguish real and fake gems.

Question 9.
What are radio waves? Give its two uses.
Answer:

1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Uses:

1. Radio waves are used for wireless communication purpose.
2. They are used for radio broadcasting and transmission of TV signals.
3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 10.
Name the most harmful radiation entering the Earth’s atmosphere from the outer space.
Answer:
Ultraviolet radiation.

Question 11.
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer:
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12.
Name the three basic units of any communication system.
Answer:
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13.
What is a carrier wave?
Answer:
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14.
Why high frequency carrier waves are used for transmission of audio signals?
Answer:
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15.
What is modulation?
Answer:
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16.
What is meant by amplitude modulation?
Answer:
When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17.
What is meant by noise?
Answer:

1. A random unwanted signal is called noise.
2. The source generating the noise may be located inside or outside the system.
3. Efforts should be made to minimize the noise level in a communication system.

Question 18.
What is meant by bandwidth?
Answer:
The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Question 19.
What is demodulation?
Answer:
The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20.
What type of modulation is required for television broadcast?
Answer:
Amplitude modulation is required for television broadcast.

Question 21.
How does the effective power radiated by an antenna vary with wavelength?
Answer:

1. To transmit a signal, an antenna or an aerial is needed.
2. Power radiated from a linear antenna of length l is, P ∝ (\(\frac {l}{λ}\))²
   where, λ is the wavelength of the signal.

Question 22.
Why should broadcasting programs use different frequencies?
Answer:
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23.
Explain the necessity of a carrier wave in communication.
Answer:

1. Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
2. The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
3. Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.

Question 24.
Why does amplitude modulation give noisy reception?
Answer:
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25.
Explain why is modulation needed.
Answer:
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1.
Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 10⁸ m/s.
Answer:
Given: λ = 250 m, c = 3 × 10⁸ m/s
To find: Frequency (v)
Formula: c = v⁸
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{250}\) = 1.2 × 10⁶ Hz
= 1.2 MHz

Question 2.
Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 10¹⁸ Hz.
Solution:
Given: c = 3 × 10⁸, v = 2 × 10¹⁸ Hz
To find: Wavelength (λ)
Formula: c = vλ
Calculation. From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{2×10^{18}}\) = 1.5 × 10^-10
= 0.15 nm

Question 3.
The speed of light is 3 × 10⁸ m/s. Calculate the frequency of red light of wavelength of 6.5 × 10^-7 m.
Answer:
Given: c = 3 × 10⁸ m/s, λ = 6.5 × 10^-7 m
To find: Frequency (v)
Formula: c = vλ
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{6.5×10^{-7}}\) = 4.6 × 10¹⁴ Hz

Question 4.
Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: v = 8 GHz = 8 × 10⁹ Hz,
c = 3 × 10⁸ m/s
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{8×10^9}\) = 3.75 × 10^-2
= 3.75 cm

Question 5.
In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 10¹⁰ What is the wavelength of the wave?
Answer:
Given: v = 2 × 10¹⁰ Hz, c = 3 × 10⁸ m
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{2×10^{10}}\) = 1.5 × 10^-2

Question 6.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?
Answer:
Given: B₀ = 5 × 10^-7 T, c = 3 × 10⁸
To find: Amplitude of electric field (E₀)
Formula: c = \(\frac {E_0}{B_0}\)
Calculation /From formula,
E₀ = c × B₀
= 3 × 10⁸ × 5 × 10^-7
= 150 V/m

Question 7.
A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 10⁶ m)
Answer:
Given: h = 200 m,
Population density (n)
= 1000/km² = 1000 × 10^-6/m² = 10^-3/m²
R = 6.4 ×10⁶ m
To find: Population covered
Formulae: i. A = πd² = π(\(\sqrt{2Rh}\))² = 2πRh
ii. Population covered = nA
Calculation /From formula (i),
A = 2πRh
= 2 × 3.142 × 6.4 × 10⁶ × 200
≈ 8 × 10⁹ m²
From formula (ii),
Population covered = nA
= 10^-3 × 8 × 10⁹
= 8 × 10⁶

Question 8.
Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: h = 600 m, R = 6.4 × 10⁶ m
To find: Range (d)
Height to get the double coverage (h’)
Formula: d = \(\sqrt{2hR}\)
Calculation: From formula,
d = \(\sqrt{2×600×6.4×10^6}\) = 87.6 × 10³ = 87.6 km
Now, for A’ = 2A
π(d’)² = 2 (πd²)
∴ (d’)² = 2d²
From formula,
h’ = \(\frac{(d’)^2}{2R}\)
= \(\frac{2d^2}{2R}\)
= 2 × h ……….. (∵ h = \(\frac{d^2}{2R}\))
= 2 × 600
=1200 m

Question 9.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 10⁶ m.
Answer:
Given: h_t = 32 m, h_r = 50 m, R = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
d_t = \(\sqrt{2Rh_t}\) = \(\sqrt{2×6.4×10^6×32}\)
= 20.238 × 10³ m
= 20.238 km
d_r = \(\sqrt{2Rh_t}\)
= \(\sqrt{2×6.4×10^6×50}\)
= 25.298 × 10³ m
= 25.298 km
Now, d = d_t + d_r
= 20.238 + 25.298
= 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1.
i. What is a wave?
Answer:
Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Question 2.
What are Lenz’s law, Ampere’s law and Faraday’s law?
Answer:
Lenz’s law:
Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]

Maharashtra State Board Class 11 Physics Textbook Solutions 

• Units and Measurements Class 11 Physics Textbook Solutions
• Mathematical Methods Class 11 Physics Textbook Solutions
• Motion in a Plane Class 11 Physics Textbook Solutions
• Laws of Motion Class 11 Physics Textbook Solutions
• Gravitation Class 11 Physics Textbook Solutions
• Mechanical Properties of Solids Class 11 Physics Textbook Solutions
• Thermal Properties of Matter Class 11 Physics Textbook Solutions
• Sound Class 11 Physics Textbook Solutions
• Optics Class 11 Physics Textbook Solutions
• Electrostatics Class 11 Physics Textbook Solutions
• Electric Current Through Conductors Class 11 Physics Textbook Solutions
• Magnetism Class 11 Physics Textbook Solutions
• Electromagnetic Waves and Communication System Class 11 Physics Textbook Solutions
• Semiconductors Class 11 Physics Textbook Solutions

Magnetism Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 12 Magnetism Textbook Exercise Questions and Answers.

Class 11 Physics Chapter 12 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 12 Exercise Solutions

1. Choose the correct option.

Question 1.
Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at r is always proportional to
(A) \(\frac {1}{r^2}\)
(B) \(\frac {1}{r^3}\)
(C) \(\frac {1}{r}\)
(D) Not necessarily \(\frac {1}{r^3}\) at all points
Answer:
(B) \(\frac {1}{r^3}\)

Question 2.
Magnetic meridian is the plane
(A) perpendicular to the magnetic axis of Earth
(B) perpendicular to geographic axis of Earth
(C) passing through the magnetic axis of Earth
(D) passing through the geographic axis
Answer:
(C) passing through the magnetic axis of Earth

Question 3.
The horizontal and vertical component of magnetic field of Earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be
(A) 30°
(B) 45°
(C) 0°
(D) 90°
Answer:
(B) 45°

Question 4.
Inside a bar magnet, the magnetic field lines
(A) are not present
(B) are parallel to the cross sectional area of the magnet
(C) are in the direction from N pole to S pole
(D) are in the direction from S pole to N pole
Answer:
(D) are in the direction from S pole to N pole

Question 5.
A place where the vertical components of Earth’s magnetic field is zero has the angle of dip equal to
(A) 0°
(B) 45°
(C) 60°
(D) 90°
Answer:
(A) 0°

Question 6.
A place where the horizontal component of Earth’s magnetic field is zero lies at
(A) geographic equator
(B) geomagnetic equator
(C) one of the geographic poles
(D) one of the geomagnetic poles
Answer:
(D) one of the geomagnetic poles

Question 7.
A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences
(A) a force but not a torque
(B) a torque but not a force
(C) both a force and a torque
(D) neither force nor a torque
Answer:
(C) both a force and a torque

2. Answer the following questions in brief.

Question 1.
What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
Answer:
i. When a magnet is cut into two pieces, then each piece behaves like an independent magnet.

ii. When a bar magnet is cut transverse to its length, the two pieces generated will behave as independent magnets of reduced magnetic length. However, the pole strength of all the four poles formed will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

iii. When the bar magnet is cut along its length, the two pieces generated will behave like an independent magnet with reduced pole strength. However, the magnetic length of both the new magnets will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

Question 2.
What could be the equation for Gauss’ law of magnetism, if a monopole of pole strength p is enclosed by a surface?
Answer:
i. According to Gauss’ law of electrostatics, the net electric flux through any Gaussian surface is proportional to net charge enclosed in it. The equation is given as,
ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\)

ii. Similarly, if a monopole of a magnet of pole strength p exists, the Gauss’ law of magnetism in S.I. units will be given as,
ø_E = ∫\(\vec{B}\) . \(\vec{dS}\) = µ₀P

3. Answer the following questions in detail.

Question 1.
Explain the Gauss’ law for magnetic fields.
Answer:
i. Analogous to the Gauss’ law for electric field, the Gauss’ law for magnetism states that, the net magnetic flux (ø_B) through a closed Gaussian surface is zero. ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0

ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.

vii. Thus, according to the Gauss’ law of electrostatics ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\), where q is the positive charge enclosed.

viii. The situation is entirely different from magnetic lines of force. Gauss’ law of magnetism can be written as ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

Question 2.
What is a geographic meridian? How does the declination vary with latitude? Where is it minimum?
Answer:
A plane perpendicular to the surface of the Earth (vertical plane) and passing through geographic axis is geographic meridian.

i. Angle between the geographic and the magnetic meridian at a place is called magnetic declination (a).
ii. Magnetic declination varies with location and over time. As one moves away from the true north the declination changes depending on the latitude as well as longitude of the place. By convention, declination is positive when magnetic north is east of true north, and negative when it is to the west. The declination is small in India. It is 0° 58′ west at Mumbai and 0° 41′ east at Delhi.

Question 3.
Define the angle of dip. What happens to angle of dip as we move towards magnetic pole from magnetic equator?
Answer:
Angle made by the direction of resultant magnetic field with the horizontal at a place is inclination or angle of dip (ø) at the place.
At the magnetic pole value of ø = 90° and it goes on decreasing when we move towards equator such that at equator value of (ø) = 0°.

4. Solve the following problems.

Question 1.
A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 Am and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.
Answer:
Given that, (q_m)₁ = 200 Am
and (2l) = 5 cm = 5 × 10^-2 m
∴ m = 200 × 5 × 10^-2 = 10 Am²
For a bar magnet, magnetic dipole moment is,
m = q_m (21)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
B_a = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac{2 \times 10}{(0.2)^{3}}\)
= 0.25 × 10^-3
= 2.5 × 10^-4 Wb/m²
The force acting on the pole will be given by,
F = q_m B_a = 100 × 2.5 × 10^-4
= 2.5 × 10^-2 N

Question 2.
A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.
Answer:
Let true value of dip be ø. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’_H = B_H cos 30° Whereas, vertical component remains unchanged.
∴ For apparent dip of 45°,
tan 45° = \(\frac{\mathrm{B}_{\mathrm{V}}^{\prime}}{\mathrm{B}_{\mathrm{H}}^{\prime}}=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}} \cos 30^{\circ}}=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}} \times \frac{1}{\cos 30^{\circ}}\)
But, real value of dip is,
tan ø = \(\frac {B_V}{B_H}\)
∴ tan 45° = \(\frac {tan ø}{cos 30°}\)
∴ tan ø = tan 45° × cos 30°
= 1 × \(\frac {√3}{2}\)
∴ ø = tan^-1 (0.866)

Question 3.
Two small and similar bar magnets have magnetic dipole moment of 1.0 Am² each. They are kept in a plane in such a way that their axes are perpendicular to each other. A line drawn through the axis of one magnet passes through the centre of other magnet. If the distance between their centres is 2 m, find the magnitude of magnetic field at the midpoint of the line joining their centres.
Answer:
Let P be the midpoint of the line joining the centres of two bar magnets. As shown in figure, P is at the axis of one bar magnet and at the equator of another bar magnet. Thus, the magnetic field on the axis of the first bar magnet at distance of 1 m from the centre will be,

B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac {2×1.0}{(1)^3}\)
= 2 × 10^-7 Wb/m²
Magnetic field on the equator of second bar magnet will be,
B_eq = \(\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\)
= 10^-7 × \(\frac {1.0}{(1)^3}\)
= 1 × 10^-7 Wb/m²
The net magnetic field at P,
B_net = \(\sqrt {B_a^2+B_{eq}^2}\)
= \(\sqrt {(2×10^{-7})^2+(1×10^{-7})^2}\)
= \(\sqrt {(10^{-7})^2×(4+1)}\)
= √5 × 10^-7 Wb/m²

Question 4.
A circular magnet is made with its north pole at the centre, separated from the surrounding circular south pole by an air gap. Draw the magnetic field lines in the gap. Draw a diagram to illustrate the magnetic lines of force between the south poles of two such magnets.
Answer:
i. For a circular magnet:

Question 5.
Two bar magnets are placed on a horizontal surface. Draw magnetic lines around them. Mark the position of any neutral points (points where there is no resultant magnetic field) on your diagram.
Answer:
The magnetic lines of force between two magnets will depend on their relative positions. Considering the magnets to be placed one besides the other as shown in figure, the magnetic lines of force will be as shown.

11th Physics Digest Chapter 12 Magnetism Intext Questions and Answers

Can you recall? (Textbook page no. 221)

Question 1.
What are the magnetic lines of force?
Answer:
The magnetic field around a magnet is shown by lines going from one end of the magnet to the other. These lines are named as magnetic lines of force.

Question 2.
What are the rules concerning the lines of force?
Answer:
i. Magnetic lines of force originate from the north pole and end at the south pole.
ii. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
iii. The direction of the net magnetic field \(\vec {B}\) at a point is given by the tangent to the magnetic line of force at that point.
iv. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec {B}\)
v. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Question 3.
What is a bar magnet?
Answer:
Bar magnet is a magnet in the shape of bar having two poles of equal and opposite pole strengths separated by certain distance (2l).

Question 4.
If you freely hang a bar magnet horizontally, in which direction will it become stable?
Answer:
A bar magnet suspended freely in air always aligns itself along geographic N-S direction.

Try this (Textbook page no. 221)

You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced.
Answer:
When a small compass needle is kept at any position near a bar magnet, the needle always aligns itself in the direction parallel to the direction of magnetic lines of force.

Hence, by placing it at different positions, A, B, C, D,… as shown in the figure, the direction of magnetic lines of force can be traced. The direction of magnetic field will be a tangent at that point.

Internet my friend: (Text book page no. 227)

https://www.ngdc.noaa.gov
[Students are expected to visit above mentioned link and collect more information about Geomagnetism.]

Maharashtra State Board Class 11 Physics Textbook Solutions 

• Units and Measurements Class 11 Physics Textbook Solutions
• Mathematical Methods Class 11 Physics Textbook Solutions
• Motion in a Plane Class 11 Physics Textbook Solutions
• Laws of Motion Class 11 Physics Textbook Solutions
• Gravitation Class 11 Physics Textbook Solutions
• Mechanical Properties of Solids Class 11 Physics Textbook Solutions
• Thermal Properties of Matter Class 11 Physics Textbook Solutions
• Sound Class 11 Physics Textbook Solutions
• Optics Class 11 Physics Textbook Solutions
• Electrostatics Class 11 Physics Textbook Solutions
• Electric Current Through Conductors Class 11 Physics Textbook Solutions
• Magnetism Class 11 Physics Textbook Solutions
• Electromagnetic Waves and Communication System Class 11 Physics Textbook Solutions
• Semiconductors Class 11 Physics Textbook Solutions

Electric Current Through Conductors Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 11 Electric Current Through Conductors Textbook Exercise Questions and Answers.

Class 11 Physics Chapter 11 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 11 Exercise Solutions

1. Choose the correct Alternative.

Question 1.
You are given four bulbs of 25 W, 40 W, 60 W, and 100 W of power, all operating at 230 V. Which of them has the lowest resistance?
(A) 25 W
(B) 40 W
(C) 60 W
(D) 100 W
Answer:
(D) 100 W

Question 2.
Which of the following is an ohmic conductor?
(A) transistor
(B) vacuum tube
(C) electrolyte
(D) nichrome wire
Answer:
(D) nichrome wire

Question 3.
A rheostat is used
(A) to bring on a known change of resistance in the circuit to alter the current.
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.
(C) to make and break the circuit at any instant.
(D) neither to alter the resistance nor the current.
Answer:
(B) to continuously change the resistance in any arbitrary manner and there by alter the current.

Question 4.
The wire of length L and resistance R is stretched so that its radius of cross-section is halved. What is its new resistance?
(A) 5R
(B) 8R
(C) 4R
(D) 16R
Answer:
(D) 16R

Question 5.
Masses of three pieces of wires made of the same metal are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratios of their resistances are
(A) 1 : 3 : 5
(B) 5 : 3 : 1
(C) 1 : 15 : 125
(D) 125 : 15 : 1
Answer:
(D) 125 : 15 : 1

Question 6.
The internal resistance of a cell of emf 2 V is 0.1 Ω, it is connected to a resistance of 0.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.8 V
(C) 1.95 V
(D) 3V
Answer:
(B) 1.8 V

Question 7.
100 cells each of emf 5 V and internal resistance 1 Ω are to be arranged so as to produce maximum current in a 25 Ω resistance. Each row contains equal number of cells. The number of rows should be
(A) 2
(B) 4
(C) 5
(D) 100
Answer:
(A) 2

Question 8.
Five dry cells each of voltage 1.5 V are connected as shown in diagram

What is the overall voltage with this arrangement?
(A) 0 V
(B) 4.5 V
(C) 6.0 V
(D) 7.5 V
Answer:
(B) 4.5 V

2. Give reasons / short answers

Question 1.
In given circuit diagram two resistors are connected to a 5V supply.

i. Calculate potential difference across the 8Q resistor.
ii. A third resistor is now connected in parallel with 6 Ω resistor. Will the potential difference across the 8 Ω resistor be larger, smaller or same as before? Explain the reason for your answer.
Answer:
Total current flowing through the circuit,
I = \(\frac {V}{R_s}\)
= \(\frac {5}{8+6}\)
= \(\frac {5}{14}\) = 0.36 A
∴ Potential difference across 8 f2 (Vi) = 0.36 × 8
= 2.88 V

ii. Potential difference across 8 Ω resistor will be larger.
Reason: As per question, the new circuit diagram will be

When any resistor is connected parallel to 6 Ω resistance. Then the resistance across that branch (6 Ω and R Ω) will become less than 6 Ω. i.e., equivalent resistance of the entire circuit will decrease and hence current will increase. Since, V = IR, the potential difference across 8 Ω resistor will be larger.

Question 2.
Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.
Answer:
i. Consider a part of conducting wire with its free electrons having the drift speed vd in the direction opposite to the electric field \(\vec{E}\).

ii. All the electrons move with the same drift speed vd and the current I is the same throughout the cross section (A) of the wire.

iii. Let L be the length of the wire and n be the number of free electrons per unit volume of the wire. Then the total number of free electrons in the length L of the conducting wire is nAL.

iv. The total charge in the length L is,
q = nALe ………….. (1)
where, e is the charge of electron.

v. Equation (1) is total charge that moves through any cross section of the wire in a certain time interval t.
∴ t = \(\frac {L}{v_d}\) ………….. (2)

vi. Current is given by,
I = \(\frac {q}{t}\) = \(\frac {nALe}{L/v_d}\) ……………. [From Equations (1) and (2)]
= n Av_de
Hence
v_d = \(\frac {1}{nAe}\)
= \(\frac {J}{ne}\) …………. (∵ J = \(\frac {1}{A}\))
Hence for constant ‘ne’, current density of a metallic conductor is directly proportional to the drift speed of electrons, J ∝ v_d.

3. Answer the following questions.

Question 1.
Distinguish between ohmic and non ohmic substances; explain with the help of example.
Answer:

Question 2.
DC current flows in a metal piece of non uniform cross-section. Which of these quantities remains constant along the conductor: current, current density or drift speed?
Answer:
Drift velocity and current density will change as it depends upon area of cross-section whereas current will remain constant.

4. Solve the following problems.

Question 1.
What is the resistance of one of the rails of a railway track 20 km long at 20°C? The cross-section area of rail is 25 cm² and the rail is made of steel having resistivity at 20°C as 6 × 10^-8 Ω m.
Answer:
Given: l = 20 km = 20 × 10³ m,
A = 25 cm² = 25 × 10^-4 m²,
ρ = 6 × 10^-8 Ω m
To find: Resistance of rail (R)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: From formula.
R = ρ\(\frac {l}{A}\)
∴ R = \(\frac {6×10^{-8}×20×10^3}{A}\) = \(\frac {6×4}{5}\) × 10^-1
= 0.48 Ω

Question 2.
A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery. Can this battery drive starting motor of car?
Answer:
E = 24 V, r = 380 Ω
i. Maximum current (I_max)
ii. Can battery start the motor?
Formula: I_max = \(\frac {E}{r}\)
Calculation:
From formula,
I_max = \(\frac {24}{380}\) = 0,063 A
As, the value of current is very small compared to required current to run a starting motor of a car, this battery cannot be used to drive the motor.

Question 3.
A battery of emf 12 V and internal resistance 3 O is connected to a resistor. If the current in the circuit is 0.5 A,
i. Calculate resistance of resistor.
ii. Calculate terminal voltage of the battery when the circuit is closed.
Answer:
Given: E = 12 V, r = 3 Ω, I = 0.5 A
To find:
i. Resistance (R)
ii. Terminal voltage (V)
Formulae:
i. E = I (r + R)
ii. V = IR
Calculation: From formula (i),
E = Ir + IR
∴ R = \(\frac {E-Ir}{l}\)
= \(\frac {12-0.5×3}{0.5}\)
= 21 Ω
From formula (ii),
V = 0.5 × 21
= 10.5 V

Question 4.
The magnitude of current density in a copper wire is 500 A/cm². If the number of free electrons per cm³ of copper is 8.47 × 10²², calculate the drift velocity of the electrons through the copper wire (charge on an e = 1.6 × 10^-19 C)
Answer:
Given: J = 500 A/cm² = 500 × 10⁴ A/m²,
n = 8.47 × 10²² electrons/cm³
= 8.47 × 10²⁸ electrons/m³
e = 1.6 × 10^-19 C
To Find: Drift velocity (v_d)
Formula: v_d = \(\frac {J}{ne}\)
Calculation:
From formula,
v_d = \(\frac {500×10^4}{8.47×10^{28}×1.6×10^{-19}}\)
= \(\frac {500}{8.47×1.6}\) × 10^-5
= {antilog [log 500 – log 8.47 – log 1.6]} × 10^-5
= {antilog [2.6990 – 0.9279 -0.2041]} × 10^-5
= {antilog [1.5670]} × 10^-5
= 3.690 × 10¹ × 10^-5
= 3.69 × 10^-4 m/s

Question 5.
Three resistors 10 Ω, 20 Ω and 30 Ω are connected in series combination.
i. Find equivalent resistance of series combination.
ii. When this series combination is connected to 12 V supply, by neglecting the value of internal resistance, obtain potential difference across each resistor.
Answer:
Given: R₁ = 10 Ω, R₂ = 20 Ω,
R₃ = 30 Ω, V = 12 V
To Find: i. Series equivalent resistance(R_s)
ii. Potential difference across each resistor (V₁, V₂, V₃)
Formula: i. R_s = R₁ + R₂ + R₃
ii. V = IR
Calculation:
From formula (i),
R_s = 10 + 20 + 30 = 60 Ω
From formula (ii),
I = \(\frac {V}{R}\) = \(\frac {12}{60}\) = 0.2 A
∴ Potential difference across R₁,
V₁ = I × R₁ = 0.2 × 10 = 2 V
∴ Potential difference across R₂,
V₂ = 0.2 × 20 = 4 V
∴ Potential difference across R₃,
V₃ = 0.2 × 30 = 6 V

Question 6.
Two resistors 1 Ω and 2 Ω are connected in parallel combination.
i. Find equivalent resistance of parallel combination.
ii. When this parallel combination is connected to 9 V supply, by neglecting internal resistance, calculate current through each resistor.
Answer:
R₁ = 1 kΩ = 10³ Ω,
R₂ = 2 kΩ = 2 × 10³ Ω, V = 9 V
To find:
i. Parallel equivalent resistance (R_p)
ii. Current through 1 kΩ and 2 kΩ (I₁ and I₂)
Formula:
i. \(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\)
ii. V = IR
Calculation: From formula (i),
\(\frac {1}{R_p}\) = \(\frac {1}{10^3}\) + \(\frac {1}{2×10^3}\)
= \(\frac {3}{2×10^3}\)
∴ R^p = \(\frac {2×10^3}{3}\) = 0.66 kΩ
From formula (ii),
I₁ = \(\frac {V}{R_1}\) + \(\frac {9}{10^3}\)
= 9 × 10^-3 A
= 3 mA
I₂ = \(\frac {V}{R_2}\) + \(\frac {9}{2×10^3}\)
= 4.5 × 10^-3 A
= 4.5 mA

Question 7.
A silver wire has a resistance of 4.2 Ω at 27°C and resistance 5.4 Ω at 100°C. Determine the temperature coefficient of resistance.
Answer:
Given: R₁ =4.2 Ω, R₂ = 5.4 Ω,
T, = 27° C, T2= 100 °C
To find: Temperature coefficient of resistance (α)
Formula: α = \(\frac {R_2-R_1}{R_1(T_2-T_1)}\)
Calculation:
From Formula
α = \(\frac {5.4-4.2}{4.2(100-27)}\) = 3.91 × 10^-3/°C

Question 8.
A 6 m long wire has diameter 0.5 mm. Its resistance is 50 Ω. Find the resistivity and conductivity.
Answer:
Given: l = 6 m, D = 0.5 mm,
r = 0.25 mm = 0.25 × 10^-3 m, R = 50 Ω
To find:
i. Resistivity (ρ)
ii. Conductivity (σ)
Formulae:
i. ρ = \(\frac {RA}{l}\) = \(\frac {Rπr^2}{l}\)
ii. σ = \(\frac {1}{ρ}\)
Calculation:
From formula (i),
ρ = \(\frac {50×3.142×(0.25×10^{-3})^2}{6}\)
= {antilog [log 50 + log 3.142 + 21og 0.25 -log 6]} × 10^-6
= {antilog [ 1.6990 + 0.4972 + 2(1.3979) -0.7782]} × 10^-6
= {antilog [2.1962 + 2 .7958 – 0.7782]} × 10^-6
= {antilog [0.9920 – 0.7782]} × 10^-6
= {antilog [0.2138]} × 10^-6
= 1.636 × 10^-6 Ω/m
From formula (ii),
σ = \(\frac {1}{1.636×10^{-6}}\)
= 0.6157 × 10⁶
….(Using reciprocal from log table)
= 6.157 × 10⁵ m/Ω

Question 9.
Find the value of resistances for the following colour code.
i. Blue Green Red Gold
ii. Brown Black Red Silver
iii. Red Red Orange Gold
iv. Orange White Red Gold
v. Yellow Violet Brown Silver
Answer:
i. Given: Blue – Green – Red – Gold
To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± T%)Ω
Calculation:

From formula,
Value of resistance = (65 × 10² ± 5%) Ω
Value of resistance = 6.5 kΩ ± 5%

ii. Given: Brown – Black – Red – Silver
To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z + T%) Ω
Calculation:

From formula,
Value of resistance = (10 × 10² ± 10%) Ω
Value of resistance = 1.0 kΩ ± 10%

iii. Given: Red – Red – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%)
Calculation:

From formula,
Value of resistance = (22 × 10³ ± 5%)Ω
Value of resistance = 22 kΩ ± 5%
[Note: The answer given above is presented considering correct order of magnitude.]

iv. Given: Orange – White – Red – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%)
Calculation:

From formula,
Value of resistance = (39 × 10² ± 5%) Ω
Value of resistance = 3.9 kΩ ± 5%

v. Given: Yellow-Violet-Brown-Silver
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%)
Calculation:

From formula,
Value of resistance = (47 × 10 ± 10%) Ω
Value of resistance = 470 Ω ± 10%
[Note: The answer given above is presented considering correct order of magnitude.]

Question 10.
Find the colour code for the following value of resistor having tolerance ± 10%.
i. 330 Ω
ii. 100 Ω
iii. 47 kΩ
iv. 160 Ω
v. 1 kΩ
Answer:

Question 11.
A current of 4 A flows through an automobile headlight. How many electrons flow through the headlight in a time of 2 hrs?
Answer:
Given: I = 4 A, t = 2 hrs = 2 × 60 × 60 s
To find: Number of electrons (N)
Formula: I = \(\frac {q}{t}\) = \(\frac {Ne}{t}\)
Calculation: As we know, e = 1.6 × 10^-19 C
From formula,
N = \(\frac {It}{e}\) = \(\frac {4×2×60×60}{1.6×10^{-19}}\) = 18 × 10^-23

Question 12.
The heating element connected to 230 V draws a current of 5 A. Determine the amount of heat dissipated in 1 hour (J = 4.2 J/cal).
Answer:
Given: V = 230 V, I = 5 A,
At = 1 hour = 60 × 60 sec
To find: Heat dissipated (H)
Formula: H = ∆U = I∆tV
Calculation: From formula,
H = 5 × 60 × 60 × 230
= 4.14 × 10⁶ J
Heat dissipated in calorie,
H = \(\frac {4.14×10^6}{4.2}\) = 985.7 × 10³ cal
= 985.7 kcal

11th Physics Digest Chapter 11 Electric Current Through Conductors Intext Questions and Answers

Can you recall? (Textbookpage no. 207)

An electric current in a metallic conductor such as a wire is due to the flow of electrons, the negatively charged particles in the wire. What is the role of the valence electrons which are the outermost electrons of an atom?
Answer:
i. The valence electrons become de-localized when a large number of atoms come together in a metal.
ii. These electrons become conduction electrons or free electrons constituting an electric current when a potential difference is applied across the conductor.

Internet my friend (Textbook page no. 218)

https://www.britannica.com/science/supercond activity physics

[Students are expected to visit the above-mentioned website and Collect more information about superconductivity.]

Maharashtra State Board Class 11 Physics Textbook Solutions 

• Units and Measurements Class 11 Physics Textbook Solutions
• Mathematical Methods Class 11 Physics Textbook Solutions
• Motion in a Plane Class 11 Physics Textbook Solutions
• Laws of Motion Class 11 Physics Textbook Solutions
• Gravitation Class 11 Physics Textbook Solutions
• Mechanical Properties of Solids Class 11 Physics Textbook Solutions
• Thermal Properties of Matter Class 11 Physics Textbook Solutions
• Sound Class 11 Physics Textbook Solutions
• Optics Class 11 Physics Textbook Solutions
• Electrostatics Class 11 Physics Textbook Solutions
• Electric Current Through Conductors Class 11 Physics Textbook Solutions
• Magnetism Class 11 Physics Textbook Solutions
• Electromagnetic Waves and Communication System Class 11 Physics Textbook Solutions
• Semiconductors Class 11 Physics Textbook Solutions

Std 10 Science Part 1 Chapter 4 Effects of Electric Current Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 4 Effects of Electric Current Notes, Textbook Exercise Important Questions and Answers.

Class 10 Science Part 1 Chapter 4 Effects of Electric Current Question Answer Maharashtra Board

Class 10 Science 1 Chapter 4 Effects Of Electric Current Exercise Question 1.
Tell the odd one out. Give proper explanation.
a. Fuse wire, bud conductor, rubber gloves, generator.
Answer:
Generator. It converts mechanical energy into electric energy, the remaining three do not.

b. Voltmeter, Ammeter, gulvanometer, thermometer.
Answer:
Thermometer. It measures temperature, the remaining three measure electrical quantities.

c. Loud speaker, microphone, electric motor, magnet.
Answer:
Magnet. It exerts a force on a magnetic material, the remaining three convert one form of energy into another.

4 Effects Of Electric Current Exercise Question 2.
Explain the construction and working of the following. Draw a neat diagram and label it.
a. Electric motor
Answer:
Figure shows the construction of an electric motor. Here, a rectangular loop ABCD of copper wire with resistive coating is placed between the north pole and south pole or a strong magnet, such as a horseshoe magnet, such that the branches AB and CD are perpendicular to the direction of the magnetic field. The ends of the loop are connected to the two halves, X and Y, of split rings X and Y have resistive coating on their inner surfaces and are tightly fitted on the axle. The outer conducting surfaces of X and Y are in contact with two stationary carbon brushes, E and F, respectively.

Working:
1. When the circuit is completed with a plug key or switch, the current flows in the direction E → A → B → C → D → F. As the magnetic field is directed from the north pole to the south pole, the force on AB is downward and that on CD is upward by Fleming’s left hand rule. Hence, AB moves downward and CD upward. These forces are equal in magnitude and opposite in direction. Therefore, as observed from the side AD, the loop ABCD and the axle start rotating in anticlockwise direction.

2. After half a rotation, X and Y come in contact with brushes F and E respectively and the current flows in the direction EDCBAF. Hence the force on CD is downward and that on AB is upward. Therefore, the loop and the axle continue to rotate in the anticlockwise direction.

3. After every half rotation, the current in the loop is reversed and the loop and the axle continue to rotate in anti clockwise direction. When the current is switched off, the loop stops rotating after some time.

b. Electric Generator (AC)
Answer:
Figure shows the construction of an AC electric generator. Here, a coil ABCD of copper wire is kept between the pole pieces (N and S) of a strong magnet. The ends of the coil are connected to the conducting rings R₁ and R₂ via carbon brushes B₁ and B₂. The rings are fixed to the axle and there is a resistive coating in between the rings and the axle. The stationary brushes are connected to a galvanometer used to show the direction of the current in the circuit.

Working:
When the axle is rotated with a machine from outside, the coil ABCD starts rotating. Suppose the coil rotates in clockwise direction, as observed from the side AD. Then as the branch AB moves upward, the branch CD moves downward. By Fleming’s right hand rule, the induced current flows in the direction A → B → C → D and in the external circuit, it flows from B2 to B, through the galvanometer. The induced current is proportional to the number of turns of the copper wire in the coil.

After half a rotation, AB and CD interchange their places. Hence, the induced current flows in the direction D → C → B → A. As AB is always in contact with B₁ and CD is in contact with B₂, the current in the external circuit flows from B₁ to B₂ through the galvanometer. Thus, the direction of the current is the external circuit is opposite to that in the previous half rotation. The process goes on repeating and alternating current is generated.

4 Effects Of Electric Current Question 3.
Electromagnetic induction means
a. Charging of an electric conductor.
b. Production of magnetic field due to a current flowing through a coil.
c. Generation of a current in a coil due to relative motion between the coil and the magnet.
d. Motion of the coil around the axle in an electric motor.
Answer:
c. Generation of a current in a coil due to relative motion between the coil and the magnet.

Electric Current Question 4.
4. Explain the difference: AC generator and DC generator.
Answer:
AC generator:

1. In an AC generator, the rings used are not split.
2. The direction of the current produced reverses after equal intervals of time.

DC generator:

1. In a DC generator, split rings are used.
2. The current produced flows in the same direction all the time.

Question 5.
Which device is used to produce electricity? Describe with a neat diagram.
(1) Electric motor
(2) Galvanometer
(3) Electric generator (DC)
(4) Voltmeter
Answer:
Electric generator (DC).

Figure shows the construction of a DC generator.
Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.

In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.

[Note In the case of a DC generator, the current is in the same direction during both the halves of the rotation of the coil. The magnitude of the current does vary periodically with time. In this respect, it differs from the current supplied by an electric cell.]

Question 6.
How does the short circuit form? What is its effect?
Answer:
If a bare live wire (phase wire) and a bare neutral wire touch each other (come in direct contact) or come very close to each other, the resistance of the circuit becomes very small and hence huge (very high) electric current flows through it. This condition is called a short circuit or short circuiting.

In this case, a large amount of heat is produced and the temperature of the components involved becomes very high. Hence, the circuit catches fire.

Question 7.
Give scientific reasons:
a. Tungsten is used to make a solenoid type coil in an electric bulb.
Answer:
1. The intensity of light emitted by the filament of a bulb depends on the temperature of the filament. It increases with the temperature.

2. The melting point of the material used to make the filament of a bulb should be very high so that the filament can be heated to a high temperature by passing a current through it, without melting it. This enables us to obtain more light. The melting point of tungsten is very high.

Hence, tungsten is used to make a solenoid type coil (filament) in an electric bulb.

b. In the electric equipment producing heat e.g. iron, electric heater, boiler, toaster, etc. an alloy such as Nichrome is used, not pure metals.
Answer:
1. The working of heating devices such as a toaster and an electric iron is based on the heating effect of electric current, i.e., conversion of electric energy into heat by passage of electric current through a metallic conductor.

2. An alloy, such as Nichrome, has high resistivity and it can be heated to a high temperature without oxidation, in contrast to pure metals. Therefore, the coils in heating devices such as a toaster and an electric iron are made of an alloy, such as Nichrome, rather than a pure metal.

c. For electric power transmission, copper or aluminium wire is used.
Answer:
1. Copper and aluminium are good conductors of electricity.

2. Copper, and aluminium have very low resistivity. Hence, when an electric current flows through a wire of copper or aluminium, heat produced is comparatively low. Therefore, for electric power transmission, copper or aluminium wire is used.

d. In practice the unit kWh is used for the measurement of electric energy, rather than the joule.
Answer:
(1) If an electric device rated 230 V, 5 A is operated for one hour, electric energy used
= VIt = 230 V × 5 A × 3600 s = 4140000 joules.

(2) If this energy is expressed in kW.h, it will be \(\frac{4140000}{3.6 \times 10^{6}}\) kW·h = 1.15 kW·h (more convenient). 3.6 × 10⁶
Hence, in practice the unit kW·h is used for the measurement of electric energy, rather than the joule.

Question 8.
Which of the statements given below correctly describes the magnetic field near a long, straight current-carrying conductor?
(1) The magnetic lines of force are in a plane, perpendicular to the conductor in the form of straight lines.
(2) The magnetic lines of force are parallel to the conductor on all the sides of conductor.
(3) The magnetic lines of force are perpendicular to the conductor going radially outward.
(4) The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
Answer:
The magnetic lines of force are in concentric circles with the wire as the centre, in a plane perpendicular to the conductor.

Question 9.
What is a solenoid? Compare the magnetic field produced by a solenoid with the magnetic field of a bar magnet. Draw neat figures and name various components.
Answer:
When a copper wire with a resistive coating is wound in a chain of loops (like a spring), it is called a solenoid.

Magnetic lines of force (magnetic field lines) due to a current carrying solenoid.
B: Battery, K: Plug key, I: Current, N: North pole, S: South pole

The magnetic field lines (magnetic lines of force) due to a current-carrying solenoid are similar to those of a bar magnet. One face of the coil acts as the south pole and the other face as the north pole.

[Note: A current-carrying coil, like a magnet, can be used to magnetise the rod of a given material such as carbon steel or chromium steel. With a strong megnetic field, permanent magnetism can be produced in these materials.]

Question 10.
Name the following diagrams and explain the concept behind them.

Answer:
(a) Fleming’s right hand rule:
Answer:
Stretch the thumb, the index finger and the middle finger of the right hand in such a way that they are perpendicular to each other. In this position, the thumb indicates the direction of the motion of the conductor, the index finger the direction of the magnetic field, and the middle finger shows the direction of the induced current.

[Note The induced current is maximum when the direction of motion of the conductor is at right angles to the magnetic field. ]

(b) Fleming’s left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor.

[Note: A magnetic field exerts a force on a current-carrying conductor. Electric current is the time rate of flow of electric charge. Thus, a magnetic field exerts a force on a moving charge. This property is used to accelerate charged particles such as protons, deuterons and alpha particles, as well as electrons, to very high energies. A machine used for this purpose is called a charged particle accelerator. It may be linear or circular in design and very big in size. Such high energy particles are used to study the structure of matter. ]

Question 11.
Identify the figures and explain their use.

Answer:
(a) Fuse:
A fuse protects electrical circuits and appliances by stopping the flow of electric current when it exceeds a specified value. For this, it is connected in series with the appliance (or circuit) to be protected. A fuse is a piece of wire made of an alloy of low melting point (e.g. an alloy of lead and tin). If a current larger than the specified value flows through the fuse, its temperature increases enough to melt it. Hence, the circuit breaks and the appliance is protected from damage.

[Note : The fuse wire is usually enclosed in a cartridge of an insulator such as glass or porcelain provided with metal caps. The current rating (such as 1 A, 2 A) may be printed on the cartridge. ]

(b) Miniature circuit breaker:
These days miniature circuit breaker (MCB) switches are used in homes. When the current in the circuit suddenly increases this switch opens and current stops. Different types of MCBs are in use. For the entire house, however the usual fuse wire is used.

(c) Figure shows the construction of a DC generator.

Here, an ammeter is shown instead of a bulb.
Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.

In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.

Question 12.
Solve the following examples.
a. Heat energy is being produced in a resistance in a circuit at the rate of 100 W. The current of 3 A is flowing in the circuit. What must be the value of the resistance?
Solution:
Data: P = 100 W, I = 3 A, R = ?, P = I²R
∴ Resistance, R = \(\frac{P}{I^{2}}=\frac{100 \mathrm{W}}{(3 \mathrm{A})^{2}}=\frac{100}{9} \Omega\) = 11.11 Ω

b. Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Solution:
Data : P₁ = 100 W, P₂ = 60 W, V = 220 V,
I = ?, ∴ I = \(\frac{P}{V}\)
P = VI
∴ I₁ = \(\frac{P_{1}}{V}\) and I₂ = \(\frac{P_{2}}{V}\)
Current in the main conductor, I = I₁ + I₂ (parallel connection)

c. Who will spend more electrical energy? 500 W TV set in 30 mins, or 600 W heater in 20 mins?
Solution:
Data : P₁ = 500 W, t₁ = 30 min = \(\frac{30}{60}\) h
= \(\frac{1}{2}\) h, P₂ = 600 W, t₂ = 20 min = \(\frac{20}{60}\) h = \(\frac{1}{3}\) h
Electrical energy used = Pt
TV set : P₁t₁ = 500 W × \(\frac{1}{2}\) h = 250 W·h
Heater : P₂t₂ = 600 W × \(\frac{1}{3}\) h = 200 W·h
Thus, the TV set will spend more electrical energy than the heater.

d. An electric iron of 1100 W is operated for 2 hours daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges ₹ 5 per unit of energy.)
Solution:
Data: P = 1100 W, t = 2 × 30 = 60 h,
₹ 5 per unit of energy, expenses = ?

∴ Electrical consumption expenses = 66 units × ₹ 5 per unit = ₹ 330.

Project:
Do it your self.
Project 1.
Under the guidance of your teachers, make a ‘free-energy generator’.

Can you recall? (Text Book Page No. 47)

Question 1.
How do we decide that a given material is a good conductor of electricity or is an insulator?
Answer:
A material which has very low electrical resistance is called a good conductor of electricity. Examples: silver, copper, aluminium.
A material which has extremely high electrical resistance is called an insulator of electricity. Examples: rubber, wood, glass.

Question 2.
Iron is a conductor of electricity, but when we pick up a piece of iron resting on the ground, why don’t we get electric shock?
Answer:
When we pick up a piece of iron resting on the ground, we don’t get electric shock because that piece does not carry any electric current at that time.

Use your brain power! (Text Book Page No. 48)

Question 1.
If in the circuit, the resistor is replaced by a motor, in which form will the energy given by the cell get transformed into?
Answer:
The energy given by the cell will get transformed into the kinetic energy of the copper coil in the motor.

Use your brain power! (Text Book Page No. 60)

Question 1.
Draw the diagram of a DC generator. Then explain as to how the DC current is obtained.
Answer:
Figure shows the construction of a DC generator.

Working:
The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.

In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.

[Note In the case of a DC generator, the current is in the same direction during both the halves of the rotation of the coil. The magnitude of the current does vary periodically with time. In this respect, it differs from the current supplied by an electric cell.]

Fill in the blanks and rewrite the completed statements:

Question 1.
Electric power = V2/…….
Answer:
Electric power = \(\frac{V^{2}}{R}\)

Question 2.
……….= 1 joule/1 second.
Answer:
1 watt = 1 joule /1 second.

Question 3.
1 kW.h =………J.
Answer:
1 kW.h = 3.6 x 10⁶ J.

Question 4.
According to Joule’s law, quantity of heat (H) produced by an electric current =……….
Answer:
According to Joule’s law, quantity of heat (H) produced by an electric current = I²Rt or VIt or \(\frac{V^{2}}{R}\)t

Question 5.
Magnetic effect of electric current was dicovered by………..
Answer:
Magnetic effect of electric current was dicovered by Hans Christian Oersted.

Question 6.
………..is expressed in oersted.
Answer:
Intensity of magnetic field is expressed in oersted.

Question 7.
Electromagnetic induction was discovered by………..
Answer:
Electromagnetic induction was discovered by Michael Faraday and independently by Joseph Henry.

Question 8.
A galvanometer is used for………
Answer:
A galvanometer is used for detecting the presence of current in a circuit, as well as for some electrical measurements.

Question 9.
In India, the frequency of alternating current is……….
Answer:
In India, the frequency of alternating current is 50 Hz or 50 cycles per second.

Question 10.
Electric motor converts electric energy into………energy.
Answer:
Electric motor converts electric energy into mechanical energy.

Question 11.
Electric generator converts………..energy into electric energy.
Answer:
Electric generator converts mechanical energy into electric energy.

Rewrite the following statements by selecting the correct options:

Question 1.
The device used for producing a current is called……….
(a) a voltmeter
(b) an ammeter
(c) a galvanometer
(d) a generator
Answer:
(d) a generator

Question 2.
At the time of short circuit, the current in the circuit………
(a) increases
(b) decreases
(c) remains the same
(d) increases in steps
Answer:
(a) increases

Question 3.
The direction of the magnetic field around a straight conductor carrying current is given by……
(a) the right hand thumb rule
(b) Fleming’s left hand rule
(c) Fleming’s right hand rule
(d) none of these
Answer:
(a) the right hand thumb rule

Question 4.
The resistance of a wire is 100 Ω. If it carries a current of 1A for 10 seconds, the heat produced will be……….
(a) 1000 J
(b) 10 J
(c) 0.1 J
(d) 10000 J
Answer:
(a) 1000 J

Question 5.
If 220 V potential difference is applied across an electric bulb, a current of 0.45 A flows in the bulb. What must be the power of the bulb? (Practice Activity Sheet – 1)
(a) 99 W
(b) 70 W
(c) 45 W
(d) 22 W
Answer:
(a) 99 W

Question 6.
Electromagnetic induction means
(a) charging of an electric conductor.
(b) production of magnetic field due to a current flowing through a coil.
(c) generation of a current in a coil due to relative motion between the coil and the magnet.
(d) motion of the coil around the axle in an electric motor.
Answer:
(c) generation of a current in a coil due to relative motion between the coil and the magnet.