Category: Class 11

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 5 Gravitation Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 5 Gravitation

1. Choose the correct option.

Question 1.
The value of acceleration due to gravity is maximum at
(A) the equator of the Earth
(B) the centre of the Earth.
(C) the pole of the Earth.
(D) slightly above the surface of the Earth.
Answer:
(C) the pole of the Earth.

Question 2.
The weight of a particle at the centre of the Earth is _________
(A) infinite.
(B) zero.
(C) same as that at other places.
(D) greater than at the poles.
Answer:
(B) zero.

Question 3.
The gravitational potential due to the Earth is minimum at
(A) the centre of the Earth.
(B) the surface of the Earth.
(C) a points inside the Earth but not at its centre.
(D) infinite distance.
Answer:
(A) the centre of the Earth.

Question 4.
The binding energy of a satellite revolving around planet in a circular orbit is 3 × 10⁹ J. Its kinetic energy is _________
(A) 6 × 10⁹ J
(B) -3 × 10⁹ J
(C) -6 × 10⁺⁹ J
(D) 3 × 10⁺⁹J
Answer:
(D) 3 × 10⁺⁹J

2. Answer the following questions.

Question 1.
State Kepler’s law equal of area.
Answer:
The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.

Question 2.
State Kepler’s law of period.
Answer:
The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.

Question 3.
What are the dimensions of the universal gravitational constant?
Answer:
The dimensions of universal gravitational constant are: [L³M^-1T^-2].

Question 4.
Define binding energy of a satellite.
Answer:
The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.

Question 5.
What do you mean by geostationary satellite?
Answer:
Some satellites that revolve around the Earth in equatorial plane have same sense of rotation as that of the Earth. The also have the same period of rotation as that of the Earth i.e.. 24 hours. Due to this, these satellites appear stationary from the Earth’s surface and are known as geostationary satellites.

Question 6.
State Newton’s law of gravitation.
Answer:
Statement:
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 7.
Define escape velocity of a satellite.
Answer:
The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (v_e) of the body.

Question 8.
What is the variation in acceleration due to gravity with altitude?
Answer:
Variation in acceleration due to gravity due to altitude is given by, g_h = g\(\left(\frac{R}{R+h}\right)^{2}\)
where,
g_h = acceleration due to gravity of an object placed at h altitude
g = acceleration due to gravity on surface of the Earth
R = radius of the Earth
h = attitude height of the object from the surface of the Earth.
Hence, acceleration due to gravity decreases with increase in altitude.

Question 9.
On which factors does the escape speed of a body from the surface of Earth depend?
Answer:
The escape speed depends only on the mass and radius of the planet.
[Note: Escape velocity does not depend upon the mass of the body]

Question 10.
As we go from one planet to another planet, how will the mass and weight of a body change?
Answer:

1. As we go from one planet to another, mass of a body remains unaffected.
2. However, due to change in mass and radius of planet, acceleration due to gravity acting on the body changes as, g ∝ \(\frac{\mathrm{M}}{\mathrm{R}^{2}}\).
   Hence, weight of the body also changes as, W ∝ \(\frac{\mathrm{M}}{\mathrm{R}^{2}}\)

Question 11.
What is periodic time of a geostationary satellite?
Answer:
The periodic time of a geostationary satellite is same as that of the Earth i.e., one day or 24 hours.

Question 12.
State Newton’s law of gravitation and express it in vector form.
Answer:

1. Statement:
   Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
2. In vector form, it can be expressed as,
   \(\overrightarrow{\mathrm{F}}_{21}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\left(-\hat{\mathrm{r}}_{21}\right)\)
   where, \(\hat{\mathbf{r}}_{21}\) is the unit vector from m₁ to m₂.
   The force \(\overrightarrow{\mathrm{F}}_{21}\) is directed from m₂ to m₁.

Question 13.
What do you mean by gravitational constant? State its SI units.
Answer:

1. From Newton’s law of gravitation,
   F = G \(\frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
   where, G = constant called universal gravitational constant Its value is 667 X 10^-11 N m²/kg².
2. G = \(\frac{\mathrm{Fr}^{2}}{\mathrm{~m}_{1} \mathrm{~m}_{2}}\)
   If m₁ = m₂ = 1 kg, r = 1 m thenF = G.
   Hence, the universal gravitational constant is the force of gravitation between two particles of unit mass separated by unit distance.
3. Unit: N m²/kg² in SI system.

Question 14.
Why is a minimum two stage rocket necessary for launching of a satellite?
Answer:

1. For the projection of an artificial satellite, it is necessary for the satellite to have a certain velocity.
2. In a single stage rocket, when the fuel in first stage of rocket is ignited on the surface of the Earth, it raises the satellite vertically.
3. The velocity of projection of satellite normal to the surface of the Earth is the vertical velocity.
4. If this vertical velocity is less than the escape velocity (v_e), the satellite returns to the Earth’s surface. While, if the vertical velocity is greater than or equal to the escape velocity, the satellite will escape from Earth’s gravitational influence and go to infinity.
5. Hence, minimum two stage rocket, one to raise the satellite to desired height and another to provide required hori7ontal velocity, is necessary for launching of a satellite.

Question 15.
State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection
Answer:
The path of the satellite depends upon the value of horizontal speed of projection v_h relative to critical velocity v_c and escape velocity v_e.
Case (I) v_h < v_c:
The orbit of satellite is an ellipse with point of projection as apogee and Earth at one of the foci. During this elliptical path, if the satellite passes through the Earth’s atmosphere. it experiences a nonconservative force of air resistance. As a result it loses energy and spirals down to the Earth.
Case (II) v_h = v_c:
The satellite moves in a stable circular orbit around the Earth.
Case (III) v_c < v_h < v_e:
The satellite moves in an elliptical orbit round the Earth with the point of projection as perigee.
Case (IV) v_h = v_e
The satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity.
Case (V) v_h > v_e:
The satellite escapes from gravitational influence of Earth traversing a hyperbolic path.

3. Answer the following questions in detail.

Question 1.
Derive an expression for critical velocity of a satellite.
Answer:
Expression for critical velocity:

1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
2. If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity v_c.
3. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
   ∴ Centripetal force = Gravitational force
   
   This is the expression for critical speed at the orbit of radius (R + h).
4. The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.

Question 2.
State any four applications of a communication satellite.
Answer:
Applications of communication satellite:

1. For the transmission of television and radiowave signals over large areas of Earth’s surface.
2. For broadcasting telecommunication.
3. For military purposes.
4. For navigation surveillance.

Question 3.
Show that acceleration due to gravity at height h above the Earth’s surface is g_h = g(\(\frac{R}{R+h}\))²
Answer:
Variation of g due to altitude:

1. Let,
   R = radius of the Earth,
   M = mass of the Earth.
   g = acceleration due to gravity at the surface of the Earth.
2. Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is given by,
   
3. The body is taken at height h above the surface of the Earth as shown in figure. The acceleration due to gravity now changes to,
   g_h = \(\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}\) …………. (2)
4. Dividing equation (2) by equation (1), we get,
   
   We can rewrite,
   
   This expression can be used to calculate the value of g at height h above the surface of the Earth as long as h<< R.

Question 4.
Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
Answer:

v_h = horizontal speed of projection
v _c = critical velocity
v_e = escape velocity

Question 5.
Derive an expression for binding energy of a body at rest on the Earth’s surface.
Answer:

1. Let,
   M = mass of the Earth
   m = mass of the satellite
   R = radius of the Earth.
2. Since the satellite is at rest on the Earth, v = 0
   ∴ Kinetic energy of satellite.
   K.E = \(\frac{1}{2}\) mv² = 0
3. Gravitational potential at the Earth’s surface
   = – \(\frac{\mathrm{GM}}{\mathrm{R}}\)
   ∴ Potential energy of satellite = Gravitational potential × mass of satellite
   = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
4. Total energy of sitellite = T.E = P.E + K.E
   ∴ T.E. = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\) + 0 = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
5. Negative sign in the energy indicates that the satellite is bound to the Earth, due to gravitational force of attraction.
6. For the satellite to be free form Earth’s gravitational influence, its total energy should become positive. That energy is the binding energy of the satellite at rest on the surface of the Earth.
   ∴ B.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\)

Question 6.
Why do astronauts in an orbiting satellite have a feeling of weightlessness?
Answer:

1. For an astronaut, in a satellite, the net force towards the centre of the Earth will always be, F = mg – N.
   where, N is the normal reaction.
2. In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.
3. In this case, the downward acceleration, a_d = g, or the satellite (along with the astronaut) is in the state of free fall.
4. Thus, the net force acting on astronaut will be, F = mg – ma_d i.e., the apparent weight will be zero, giving the feeling of total weightlessness.

Question 7.
Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth’s surface.
Answer:

Question 8.
At which place on the Earth’s surface is the gravitational acceleration maximum? Why?
Answer:

1. Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
2. At poles, latitude θ = 90°.
   ∴ g’ = g
   i.e., there is no reduction in acceleration due to gravity at poles.
3. Also, shape of the Earth is actually an ellipsoid, bulged at equator. The polar radius of the Earth is 6356 km which minimum. As g ∝ \(\frac{1}{\mathrm{R}^{2}}\), acceleration due to gravity is maximum at poles i.e., 9.8322 m/s².

Question 9.
At which place on the Earth surface the gravitational acceleration minimum? Why?
Answer:

1. Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
2. At equator, latitude θ = 0°.
   ∴ g’ = g – Rω²
   i.e., the acceleration due to gravity ¡s reduced by amount Rω²(≈ 0.034 m/s²) at equator.
3. Also, shape of the Earth is actually an ellipsoid, bulged at equator. The equatorial radius of the Earth is 6378 km, which is maximum. As g ∝ \(\frac{1}{\mathrm{R}^{2}}\) acceleration due to gravity is minimum on equator i.e., 9.7804 m/s².

Question 10.
Define the binding energy of a satellite. Obtain an expression for binding energy of a satellite revolving around the Earth at certain attitude.
Answer:
The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.
Expression for binding energy of satellite revolving in circular orbit round the Earth:

1. Consider a satellite of mass m revolving at height h above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy.
2. Let M be the mass of the Earth, R be the Radius of the Earth, v_c be critical velocity of satellite, r = (R + h) be thc radius of the orbit.
3. Kinetic energy of satellite = \(\frac{1}{2} \mathrm{mv}_{\mathrm{c}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\)
4. The gravitational potential at a distance r from the centre of the Earth is –\(\frac{\mathrm{GM}}{\mathrm{r}}\)
   ∴ Potential energy of satellite = Gravitational potential × mass of satellite
   = –\(\frac{\mathrm{GMm}}{\mathrm{r}}\)
5. The total energy of satellite is given as T.E. = KF. + P.E.
   = \(\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}-\frac{\mathrm{GMm}}{\mathrm{r}}=-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\)
6. Total energy of a circularly orbiting satellite is negative. Negative sign indicates that the satellite is bound to the Earth, due to gravitational force of attraction. For the satellite to be free from the Earth’s gravitational influence its total energy should become zero or positive.
7. Hence the minimum energy to be supplied to unbind the satellite is +\(\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\). This is the binding energy of a satellite.

Question 11.
Obtain the formula for acceleration due to gravity at the depth ‘d’ below the Earth’s surface.
Answer:

1. The Earth can be considered to be a sphere made of large number of concentric uniform spherical shells.
2. When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.
3. The acceleration due to gravity on the surface of the Earth is, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
4. Assuming that the density of the Earth is uniform, mass of the Earth is given by
   M = volume x density = \(\frac{4}{3}\) πR³ρ
   ∴ g = \(\frac{\mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}^{2}}\) = \(\frac{4}{3}\) πRρG ………….. (1)
5. Consider a body at a point P at the depth d below the surface of the Earth as shown in figure.
   
   Here the force on a body at P due to outer spherical shell shown by shaded region, cancel out due to symmetry.
   The net force on P is only due to the inner sphere of radius OP = R – d.
6. Acceleration due to gravity because of this sphere is,
   
   This equation gives acceleration due to gravity at depth d below the Earth’s surface.

Question 12.
State Kepler’s three laws of planetary motion.
Answer:

• All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.
• The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
• The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.

Question 13.
State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’ Hence show that their ratio is equal to \(\left(\frac{R-d}{R-2 h}\right)\) by assuming that the altitude is very small as compared to the radius of the Earth.
Answer:

1. For an object at depth d, acceleration due to gravity of the Earth is given by,
   g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) ………………. (1)
2. Also, the acceleration due to gravity at smaller altitude h is given by,
   g_h = g(1 – \(\frac{2 \mathrm{~h}}{\mathrm{R}}\)) ……………. (2)
3. Hence, dividing equation (1) by equation (2),
   we get,
   

Question 14.
What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?
Answer:
The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbIt round the Earth is called the critical velocity or orbital velocity (v_c).
Expression for critical velocity:

1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
2. If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity v_c.
3. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
   ∴ Centripetal force = Gravitational force
   
   This is the expression for critical speed at the orbit of radius (R + h).
4. The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.

Question 15.
Define escape speed. Derive an expression for the escape speed of an object from the surface of the earth.
Answer:

1. The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (v_e) of the body.
2. As the gravitational force due to Earth becomes zero at infinite distance, the object has to reach infinite distance in order to escape.
3. Let us consider the kinetic and potential energies of an object thrown vertically upwards with escape velocity v_e.
4. On the surface of the Earth,
   K.E.= \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}\)
   P.E. = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
   Total energy = P.E. + K.E.
   ∴ T.E. = \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}\) ………………. (1)
5. The kinetic energy of the object will go on decreasing with time as it is pulled back by Earth’s gravitational force. It will become zero when it reaches infinity. Thus, at infinite distance from the Earth,
   K.E. = 0
   Also,
   P.E. = –\(\frac{\mathrm{GMm}}{\infty}\) = 0
   ∴ Total energy = P.E. + K.E. = 0
6. As energy is conserved
   \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}=0\) ……[From(1)]
   or, v_e = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)

Question 16.
Describe how an artificial satellite using two stage rocket is launched in an orbit around the Earth.
Answer:

1. Launching of a satellite in an orbit around the Earth cannot take place by use of single stage rocket. It requires minimum two stage rocket.
2. With the help of first stage of rocket, satellite can be taken to a desired height above the surface of the Earth.
3. Then the launcher is rotated in horizontal direction i.e.. through 900 using remote control and the first stage of the rocket is detached.
4. With the help of second stage of rocket, a specific horizontal velocity (v_h) is given to satellite so that it can revolve in a circular path around the Earth.
5. The satellite follows different paths depending upon the horizontal velocity provided to it.

4. Solve the following problems.

Question 1.
At what distance below the surface of the Earth, the acceleration due to gravity decreases by 10% of its value at the surface, given radius of Earth is 6400 km.
Solution:
Given: g_d = 90% of g i.e., \(\frac{\mathrm{g}_{\mathrm{d}}}{\mathrm{g}}\) = 0.9,
R = 6400km = 6.4 × 10⁶ m
To find: Distance below the Earth’s surface (d)

At distance 640 km below the surface of the Earth, value of acceleration due to gravity decreases by 10%.

Question 2.
If the Earth were made of wood, the mass of wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate escape speed from the surface of this Earth.
Solution:

As, we know that the escape speed from surface of the Earth is 11.2 km/s, Substituting value of v_e = 11.2 km/s
V_ew = 11.2 × \(\frac{1}{\sqrt{10}}=\frac{11.2}{3.162}\)
= 11.2 × \(\frac{1}{3.162}\)
…………… [Taking square root value]
= antilog {log(1 1.2) –  Log(3.162)}
= antilog {1.0492 – 0.5000}
= antilog {0.5492} = 3.542
∴ V_ew = 3.54km/s
The escape velocity from the surface of wooden Earth is 3.54 km/s.

Question 3.
Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth.
Given:- G = 6.67 × 10^-11 Nm²/kg²
R = 6400 km
M = 6 × 10²⁴ kg
Solution:
Given:- m = 2000 kg, h = 3600 km = 3.6 × 10⁶ m,
G = 6.67 × 10^-11 Nm²/kg²
R = 6400 km
M = 6 × 10²⁴ kg

To find: i) Kninetic energy (K.E.)
ii) Potential Energy (P.E.)
iii) Total Energy (T.E.)
iv) Binding Energy (B.E.)

From formula (ii),
P.E. = -2 × 40.02 × 10⁹
= -80.04 × 10⁹ J
From formula (iii),
T.E. = (40.02 × 10⁹) + (-80.02 × 10⁹)
= -40.02 × 10⁹ J
From formula (iv),
B.E.= -(-40.02 × 10⁹)
= 40.02 × 10⁹ J
Kinetic energy of the satellite is 40.02 × 10⁹ J, potential energy is -80.04 × 10⁹ J, total energy is -40.02 × 10⁹ J and binding energy is 40.02 × 10⁹ J.
[Note: Total energy of orbiting satellite is negative.]

Question 4.
Two satellites A and B are revolving around a planet. Their periods of revolution are 1 hour and 8 hours respectively. The radius of orbit of satellite B is 4 × 10⁴ km. find radius of orbit of satellite A .
Solution:
Given: T_A = 1 hour, T_B = 8 hour,
r_B = 4 × 10⁴ km
To find: Radius of orbit of satellite A (r_A)

Radius of orbit of satellite A will be 1 × 10⁴ km.

Question 5.
Find the gravitational force between the Sun and the Earth.
Given Mass of the Sun = 1.99 × 10³⁰ kg
Mass of the Earth = 5.98 × 10²⁴ kg
The average distance between the Earth and the Sun = 1.5 × 10¹¹ m.
Solution:
Given: M_S = 1.99 × 10³⁰ kg
M_E = 5.98 × 10²⁴ kg, R = 1.5 × 10¹¹ m.
To find: Gravitational force between the Sun and the Earth (F)
Formula: F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
Calculation:As, we know, G = 6.67 × 10^-11 N m²/kg²
From formula,

= antilog {(log(6.67) + log( 1.99) + log(5.98) – log(2.25)} × 10²¹
= antilog {(0.8241) + (0.2989) + (0.7767) – (0.3522)} × 10²¹
= antilog {1.5475} × 10²¹
= 35.28 × 10²¹
= 3.5 × 10²² N
The gravitational force between the Sun and the Earth is 3.5 × 10²² N.

Question 6.
Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 × 10²⁴ kg, R = 6400 km).
Solution:
Given: h = 300 km = 0.3 × 10⁶ m,
M = 5.98 × 10²⁴ kg,
R = 6400km = 6.4 × 10⁶ m
G = 6.67 × 10^-11 Nm²/kg²
To find: Acceleration due to gravity at height (g_h)
Formula: g_h = \(\frac{G M}{(R+h)^{2}}\)

Calculation: From formula,
g_h = \(\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{\left[\left(6.4 \times 10^{6}\right)+\left(0.3 \times 10^{6}\right)\right]^{2}}\)
= \(\frac{6.67 \times 5.98 \times 10^{13}}{(6.7)^{2} \times 10^{12}}\)
6.67 X 10” x 5.98 X iO
= antilog {log(6.67) + log(5.98) – 2log(6.7)} × 10
= antilog{0.8241 + 0.7767 – 2(0.8261)} × 10
= antilog {1.6008 – 1.6522} × 10
= antilog {\(\overline{1}\) .9486} × 10
= 0.8884 × 10 = 8.884 m/s²
Acceleration due to gravity at 300 km will be 8.884 m/s².

Question 7.
Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface. M_E = 5.98 × 10²⁴ kg, R = 6.4 × 10⁶ m.
Solution:
Given: h = 1000 km = 1 × 10⁶ m,
ME = 5.98 × 10²⁴ kg, R = 6.4 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²
To find: Speed of satellite (v_c)

Speed of the satellite at height 1000 km is 7.34 × 10³ m/s.

Question 8.
Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars = 3.4 × 10³ km and its mass is 6.4 × 10²³ kg.
Solution:
Given:
M = 6.4 × 10²³ kg
R = 3.4 × 10³ = 3.4 × 10⁶ m,
To find: Acceleration due to gravity on the surface of the Mars (g_M)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: As, G = 6.67 × 10^-11 N m²/kg²
From formula,
g_M = \(\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{\left(3.4 \times 10^{6}\right)^{2}}=\frac{6.67 \times 6.4}{3.4 \times 3.4}\)
= antilog {log(6.67) + log(6.4) – log(3.4) – log(3.4)}
= antilog {(0.8241) + (0.8062) – (0.5315) – (0.53 15)}
= antilog {0.5673}
= 3.693 m/s²
Acceleration due to gravity on the surface of Mars is 3.693 m/s².

Question 9.
A planet has mass 6.4 × 10²⁴ kg and radius 3.4 × 10⁶ m. Calculate energy required to remove on object of mass 800 kg from the surface of the planet to infinity.
Solution:
Given: M = 6.4 × 10²⁴ kg, R = 3.4 × 10⁶ m, m = 800 kg
To find:   Energy required to remove the object from surface of planet to infinity = B.E.
Formula:    B.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Calculation: We know that,
G = 6.67 × 10^-11 N m²/kg²
From formula,

= antilog{log(6.67) + log(51.2) – log(3.4)} × 10⁹
= antilog{0.8241 + 1.7093 – 0.5315} × 10⁹
= antilog {2.0019} × 10⁹
= 1.004 × 10² × 10⁹
= 1.004 × 10¹¹ J
Energy required to remove the object from the surface of the planet is 1.004 × 10¹¹ J.
[Note: Answer calculated above ¡s in accordance with retual methods of calculation.]

Question 10.
Calculate the value of the universal gravitational constant from the given data. Mass of the Earth = 6 × 10²⁴ kg, Radius of the Earth = 6400 km and the acceleration due to gravity on the surface = 9.8 m/s²
Solution:
Given: M = 6 × 10²⁴ kg,
R = 6400km = 6.4 × 10⁶ m,
g = 9.8 m/s²
To find: Gravitational constant (G)
Formula. g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,
G = \(\frac{\mathrm{gR}^{2}}{\mathrm{M}}\)
G = \(\frac{9.8 \times\left(6.4 \times 10^{6}\right)^{2}}{6 \times 10^{24}}=\frac{401.4 \times 10^{12}}{6 \times 10^{24}}\)
∴ G = 6.69 × 10^-11 N m²/kg²
The value of gravitational constant is 6.69 × 10^-11 N m²/kg².

Question 11.
A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 1/7 times the mass of the Earth and radius twice that of the Earth’s radius.
Solution:
Given: W_E = 5.6 kg-wt.,
\(\frac{\mathrm{M}_{\mathrm{p}}}{\mathrm{M}_{\mathrm{E}}}=\frac{1}{7}, \frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{E}}}\) = 2
To find: Weight of the body on the surface of planet (W_p)
Formula: W = mg = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
Calculation: From formula,

Weight of the body on the surface of a planet will be 0.2 kg-wt.
[Note: The answer given above is calculated in accordance with textual method considering the given data].

Question 12.
What is the gravitational potential due to the Earth at a point which is at a height of 2R_E above the surface of the Earth, Mass of the Earth is 6 × 10²⁴ kg, radius of the Earth = 6400 km and G = 6.67 × 10^-11 Nm² kg^-2.
Solution:
Given: M = 6 × 10²⁴ kg,
R_E = 6400km = 6.4 × 10⁶ m,
G = 6.67 × 10^-11 Nm²/kg²,
h = 2R_E
To find: Gravitational potential (V)
Formula: V = – \(\frac{\mathrm{GM}}{\mathrm{r}}\)
Calculation: From formula,

= -2.08 × 10⁷ J kg^-1
Negative sign indicates the attractive nature of gravitational potential.
Gravitational potential due to Earth will be 2.08 × 10⁷ J kg^-1 towards the centre of the Earth.
[Note: According lo definition of gravitational potential its SI unit is J/kg.]

11th Physics Digest Chapter 5 Gravitation Intext Questions and Answers

Can you recall? (Textbook Page No. 78)

Question 1.
i) What are Kepler’s laws?
ii)What is the shape of the orbits of planets?
Answer:

1. The Kepler’s laws are:
   • Kepler’s first law: The orbit of a planet is an ellipse with the Sun at one of the foci.
   • Kepler’s second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
   • Kepler’s third law: The square of orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
2. The orbits of the planet are elliptical in shape.

Question 2.
When released from certain height why do objects tend to fall vertically downwards?
Answer:
When released from certain height, objects tend to fall vertically downwards because of the gravitational force exerted by the Earth.

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 4 Laws of Motion Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 4 Laws of Motion

1. Choose the correct answer.

Question 1.
Consider the following pair of forces of equal magnitude and opposite directions:
(P) Gravitational forces are exerted on each other by two-point masses separated by a distance.
(Q) Couple of forces are used to rotate a water tap.
(R) Gravitational force and normal force are experienced by an object kept on a table.
For which of these pair/pairs the two forces do NOT cancel each other’s translational
effect?
(A) Only P
(B) Only P and Q
(C) Only R
(D) Only Q and R
Answer:
(A) Only P

Question 2.
Consider the following forces: (w) Force due to tension along a string, (x) Normal force given by a surface, (y) Force due to air resistance, and (z) Buoyant force or upthrust given by a fluid.
Which of these are electromagnetic forces?
(A) Only w, y, and z
(B) Only w, x, and y
(C) Only x, y and z
(D) All four.
Answer:
(D) All four.

Question 3.
At a given instant three-point masses m, 2m, and 3m are equidistant from each other. Consider only the gravitational forces between them. Select correct statement/s for this instance only:
(A) Mass m experiences maximum force.
(B) Mass 2m experiences a maximum force.
(C) Mass 3m experiences a maximum force.
(D) All masses experience a force of the same magnitude.
Answer:
(C) Mass 3m experiences a maximum force.

Question 4.
The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along with the table, which is proportional to the resultant normal force given by the table. Forces F₁ and F₂ act at the same angle θ with the horizontal and both are just initiating the sliding motion of the block along with the table. Force F₁ is a pulling force while the force F₂ is a pushing force. F₂ > F₁, because
(A) Component of F₂ adds up to weight to increase the normal reaction.
(B) Component of F₁ adds up to weight to increase the normal reaction.
(C) Component of F₂ adds up to the opposing force.
(D) Component of F₁ adds up to the opposing force.
Answer:
(A) Component of F2 adds up to weight to increase the normal reaction.

Question 5.
A mass 2m moving with some speed is directly approaching another mass m moving with double speed. After some time, they collide with coefficient of restitution 0.5. Ratio of their respective speeds after collision is
(A) 2/3
(B) 3/2
(C) 2
(D) ½
Answer:
(B) 3/2

Question 6.
A uniform rod of mass 2m is held horizontal by two sturdy, practically inextensible vertical strings tied at its ends. A boy of mass 3m hangs himself at one third length of the rod. Ratio of the tension in the string close to the boy to that in the other string is
(A) 2
(B) 1.5
(C) 4/3
(D) 5/3
Answer:
(B) 1.5

Question 7.
Select WRONG statement about centre of mass:
(A) Centre of mass of a ‘C’ shaped uniform rod can never be a point on that rod.
(B) If the line of action of a force passes through the centre of mass, the moment of that force is zero.
(C) Centre of mass of our Earth is not at its geometrical centre.
(D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.
Answer:
(D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.

Question 8.
For which of the following objects will the centre of mass NOT be at their geometrical centre?
(I) An egg
(II) a cylindrical box full of rice
(III) a cubical box containing assorted sweets
(A) Only (I)
(B) Only (I) and (II)
(C) Only (III)
(D) All, (I), (II) and (III).
Answer:
(D) All, (I), (II) and (III).

2. Answer the following questions.

Question 1.
In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write respectively against A, B, C and D.

Answer:

1. Force due to tension in string: Electromagnetic (EM) force, reaction force, non-conservative force.
2. Normal force: Electromagnetic (EM) force, non-conservative force. Reaction force
3. Frictional force: Electromagnetic (EM) force, reaction force, non-conservative force.
4. Resistive force offered by air or water for objects moving through it: Electromagnetic (EM) force, non-conservative force.

Question 2.
In real life objects, never travel with uniform velocity, even on a horizontal surface, unless something is done? Why
is it so? What is to be done?
Answer:

1. According to Newton’s first law, for a body to achieve uniform velocity, the net force acting on it should be zero.
2. In real life, a body in motion is constantly being acted upon by resistive or opposing force like friction, in the direction opposite to that of the motion.
3. To overcome these opposing forces, an additional external force is required. Thus, the net force is not maintained at zero, making it hard to achieve uniform velocity.

Question 3.
For the study of any kind of motion, we never use Newton’s first law of motion directly. Why should it be studied?
Answer:

1. Newton’s first law shows an equivalence between the ‘state of rest’ and ‘state of uniform motion along a straight line.’
2. Newton’s first law of motion defines force as a physical quantity that brings about a change in ‘state of rest’ or ‘state of .uniform motion along a straight line’ of a body.
3. Newton’s first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line. Due to all these reasons, Newton’s first law should be studied.

Question 4.
Are there any situations in which we cannot apply Newton’s laws of motion? Is there any alternative for it?
Answer:

1. Limitation: Newton’s laws of motion cannot be applied for objects moving in non-inertial (accelerated) frame of reference.
   Alternative solution: For non-inertial (accelerated) frame of reference, pseudo force needs to be considered along with all the other forces.
2. Limitation: Newton’s laws of motion are applicable to point objects and rigid bodies. Alternative solution: Body needs to be approximated as a particle as the laws can be applied to individual particles in a rigid body and then summed up over the body.
3. Limitation: Newton’s laws of motion cannot be applied for objects moving with speeds comparable to that of light.
   Alternative solution: Einstein’s special theory of relativity has to be used.
4. Limitation: Newton’s laws of motion cannot be applied for studying the behaviour and interactions of objects having atomic or molecular sizes.
   Alternative solution: Quantum mechanics has to be used.

Question 5.
You are inside a closed capsule from where you are not able to see anything about the outside world. Suddenly you feel that you are pushed towards your right. Can you explain the possible cause (s)? Is it a feeling or a reality? Give at least one more situation like this.
Answer:

1. In a capsule, if we suddenly feel a push towards the right it is because the capsule is in motion and taking a turn towards the left.
2. The push towards the right is a feeling. In reality, when the capsule is beginning its turning motion towards the left, we continue in a straight line.
3. This happens because we try to maintain our direction of motion while the capsule takes a turn towards the left.
4. An external force is required to change our direction of motion. In accordance with one of the inferences from Newton’s first law of motion, in the absence of any external force, we continue to move in a straight line at constant speed and feel the sudden push in the direction opposite to the motion of the capsule.
5. Example: While travelling by bus, when the bus takes a sudden turn we feel the push in the opposite direction.

Question 6.
Among the four fundamental forces, only one force governs your daily life almost entirely. Justify the statement by stating that force.
Answer:

1. Electromagnetic force is the attractive and repulsive force between electrically charged particles.
2. Since electromagnetic force is much stronger than the gravitational force, it dominates all the phenomena on atomic and molecular scales.
3. Majority of the forces experienced in our daily life like friction, normal reaction, tension in strings, elastic forces, viscosity etc. are electromagnetic in nature.
4. The structure of atoms and molecules, the dynamics of chemical reactions etc. are governed by electromagnetic forces.

Thus, out of the four fundamental forces, electromagnetic force governs our daily life almost entirely.

Question 7.
Find the odd man out:
(i) Force responsible for a string to become taut on stretching
(ii) Weight of an object
(iii) The force due to which we can hold an object in hand.
Answer:
Weight of an object.
Reason: Weight of an object (force due to gravity) is a non-contact force while force responsible for a string to become taut (tension force) and force due to which we can hold an object in hand (normal force) are contact forces.

Question 8.
You are sitting next to your friend on ground. Is there any gravitational force of attraction between you two? If so, why are you not coming together naturally? Is any force other than the gravitational force of the earth coming in picture?
Answer:

1. Yes, there exists a gravitational force between me and my friend sitting beside each other.
2. The gravitational force between any two objects is given by, \(\overrightarrow{\mathrm{F}}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\) Where,
   G = universal gravitational constant, m₁ and m₂ = mass of the two objects, r = distance between centres of the two objects
3. Thus, me and my friend attract each other. But due to our small masses, we exert a force on each other, which is too small as compared to the gravitational force of the earth. Hence, me and my friend don’t move towards each other.
4. Apart from gravitational force of the earth, there is the normal force and frictional force acting on both me and my friend.

In Chapter 5, you will study about force of gravitation in detail.

Question 9.
Distinguish between:
(A) Real and pseudo forces,
(B) Conservative and non-conservative forces,
(C) Contact and non-contact forces,
(D) Inertial and non-inertial frames of reference.
Answer:
(A) Real and pseudo forces,

No	Real force	Pseudo Force
i.	A force which is produced due to interaction between the objects is called real force.	A pseudo force is one which arises due to the acceleration of the observer’s frame of reference.
ii.	Real forces obey Newton’s laws of motion.	Pseudo forces do not obey Newton’s laws of motion.
iii.	Real forces are one of the four fundamental forces.	Pseudo forces are not among any of the four fundamental forces.
	Example: The earth revolves around the sun in circular path due to gravitational force of attraction between the sun and the earth.	Example: Bus is moving with an acceleration (a) on a straight road in forward direction, a person of mass ‘m’ experiences a backward pseudo force of magnitude ‘ma’.

(B) Conservative and non-conservative forces,

No	Conservative	Non-conservative forces
i.	If work done by or against a force is independent’ of the actual path, the force is said to be a conservative force.	If work done by or against a force is dependent of the actual path, the force is said to be a non- conservative force.
ii.	During work done by a conservative force, the mechanical energy is conserved.	During work done by a non­ conservative force, the mechanical energy may not be conserved.
iii.	Work done is completely recoverable.	Work done is not recoverable.
	Example: gravitational force, magnetic force etc.	Example: Frictional force, air drag etc.

(C) Contact and non-contact forces,

No	Contact forces	Non-contact forces
i.	The forces experienced by a body due to physical contact are called contact forces.	The forces experienced by a body without any physical contact are called non-contact forces.
ii.	Example: gravitational force, electrostatic force, magnetostatic force etc.	Example: Frictional force, force exerted due to collision, normal reaction etc.

(D) Inertial and non-inertial frames of reference.

No.	Inertial frame of reference	Non-inertial frame of reference
i	The body moves with a constant velocity (can be zero).	The body moves with variable velocity.
ii.	Newton’s laws are	Newton’s laws are
iii.	The body does not accelerate.	The body undergoes acceleration.
iv.	In this frame, force acting on a body is a real force.	The acceleration of the frame gives rise to a pseudo force.
	Example: A rocket in inter-galactic space (gravity free space between galaxies) with all its engine shut.	Example: If a car just starts its motion from rest, then during the time of acceleration the car will be in a non- inertial frame of reference.

Question 10.
State the formula for calculating work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly. Is there any mathematical way out for it? Explain.
Answer:

1. Suppose a constant force \(\overrightarrow{\mathrm{F}}\) acting on a body produces a displacement \(\overrightarrow{\mathrm{S}}\) in the body along the positive X-direction. Then the work done by the force is given as,
   W = F.s cos θ
   Where θ is the angle between the applied force and displacement.
2. If displacement is in the direction of the force applied, θ = 0°
   W = \(\overrightarrow{\mathrm{F}}\).\(\overrightarrow{\mathrm{s}}\)

Conditions/limitations for application of work formula:

1. The formula for work done is applicable only if both force \(\overrightarrow{\mathrm{F}}\) and displacement \(\overrightarrow{\mathrm{s}}\) are constant and finite i.e., it cannot be applied when the force is variable.
2. The formula is not applicable in several real- life situations like lifting an object through several thousand kilometres since the gravitational force is not constant. It is not applicable to viscous forces like fluid resistance as they depend upon speed and thus are often not constant with time.
3. The method of integration has to be applied to find the work done by a variable force.

Integral method to find work done by a variable force:

1. Let the force vary non-linearly in magnitude between the points A and B as shown in figure (a).
2. In order to calculate the total work done during the displacement from s₁ to s₂, we need to use integration. For integration, we need to divide the displacement into large numbers of infinitesimal (infinitely small) displacements.
3. Let at P₁, the magnitude of force be F = P₁P₁‘. Due to this force, the body displaces through infinitesimally small displacement ds, in the direction of force.
   It moves from P₁ to P₂.
   ∴ \(\mathrm{d} \overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{P}_{1} \mathrm{P}_{2}}\)
4. But direction of force and displacement are same, we have
   \(\mathrm{d} \overrightarrow{\mathrm{s}}=\mathrm{P}_{1}{ }^{\prime} \mathrm{P}_{2}^{\prime}\)
5. \(\mathrm{d} \overrightarrow{\mathrm{s}}\) is so small that the force F is practically constant for the displacement. As the force is constant, the area of the strip \(\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}\) is the work done dW for this displacement.
6. Hence, small work done between P₁ to P₂ is dW and is given by
   dW = \(\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}\) = \(\mathrm{P}_{1} \mathrm{P}_{1}^{\prime} \times \mathrm{P}_{1}^{\prime} \mathrm{P}_{2}^{\prime}\)
   = Area of the strip P₁P₂P₂‘P₁‘.
7. The total work done can be found out by dividing the portion AB into small strips like P₁P₂P₂‘P₁‘ and taking sum of all the areas of the strips.
   ∴ W = \int_{s_{1}}^{s 2} \vec{F} \cdot d \vec{s}=\text { Area } A B B^{\prime} A^{\prime}\(\)
8. Method of integration is applicable if the exact way of variation in \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{s}}\) is known and that function is integrable.
9. The work done by the non-linear variable force is represented by the area under the portion of force-displacement graph.
10. Similarly, in case of a linear variable force, the area under the curve from s₁ to s₂ (trapezium APQB) gives total work done W [figure (b)].
    

Question 11.
Justify the statement, “Work and energy are the two sides of a coin”.
Answer:

1. Work and energy both are scalar quantities.
2. Work and energy both have the same dimensions i.e., [M¹L²T^-2].
3. Work and energy both have the same units i.e., SI unit: joule and CGS unit: erg.
4. Energy refers to the total amount of work a body can do.
5. A body capable of doing more work possesses more energy and vice versa.
6. Work done on a body by a conservative force is equal to the change in its kinetic energy.

Thus, work and energy are the two sides of the same

Question 12.
From the terrace of a building of height H, you dropped a ball of mass m. It reached the ground with speed v. Is the relation mgH = \(\frac{1}{2} m \mathrm{v}^{2}\) applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped?
Answer:

1. Let the ball dropped from the terrace of a building of height h have mass m. During free fall, the ball is acted upon by gravity (accelerating conservative force).
2. While coming down, the work that is done is equal to the decrease in the potential energy.
3. This work done however is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance (retarding non-conservative force). This part of energy appears in some other forms such as heat, sound, etc.
4. Thus, in this case of an accelerating conservative force along with a retarding non-conservative force, the work-energy theorem is given as, Decrease in the gravitational
   P.E. = Increase in the kinetic energy + work done against non-conservative forces.
5. Thus, the relation mgh = \(\frac{1}{2} \mathrm{mv}^{2}[latex] is not applicable when non-conservative forces are considered. The part of the energy converted to heat, sound etc also needs to be added to the equation,
6. The ball will not bounce to the same height from where it was dropped due to the loss in kinetic energy during the collision making it an inelastic collision.

Question 13.
State the law of conservation of linear momentum. It is a consequence of which law? Given an example from our daily life for conservation of momentum. Does it hold good during burst of a cracker?
Answer:

1. Statement: The total momentum of an isolated system is conserved during any interaction.
2. The law of conservation of linear momentum is a consequence of Newton’s second law of motion, (in combination with Newton’s third law)
3. Example: When a nail is driven into a wall by striking it with a hammer, the hammer is seen to rebound after striking the nail. This is because the hammer imparts a certain amount of momentum to the nail and the nail imparts an equal and opposite amount of momentum to the hammer.
   Linear momentum conservation during the burst of a cracker:
   • The law of conservation of linear momentum holds good during bursting of a cracker.
   • When a cracker is at rest before explosion, the linear momentum of the cracker is zero.
   • When cracker explodes into number of pieces, scattered in different directions, the vector sum of linear momentum of these pieces is also zero. This is as per the law of conservation of linear momentum.

Question 14.
Define coefficient of restitution and obtain its value for an elastic collision and a perfectly inelastic collision.
Answer:

i. For two colliding bodies, the negative of ratio of relative velocity of separation to relative velocity of approach is called the coefficient of restitution.

ii. Consider an head-on collision of two bodies of masses m₁ and m₂ with respective initial velocities u₁ and u₂. As the collision is head on, the colliding masses are along the same line before and after the collision. Relative velocity of approach is given as,
u_a = u₂ – u₁
Let v₁ and v₂ be their respective velocities after the collision. The relative velocity of recede (or separation) is then v_s = v₂ – v₁

iii. For a head on elastic collision, According to the principle of conservation of linear momentum,
Total initial momentum = Total final momentum

iv. For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e., v₁ = v₂
∴ v₁ – v₂ = 0
Substituting this in equation (1), e = 0.

Question 15.
Discuss the following as special cases of elastic collisions and obtain their exact or approximate final velocities in terms
of their initial velocities.
(i) Colliding bodies are identical.
(ii) A very heavy object collides on a lighter object, initially at rest.
(iii) A very light object collides on a comparatively much massive object, initially at rest.
Answer:
The final velocities after a head-on elastic collision is given as,

i. Colliding bodies are identical
If m₁ = m₂, then v₁ = u₂ and v₂ = u₁
Thus, objects will exchange their velocities after head on elastic collision.

ii. A very heavy object collides with a lighter object, initially at rest.
Let m₁ be the mass of the heavier body and m₂ be the mass of the lighter body i.e., m₁ >> m₂; lighter particle is at rest i.e., u₂ = 0 then,

i.e., the heavier colliding body is left unaffected and the lighter body which is struck, travels with double the speed of the massive striking body.

iii. A very light object collides on a comparatively much massive object, initially at rest.
If m₁ is mass of a light body and m₂ is mass of heavy body i.e., m₁ << m₂ and u₂ = 0. Thus, m₁ can be neglected.
Hence v₁ ≅ -u₁, and v₂ ≅ 0.
i.e., the tiny (lighter) object rebounds with same speed while the massive object is unaffected.

Question 16.
A bullet of mass m₁ travelling with a velocity u strikes a stationary wooden block of mass m₂ and gets embedded into it. Determine the expression for loss in the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss?
Answer:

1. A bullet of mass m₁ travelling with a velocity u, striking a stationary wooden block of mass m₂ and getting embedded into it is a case of perfectly inelastic collision.
2. In a perfectly inelastic collision, although there is a loss in kinetic energy, the principle of conservation of energy is not violated as the total energy of the system is conserved.

Loss in the kinetic energy during a perfectly inelastic head on collision:

1. Let two bodies A and B of masses m₁ and m₂ move with initial velocity [latex]\overrightarrow{\mathrm{u}_{1}}\), and \(\overrightarrow{\mathrm{u}_{2}}\) respectively such that particle A collides head- on with particle B i.e., u₁ > u₂.
2. If the collision is perfectly inelastic, the particles stick together and move with a common velocity \(\overrightarrow{\mathbf{V}}\) after the collision along the same straight line.
   loss in kinetic energy = total initial kinetic energy – total final kinetic energy,
3. By the law of conservation of momentum, m₁u₁ + m₂u₂ = (m₁ + m₂) v
   ∴ v = \(\frac{\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
4. Loss of Kinetic energy,
   
5. Both the masses and the term (u₁ – u₂)² are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as e = 0, the loss is maximum.

Question 17.
One of the effects of a force is to change the momentum. Define the quantity related to this and explain it for a variable
force. Usually when do we define it instead of using the force?
Answer:

1. Impulse is the quantity related to change in momentum.
2. Impulse is defined as the change of momentum of an object when the object is acted upon by a force for a given time interval.

Need to define impulse:

1. In cases when time for which an appreciable force acting on an object is extremely small, it becomes difficult to measure the force and time independently.
2. In such cases, however, the effect of the force i.e, the change in momentum due to the force is noticeable and can be measured.
3. For such cases, it is convenient to define impulse itself as a physical quantity.
4. Example: Hitting a ball with a bat, giving a kick to a foot-ball, hammering a nail, bouncing a ball from a hard surface, etc.

Impulse for a variable force:

1. Consider the collision between a bat and ball. The variation of the force as a function of time is shown below. The force axis is starting from zero.
2. From the graph, it can be seen that the force is zero before the impact. It rises to a maximum during the impact and decreases to zero after the impact.
3. The shaded area or the area under the curve of the force -time graph gives the product of force against the corresponding time (∆t) which is the impulse of the force.
   Area of ABCDE = F. ∆t = impulse of force
4. For a constant force, the area under the curve is a rectangle.
5. In case of a softer tennis ball, the collision time becomes larger and the maximum force becomes less keeping the area under curve of the (F – t) graph same.
   Area of ABCDE = Area of PQRST

In chapter 3, you have studied the concept of using area under the curve.

Question 18.
While rotating an object or while opening a door or a water tap we apply a force or forces. Under which conditions is this process easy for us? Why? Define the vector quantity concerned. How does it differ for a single force and for two opposite forces with different lines of action?
Answer:

1. Opening a door can be done with ease if the force applied is:
   • proportional to the mass of the object
   • far away from the axis of rotation and the direction of force is perpendicular to the line joining the axis of rotation with the point of application of force.
2. This is because, the rotational ability of a force depends not only upon the magnitude and direction of force but also on the point where the force acts with respect to the axis of rotation.
3. Rotating an object like a water tap can be done with ease if the two forces are equal in magnitude but opposite in direction are applied along different lines of action.
4. The ability of a force to produce rotational motion is measured by its turning effect called ‘moment of force’ or ‘torque’.
5. However, a moment of couple or rotational effect of a couple is also called torque.
6. For differences in the two vector quantities.

No.	Moment of a force	Moment of a couple
i.	Moment of a force is given as, \(\vec{\tau}=\vec{r} \times \vec{F}\)	Moment of a couple is given as, \(\vec{\tau}=\vec{r}_{12} \times \vec{F}_{1}=\vec{r}_{21} \times \vec{F}_{2}\)
ii.	It depends upon the axis of rotation and the point of application of the force.	It depends only upon the two forces, i.e., it is independent of the axis of rotation or the points of application of forces.
iii.	It can produce translational acceleration also, if the axis of rotation is not fixed or if friction is not enough.	Does not produce any translational acceleration, but produces only rotational or angular acceleration.
iv.	Its rotational effect can be balanced by a proper single force or by a proper couple.	Its rotational effect can be balanced only by another couple of equal and opposite torque.

Question 19.
Why is the moment of a couple independent of the axis of rotation even if the axis is fixed?
Answer:

1. Consider a rectangular sheet free to rotate only about a fixed axis of rotation, perpendicular to the plane.
2. A couple of forces \(\overrightarrow{\mathrm{F}}\) and –\(\overrightarrow{\mathrm{F}}\) is acting on the sheet at two different locations.
3. Consider the torque of the couple as two torques due to individual forces causing rotation about the axis of rotation.
4. Case 1: The axis of rotation is between the lines of action of the two forces constituting the couple. Let x and y be the perpendicular distances of the axis of rotation from the forces \(\overrightarrow{\mathrm{F}}\) and –\(\overrightarrow{\mathrm{F}}\) respectively.
   In this case, the pair of forces cause anticlockwise rotation. As a result, the direction of individual torques due to the two forces is the same.
   
5. Case 2: Lines of action of both the forces are on the same side of the axis of rotation. Let q and p be the perpendicular distances of the axis of rotation from the forces \(\overrightarrow{\mathrm{F}}\) and –\(\overrightarrow{\mathrm{F}}\)

Question 20.
Explain balancing or mechanical equilibrium. Linear velocity of a rotating fan as a whole is generally zero. Is it in
mechanical equilibrium? Justify your answer.
Answer:

1. The state in which the momentum of a system is constant in the absence of an external unbalanced force is called mechanical equilibrium.
2. A particle is said to be in mechanical equilibrium, if no net force is acting upon it.
3. In case of a system of bodies to be in mechanical equilibrium, the net force acting on any part of the system should be zero i.e., the velocity or linear momentum of all parts of the system must be constant or zero. There should be no acceleration in any part of the system.
4. Mathematically, for a system in mechanical equilibrium, \(\sum \vec{F}\) = 0.
5. In case of rotating fan, if linear velocity is zero, then the linear momentum is zero. That means there is no net force acting on the fan. Hence, the fan is in mechanical equilibrium.

Question 21.
Why do we need to know the centre of mass of an object? For which objects, its position may differ from that of the centre of gravity?
Use g = 10 m s^-2, unless, otherwise stated.
Answer:

1. Centre of mass of an object allows us to apply Newton’s laws of motion to finite objects (objects of measurable size) by considering these objects as point objects.
2. For objects in non-uniform gravitational field or whose size is comparable to that of the Earth (size at least few thousand km), the position of centre of mass will differ than that of centre of gravity.

3. Solve the following problems.

Question 1.
A truck of mass 5 ton is travelling on a horizontal road with 36 km hr^-1 stops on travelling 1 km after its engine fails suddenly. What fraction of its weight is the frictional force exerted by the road? If we assume that the story repeats for a car of mass 1 ton i.e., can moving with same speed stops in similar distance same how much will the fraction be?
[Ans: \(\frac{1}{200}\) in the both]
Solution:
Given: m_truck = 5 ton = 5000 kg,
m_car = 1 ton = 1000 kg,
u = 36km/hr = 10 m/s,
v = 0 m/s, s = 1 km = 1000 m
To find: Ratio of force of friction to the weight of vehicle
Formulae:
i. v² = u² + 2as
ii. F = ma
Calculation: From formula (i),
2 × a_truck × s = v² – u²
∴ 2 × a_truck × 1000 = 0² – 10²
∴ 200a_truck = -100
∴ a_truck = -0.05 m/s²
Negative sign indicates that velocity is decreasing
From formula (ii),
F_truck = m_truck × a_truck = 5000 × 0.05
= 250 N

Answer:
The frictional force acting on both the truck and the car is \(\frac{1}{200}\) of their weight.

Question 2.
A lighter object A and a heavier object B are initially at rest. Both are imparted the same linear momentum. Which will start with greater kinetic energy: A or B or both will start with the same energy?
[Ans: A]
Solution:

1. Let m₁ and m₂ be the masses of light object A and heavy object B and v₁ and v₂ be their respective velocities.
2. Since both are imparted with the same linear momentum,
   m₁ v₁ = m₂ v₂
3. Kinetic energy of the lighter object A
   
4. As m₁ < m₂, therefore K.E._A > K.E._B, i.e, the lighter body A has more kinetic energy.

Question 3.
As i was standing on a weighing machine inside a lift it recorded 50 kg wt. Suddenly for few seconds it recorded 45 kg wt. What must have happened during that time? Explain with complete numerical analysis. [Ans: Lift must be coming down with acceleration \(\frac{\mathrm{g}}{10}\) = 1 ms^-2]
Solution:
The weight recorded by weighing machine is always apparent weight and a measure of reaction force acting on the person. As the apparent weight (45 kg-wt) in this case is less than actual weight (50 kg-wt) the lift must be accelerated downwards during that time.

Numerical Analysis

1. Weight on the weighing machine inside the lift is recorded as 50 kg-wt
   ∴ mg = 50 kg-wt
   
2. This weight acts on the weighing machine which offers a reaction R given by the reading of the weighing machine
   ∴ R = 45kg-wt = \(\frac{9}{10}\)mg
3. The forces acting on person inside lift are as follows:
   • Weight mg downward (exerted by the earth)
   • Normal reaction (R) upward (exerted by the floor)
4. As, R < mg, the net force is in downward direction and given as,
   mg – R=ma
   But R = \(\frac{9}{10}\)mg.
   ∴ mg – \(\frac{9}{10}\)mg = ma
   ∴ \(\frac{mg}{10}\) = ma
   ∴ a = g/10
   ∴ a = 1 m/s² (∵ g = 1 m/s²)
5. Therefore, the elevator must be accelerated downwards with an acceleration of 1 m/s² at that time.

Question 4.
Figure below shows a block of mass 35 kg resting on a table. The table is so rough that it offers a self adjusting resistive force 10% of the weight of the block for its sliding motion along the table. A 20 kg wt load is attached to the block and is passed over a pulley to hang freely on the left side. On the right side there is a 2 kg wt pan attached to the block and hung freely. Weights of 1 kg wt each, can be added to the pan. Minimum how many and maximum how many such weights can be added into the pan so that the block does not slide along the table? [Ans: Min 15, maximum 21].

Solution:
Frictional (resistive) force f = 10% (weight)
= \(\frac{10}{100}\) × 35 × 10 = 35N 100

i. Consider FBD for 20 kg-wt load. Initially, the block kept on the table is moving towards left, because of the movement of block of mass 20 kg in downward direction.
Thus, for block of mass 20 kg,
ma = mg – T₁ …. (1)
Consider the forces acting on the block of mass 35 kg in horizontal direction only as shown in figure (b). Thus, the force equation for this block is, m₁a = T₁ – T₂ – f ….(2)
To prevent the block from sliding across the table,
m₁a = ma = 0
∴ T₁ = mg = 200 N ….[From (1)]
T₁ = T₂ + f ….[From (2)]
∴ T₂ + f = 200
∴ T₂ = 200 – 35 = 165 N
Thus, the total force acting on the block from right hand side should be 165 N.
∴ Total mass = 16.5 kg
∴ Minimum weight to be added = 16.5 – 2 = 14.5 kg
≈ 15 weights of 1 kg each

ii. Now, considering motion of the block towards right, the force equations for the masses in the pan and the block of mass 35 kg can be determined from FBD shown

From figure (c)
m₁a = T₂ – T₁ – f ….(iii)
From figure (d),
m₂a = m₂g – T₂ … .(iv)
To prevent the block of mass 35 kg from sliding across the table, m₁a = m₂a = 0
From equations (iii) and (iv),
T₂ = T₁ + f
T₂ = m₂g
∴ m₂g = 200 + 35 = 235 N
∴ The maximum mass required to stop the sliding = 23.5 – 2 = 21.5kg ≈ 21 weights of 1 kg
Answer:
The minimum 15 weights and maximum 21 weights of 1 kg each are required to stop the block from sliding.

Question 5.
Power is rate of doing work or the rate at which energy is supplied to the system. A constant force F is applied to a body
of mass m. Power delivered by the force at time t from the start is proportional to
(a) t
(b) t²
(c) \(\sqrt{t}\)
(d) t⁰
Derive the expression for power in terms of F, m and t.
[Ans: p = \(\frac{F^{2} t}{m}\), ∴ p ∝ t]
Solution:
Derivation for expression of power:

i. A constant force F is applied to a body of mass (m) initially at rest (u = 0).

ii. We have,
v = u + at
∴ v = 0 + at
∴ v = at …. (1)

iii. Now, power is the rate of doing work,
∴ P = \(\frac{\mathrm{d} \mathrm{W}}{\mathrm{d} \mathrm{t}}\)
∴ P = F. \(\frac{\mathrm{d} \mathrm{s}}{\mathrm{dt}}\) [∵ dW = F. ds]

iv. But \(\frac{\mathrm{d} \mathrm{s}}{\mathrm{dt}}\) = v, the instantaneous velocity of the particle.
∴ P = F.V … (2)

v. According to Newton’s second law,
F = ma … (3)

vi. Substituting equations (1) and (3) in equation (2)
P = (ma) (at)
∴ P = ma²t
∴ P = \(\frac{m^{2} a^{2}}{m}\) × t
∴ P = \(\frac{\mathrm{F}^{2}}{\mathrm{~m}} \mathrm{t}\)

vii. As F and m are constant, therefore, P ∝ t.

Question 6.
40000 litre of oil of density 0.9 g cc is pumped from an oil tanker ship into a storage tank at 10 m higher level than the ship in half an hour. What should be the power of the pump? [Ans: 2 kW]
Solution:
h = 10 m, ρ = 0.9 g/cc = 900 kg/m³, g = 10 m/s²,
V = 40000 litre = 40000 × 10³ × 10^-6 m³
= 40 m³
T = 30 min = 1800 s
To find: Power(P)
Formula: P = \(\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{h} \rho \mathrm{gV}}{\mathrm{t}}\)
Calculation: From formula,
P = \(\frac{10 \times 900 \times 10 \times 40}{1800}\)
∴ P = 2000 W
∴ P = 2 kW
Answer:
The power of the pump is 2 kW.

Question 7.
Ten identical masses (m each) are connected one below the other with 10 strings. Holding the topmost string, the system is accelerated upwards with acceleration g/2. What is the tension in the 6th string from the top (Topmost string being the first string)? [Ans: 6 mg]
Solution
Consider the 6^th string from the top. The number of masses below the 6^th string is 5. Thus, FBD for the 6^th mass is given in figure (b).

Answer:
Tension in the 6^th string is 7.5 mg.
[Note: The answer given above is modified considering the correct textual concepts.]

Question 8.
Two galaxies of masses 9 billion solar mass and 4 billion solar mass are 5 million light-years apart. If, the Sun has to cross the line joining them, without being attracted by either of them, through what point it should pass? [Ans: 3 million light-years from the 9 billion solar mass]
Solution:
The Sun can cross the line joining the two galaxies without being attracted by either of them if it passes from a neutral point. A neutral point is a point on the line joining two objects where the effect of gravitational forces acting due to both the objects is nullified.
Given that;
m₁ = 9 × 10⁹ M_s
m₂ = 4 × 10⁹ M_s
r = 5 × 10⁶ light years
Let the neutral point be at distance x from mi. If sun is present at that point,

Answer: The Sun has to cross the line from a point at a distance 3 million light years from the galaxy of mass 9 billion solar mass.

Question 9.
While decreasing linearly from 5 N to 3 N, a force displaces an object from 3 m to 5 m. Calculate the work done by this force during this displacement. [Ans: 8 N]
Solution:
For a variable force, work done is given by area under the curve of force v/s displacement graph. From given data, graph can be plotted as follows:

= 8 J
Ans: Work done is 8 J.
[Note: According to the definition of work done, S.J. unit of wõrk done is joule (J)]

Alternate solution:
Work done, w = Area of trapezium ADCB
∴ W = \(\frac{1}{2}\)(AD + CB) × DC
∴ W = 1 (5N + 3N) × (5m – 3m)
= \(\frac{1}{2}\) × 8 × 2 = 8J

Question 10.
Variation of a force in a certain region is given by F = 6x² – 4x – 8. It displaces an object from x = 1 m to x = 2 m in this region. Calculate the amount of work
done. [Ans: Zero]
Solution:

Answer:
The work done is zero.

Question 11.
A ball of mass 100 g dropped on the ground from 5 m bounces repeatedly. During every bounce 64% of the potential energy is converted into kinetic energy. Calculate the following:
(a) Coefficient of restitution.
(b) Speed with which the ball comes up from the ground after third bounce.
(c) Impulse given by the ball to the ground during this bounce.
(d) Average force exerted by the ground if this impact lasts for 250 ms.
(e) Average pressure exerted by the ball on the ground during this impact if the contact area of the ball is 0.5 cm².
[Ans: 0.8, 5.12 m/s, 1.152N s, 4.608 N, 9.216 × 104 N/m²]
Solution:
Given that, for every bounce, 64% of the initial energy is converted to final energy.
i. Coefficient of restitution in case of inelastic collision is given by,

ii. From equation (1),
∴ v = – eu
∴ After first bounce,
v₁ = – eu
after second bounce,
v₂ = -ev₁ = -e(-eu)= e²u
and after third bounce,
v₃ = – ev₂ = – e(e²u) = – e³u
But u = \(\sqrt{2 \mathrm{gh}}\)

iii. Impulse given by the ball during third bounce, is,

iv. Average force exerted in 250 ms,

v. Average pressure for area
0.5 cm² = 0.5 × 10^-4m²
P = \(\frac{\mathrm{F}}{\mathrm{A}}=\frac{4.608}{0.5 \times 10^{-4}}\) = 9.216 × 10⁴ N/m²

Question 12.
A spring ball of mass 0.5 kg is dropped from some height. On falling freely for 10 s, it explodes into two fragments of
mass ratio 1:2. The lighter fragment continues to travel downwards with speed of 60 m/s. Calculate the kinetic energy supplied during explosion. [Ans: 200 J]
Solution:
m₁ + m₂ = 0.5 kg, m₁ : m₂ = 1 : 2,
m₁ = \(\frac{1}{6}\) kg,
∴ m₂ = \(\frac{1}{3}\) kg
Initially, when the ball is falling freely for 10s,
v = u + at = 0 + 10(10)
∴ v = 100 m/s = u₁ = u₂
(m₁ + m₂)v = m₁v₁ + m₂v₂
∴ 0.5 × 100 = \(\frac{1}{6}\)(60) + \(\frac{1}{3}\)v₂
∴ 50 = 10 + \(\frac{1}{3}\)v₂
∴ 40 = \(\frac{1}{3}\)v₂
∴ v₂ = 120m/s

∴ K.E. = 200J.
Ans: Kinetic energy supplied is 200 J.

Question 13.
A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m with double speed. After collision, the heavier one travels with the average initial speed of the two. Calculate the coefficient of restitution. [Ans: 0.5]
Solution:
Given: m₁ = 2m, m₂ = m, u₁ = 6 cm/s,
u₂ = 2u₁ = 12 cm/s,
v₁ = \(\frac{\mathrm{u}_{1}+\mathrm{u}_{2}}{2}\) = 9cm/s
To find: Coefficient of restitution(e)
Formulae:

i. m₁u₁ +m₂u₂ = m₁v₁ + m₂v₂
ii. e = \(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
Calculation: From formula (i),
[(2m) × 6] + (m × 12) = (2m × 9) + mv₂
∴ v₂ = 6cm/s

From formula (ii),
e = \(\frac{6-9}{6-12}\) = \(\frac{-3}{-6}\) = 0.5

Answer: The coefficient of restitution is 0.5

Question 14.
A 2 m long wooden plank of mass 20 kg is pivoted (supported from below) at 0.5 m from either end. A person of mass 40 kg starts walking from one of these pivots to the farther end. How far can the person walk before the plank topples? [Ans: 1.25 m]
Solution:
Let the person starts walking from pivot P₂ as shown in the figure.

Assume the person can walk up to distance x from P₁ before the plank topples. The plank will topple when the moment exerted by the person about P₁ is not balanced by a moment of force due to plank about P₂.

∴ For equilibrium,
40 × x = 20 × 0.5
∴ x = \(\frac{1}{4}\) = 0.25 m
Hence, the total distance walked by the person is 1.25 m.

Question 15.
A 2 m long ladder of mass 10 kg is kept against a wall such that its base is 1.2 m away from the wall. The wall is smooth but the ground is rough. The roughness of the ground is such that it offers a maximum horizontal resistive force (for sliding motion) half that of normal reaction at the point of contact. A monkey of mass 20 kg starts climbing the ladder. How far can it climb along the ladder? How much is the horizontal reaction at the wall? [Ans: 1.5 m, 15 N]
Solution:

From the figure,
Given that, AC = length of ladder = 2 m
BC= 1.2m
From Pythagoras theorem,
AB = \(\sqrt{\mathrm{AC}^{2}-\mathrm{BC}^{2}}\) = 1.6 m … (i)
Also, ∆ABC ~ ∆DD’C
∴ \(\frac{\mathrm{AB}}{\mathrm{DD}^{\prime}}\) = \(\frac{\mathrm{BC}}{\mathrm{D}^{\prime} \mathrm{C}}\) = \(\frac{\mathrm{AC}}{\mathrm{DC}}\)
∴ \(\frac{1.2}{\mathrm{D}^{\prime} \mathrm{C}}=\frac{2}{1}\)
∴ D’C = 0.6 m … (ii)

The ladder exerts horizontal force \(\overrightarrow{\mathrm{H}}\) on the wall at A and \(\overrightarrow{\mathrm{F}}\) is the force exerted on the ground at C.
As |\(\overrightarrow{\mathrm{F}}\)| = \(|\overrightarrow{\mathrm{H}}|=|\overrightarrow{\mathrm{F}}|=\frac{\mathrm{N}}{2}\) … (iii)

Let monkey climb upto distance x along BC (Horizontal) i.e., CM’ = x .. . .(iv)
Then, the net normal reaction at point C will be, N = 100 + 200 = 300N
From equation (iii),
H = \(\frac{\mathrm{N}}{2}=\frac{300}{2}\) = 150N
By condition of equilibrium, taking moments about C,
(-H × AB) + (W₁ × CD’) + (W₂ × CM’) + (F × 0)’0
∴ (-150 × 1.6) + (100 × 0.6) + (200 × x) = 0
∴ 60 + 200x = 240
∴ 200x = 180
∴ x = 0.9

From figure, it can be shown that,
∆ABC ~ ∆MM’C
∴ \(\frac{\mathrm{BC}}{\mathrm{CM}^{\prime}}\) = \(\frac{\mathrm{AC}}{\mathrm{CM}^{\prime}}\) ∴ \(\frac{\mathrm{1.2}}{\mathrm{0.9}^{\prime}}\) = \(\frac{\mathrm{2}}{\mathrm{CM}^{\prime}}\)
∴ CM = 1.5 m

Answer:

1. The monkey can climb upto 1.5 m along the ladder.
2. The horizontal reaction at wall is 150 N.

Question 16.
Four uniform solid cubes of edges 10 cm, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order. Locate centre of mass of their system. [Ans: 65 cm, 17.7 cm]
Solution:

The given cubes are arranged as shown in figure. Let one of the comers of smallest cube lie at the origin.
As the cubes are uniform, let their centre of masses lie at their respective centres.
r_A \(\equiv\) (5, 5), r_B \(\equiv\) (20, 10), r_C \(\equiv\) = (45, 15) and r_D \(\equiv\) – (80, 20)
Also, masses of the cubes are,
m_A = \(l_{\mathrm{A}}^{3} \times \rho=10^{3} \rho\)
m_B = (20)³ρ
m_C = (30)³ρ
m_D = (40)³ρ
As the cubes are uniform, p is same for all of them.
∴ For X – co-ordinate of centre of mass of the system,

Similarly,
Y – co-ordinate of centre of mass of system is,

Answer: Centre of mass of the system is located at point (65 cm, 17.7 cm)

Question 17.
A uniform solid sphere of radius R has a hole of radius R/2 drilled inside it. One end of the hole is at the centre of the
sphere while the other is at the boundary. Locate centre of mass of the remaining sphere. [Ans: -R/14 ]
Solution:

Let the centre of the sphere be origin O. Then, r₁ be the position vector of centre of mass of uniform solid sphere and r₂ be the position vector of centre of mass of the cut-out part of the sphere.
Now, mass of the sphere is given as,

∴ Position vector of centre of mass of remaining part,

r_CM = \(\frac{-\mathbf{R}}{14}\)
(Negative sign indicates the distance is on left side of the origin.)
Ans: Position of centre of mass of remaining sphere \(\frac{-\mathbf{R}}{14}\)

Question 18.
In the following table, every item on the left side can match with any number of items on the right hand side. Select all those.

Answer:
i. Elastic collision: (b)
ii. Inelastic collision: (a), (c), (f), (e)
iii. Perfectly inelastic collision: (d)
iv. Head-on collision: (c), (f)

11th Physics Digest Chapter 4 Laws of Motion Intext Questions and Answers

Can you recall? (Textbook Page No. 47)

Question 1.
What are different types of motions?
Answer:
The various types of motion are linear, uniform linear, non-uniform linear, oscillatory, circular, periodic, and random motion.

Question 2.
What do you mean by kinematical equations and what are they?
Answer:
A set of three equations that analyses rectilinear motion of the uniformly accelerated body and help to predict the position of body are called kinematical equations.

1. Equation for velocity-time relation: v = u + at
2. Equation for position-time relation: s = ut + \(\frac{1}{2}\)at²
3. Position-velocity relation: v² = u² + 2as

Can you tell? (Textbook Page No. 48)

Question 1.
Was Aristotle correct? If correct, explain his statement with an illustration.
Answer:
Aristotle was not correct in stating that an external force is required to keep a body in uniform motion.

Question 2.
If wrong, give the correct modified version of his statement.
Answer:
For an uninterrupted motion of a body, an additional external force is required for overcoming opposing/resistive forces.

Can you tell? (Textbook Page No. 48)

Question 1.
What is then special about Newton’s first law if it is derivable from Newton’s second law?
Answer:

1. Newton’s first law shows an equivalence between the ‘state of rest’ and ‘state of uniform motion along a straight line.’
2. Newton’s first law of motion defines force as a physical quantity that brings about a change in ‘state of rest’ or ‘state of .uniform motion along a straight line’ of a body.
3. Newton’s first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line.
4. Due to all these reasons, Newton’s first law should be studied.

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 3 Motion in a Plane Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 3 Motion in a Plane

1. Choose the correct option.

Question 1.
An object thrown from a moving bus is on example of __________
(A) Uniform circular motion
(B) Rectilinear motion
(C) Projectile motion
(D) Motion in one dimension
Answer:
(C) Projectile motion

Question 2.
For a particle having a uniform circular motion, which of the following is constant ____________.
(A) Speed
(B) Acceleration
(C) Velocity
(D) Displacement
Answer:
(A) Speed

Question 3.
The bob of a conical pendulum undergoes ___________
(A) Rectilinear motion in horizontal plane
(B) Uniform motion in a horizontal circle
(C) Uniform motion in a vertical circle
(D) Rectilinear motion in vertical circle
Answer:
(B) Uniform motion in a horizontal circle

Question 4.
For uniform acceleration in rectilinear motion which of the following is not correct?
(A) Velocity-time graph is linear
(B) Acceleration is the slope of velocity time graph
(C) The area under the velocity-time graph equals displacement
(D) Velocity-time graph is nonlinear
Answer:
(D) Velocity-time graph is nonlinear

Question 5.
If three particles A, B and C are having velocities \(\overrightarrow{\mathrm{v}}_{A}\), \(\overrightarrow{\mathrm{v}}_{B}\) and \(\overrightarrow{\mathrm{v}}_{C}\) which of the following formula gives the relative velocity of A with respect to B
(A) \(\overrightarrow{\mathrm{v}}_{A}\) + \(\overrightarrow{\mathrm{v}}_{B}\)
(B) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{C}\) + \(\overrightarrow{\mathrm{v}}_{B}\)
(C) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{B}\)
(D) \(\overrightarrow{\mathrm{v}}_{C}\) – \(\overrightarrow{\mathrm{v}}_{A}\)
Answer:
(C) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{B}\)

2. Answer the following questions.

Question 1.
Separate the following in groups of scalar and vectors: velocity, speed, displacement, work done, force, power, energy, acceleration, electric charge, angular velocity.
Answer:
Scalars
Speed, work done, power, energy, electric charge.

Vectors
Velocity, displacement, force, acceleration, angular velocity (pseudo vector).

Question 2.
Define average velocity and instantaneous velocity. When are they same?
Answer:
Average velocity:

1. Average velocity (\(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\)) of an object is the displacement (\(\Delta \overrightarrow{\mathrm{x}}\)) of the object during the time interval (∆t) over which average velocity is being calculated, divided by that time interval.
2. Average velocity = (\(\frac{\text { Displacement }}{\text { Time interval }}\))
   \(\overrightarrow{\mathrm{V}_{\mathrm{av}}}=\frac{\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}\)
3. Average velocity is a vector quantity.
4. Its SI unit is m/s and dimensions are [M⁰L¹T^-1]
5. For example, if the positions of an object are x +4 m and x = +6 m at times t = O and t = 1 minute respectively, the magnitude of its average velocity during that time is V_av = (6 – 4)1(1 – 0) = 2 m per minute and its direction will be along the positive X-axis.
   ∴ \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\) = 2 i m/min
   Where, i = unit vector along X-axis.

Instantaneous velocity:

1. The instantaneous velocity (\(\overrightarrow{\mathrm{V}}\)) is the limiting value of ¡he average velocity of the object over a small time interval (∆t) around t when the value of lime interval goes to zero.
2. It is the velocity of an object at a given instant of time.
3. \(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
   where \(\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) derivative of \(\overrightarrow{\mathrm{x}}\) with respect to t.

In case of uniform rectilinear motion, i.e., when an object is moving with constant velocity along a straight line, the average and instantaneous velocity remain same.

Question 3.
Define free fall.
Answer:
The motion of any object under the influence of gravity alone is called as free fall.

Question 4.
If the motion of an object is described by x = f(t) write formulae for instantaneous velocity and acceleration.
Answer:

1. Instantaneous velocity of an object is given as,
   \(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
2. Motion of the object is given as, x = f(t)
3. The derivative f ‘(f) represents the rate of change of the position f (t) at time t, which is the instantaneous velocity of the object.
   ∴ \(\vec{v}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) = f'(t)
4. Acceleration is defined as the rate of change of velocity with respect to time.
5. The second derivative of the position function f “(t) represents the rate of change of velocity i.e., acceleration.
   ∴ \(\overrightarrow{\mathrm{a}}=\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta \mathrm{t}}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = f”(t)

Question 5.
Derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutually perpendicular directions.
Answer:

1. Consider an object moving in an x-y plane. Let the initial velocity of the object be \(\overrightarrow{\mathrm{u}}\) at t = 0 and its velocity at time t be \(\overrightarrow{\mathrm{v}}\).
2. As the acceleration is constant, the average acceleration and the instantaneous acceleration will be equal.
   
   This is the first equation of motion in vector form.
3. Let the displacement of the object from time t
   = 0 to t be \(\overrightarrow{\mathrm{s}}\)
   For constant acceleration, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{u}}}{2}\)
   
   This is the second equation of motion in vector form.
4. Equations (1) and (2) can be resolved into their x and y components so as to get corresponding scalar equations as follows.
   v_x = u_x + a_xt ………….. (3)
   v_y = u_y + a_y t …………… (4)
   s_x = u_xt + \(\frac{1}{2}\) a_xt² ………….. (5)
   s_y = u_yt + \(\frac{1}{2}\) a_yt² ………..(6)
5. It can be seen that equations (3) and (5) involve only the x components of displacement, velocity and acceleration while equations (4) and (6) involve only the y components of these quantities.
6. Thus, the motion along the x direction of the object is completely controlled by the x components of velocity and acceleration while that along the y direction is completely controlled by the y components of these quantities.
7. This shows that the two sets of equations are independent of each other and can be solved independently.

Question 6.
Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line.
Answer:

1. Consider an object starting from position x = 0 at time t = 0. Let the velocity at time (t = 0) and t be u and v respectively.
2. The slope of line PQ gives the acceleration. Thus
   ∴ a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}-0}=\frac{\mathrm{v}-\mathbf{u}}{\mathrm{t}}\)
   ∴ v = u + at …………… (1)
   This is the first equation of motion.
3. The area under the curve in velocity-time graph gives the displacement of the object.
   ∴ s = area of the quadrilateral OPQS = area of rectangle OPRS + area of triangle PQR.
   = ut + \(\frac{1}{2}\) (v – u) t
   But, from equation (1)
   at = v – u
   ∴ s = ut + \(\frac{1}{2}\) at²
   This is the second equation of motion,
4. The velocity is increasing linearly with time as acceleration is constant. The displacement is given as,
   
   ∴ s = (v² – u²) / (2a)
   ∴ v² – u² = 2as
   This is the third equation of motion.

Question 7.
Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity \(\vec{u}\) at an angel θ to the horizontal.
Answer:
Expression for range:

1. Consider a body projected with velocity \(\vec{u}\), at an angle θ of projection from point O in the co-ordinate system of the XY- plane, as shown in figure.
2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
   
   u_x = u cos θ (Horizontal component)
   u_y = u sin θ (Vertical component)
3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to
   v_y = u_y + a_yt
   with a_y = -g and u_y = u sinθ
4. Thus, the components of velocity of the projectile at time t are given by,
   v_x = u_x = u cos θ
   v_y = u_x – gt = usin θ – gt
5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
   s = (u cosθ) t
   s_y = (usinθ)t – \(\frac{1}{2}\) gt²
6. At the highest point, the time of ascent of the projectile is given as,
   t_A = \(\frac{u \sin \theta}{g}\) …………..(2)
7. The total time in air i.e., time of flight is given as, T = 2t_A = \(\frac{2u \sin \theta}{g}\) …… (3)
8. The total horizontal distance travelled by the particle in this time T is given as,
   R = u_x ∙ T
   R = u cos θ ∙ (2t_A)
   R = u cos θ ∙ \(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) ……………[From (3)]
   R = \(\frac{u^{2}(2 \sin \theta \cdot \cos \theta)}{g}\)
   R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) ………..[∵ sin 2θ = 2sin∙cosθ]
   This is required expression for horizontal range of the projectile.

Expression for maximum height of a projectile:
The maximum height H reached by the projectile is the distance travelled along the vertical (y) direction in time t_A.

Substituting s_y = H and t = t_a in equation (1),
we have,
H = (u sin θ)t_A – \(\frac{1}{2}\) gt_A²

This equation represents maximum height of projectile.

Question 8.
Show that the path of a projectile is a parabola.
Answer:

1. Consider a body projected with velocity initial velocity \(\vec{u}\) , at an angle θ of projection from point O in the co-ordinate system of the XY-plane. as shown in figure.
   
2. The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
   u_x = u cos θ (Horizontal component)
   u_y = u sin θ (Vertical component)
3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to,
   v_y = u_y + a_y t
   with a_y, = -g and u_y = u sinθ
4. Thus, the components of velocity of the projectile at time t are given by,
   v_x = u_x = u cosθ
   v_y = u_y – gt = u sinθ – gt
5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
   s_x = (u cosθ)t …………..(1)
   s_y = (u sinθ)t – \(\frac{1}{2}\) gt² ………………. (2)
6. As the projectile starts from x = O, we can use
   s_x = x and s_y = y.
   Substituting s_x = x in equation (1),
   x = (u cosθ) t
   ∴ t = \(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\) …………….. (3)
   Substituting, s_y in equation (2),
   y = (u sinθ)t – \(\frac{1}{2}\) gt² …………… (4)
   Substituting equation (3) in equation (4), we have,
   y = u sin θ (\(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\)) – \(\frac{1}{2}\) (\(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\))² g
   ∴ y = x (tan θ) – (\(\frac{g}{2 u^{2} \cos ^{2} \theta}\))x² ………………. (5)
   Equation (5) represents the path of the projectile.
7. If we put tan θ = A and g/2u²cos²θ = B then equation (5) can be written as y = Ax – Bx² where A and B are constants. This is equation of parabola. Hence, path of projectile is a parabola.

Question 9.
What is a conical pendulum? Show that its time period is given by 2π \(\sqrt{\frac{l \cos \theta}{g}}\), where l is the length of the string, θ is the angle that the string makes with the vertical and g is the acceleration due to gravity.
Answer:
A simple pendulum, Ch i given such a motion that the bob describes a horizontal circle and the string making a constant angle with the vertical describes a cone, is called a conical pendulum.

1. Consider a bob of mass m tied to one end of a string of length ‘P and other end is fixed to rigid support.
2. Let the bob be displaced from its mean position and whirled around a horizontal circle of radius ‘r’ with constant angular velocity ω, then the bob performs U.C.M.
3. During the motion, string is inclined to the vertical at an angle θ as shown in the figure above.
4. In the displaced position, there are two forces acting on the bob.
   • The weight mg acting vertically downwards.
   • The tension T acting upward along the string.
5. The tension (T) acting in the string can be resolved into two components:
   • T cosθ acting vertically upwards.
   • T sinθ acting horizontally towards centre of the circle.
6. Since, there is no net force, vertical component T cosθ balances the weight and horizontal component T sinθ provides the necessary centripetal force.
   ∴ T cos θ = mg ……………. (1)
   T sin θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mrω² ……….. (2)
7. Dividing equation (2) by (1),
   tan θ = \(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) ……….. (3)
   Therefore, the angle made by the string with the vertical is θ = tan^-1 (\(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\))
8. Since we know v = \(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)
   
   where l is length of the pendulum and h is the vertical distance of the horizontal circle from the fixed point O.

Question 10.
Define angular velocity. Show that the centripetal force on a particle undergoing uniform circular motion is -mω²\(\vec{r}\).
Answer:
Angular velocity of a particle is the rate of change of angular displacement.

Expression for centripetal force on a particle undergoing uniform circular motion:
i) Suppose a particle is performing U.C.M in anticlockwise direction.
The co-ordinate axes are chosen as shown in the figure.
Let,
A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
\(\overrightarrow{\mathrm{r}}\) = instantaneous position vector at time t

ii) From the figure,
\(\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}} \mathrm{x}+\hat{\mathrm{j}} \mathrm{y}\)
where, \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) are unit vectors along X-axis and Y-axis respectively.


Negative sign shows that direction of acceleration is opposite to the direction of position vector. Equation (3) is the centripetal acceleration.
vii) Magnitude of centripetal acceleration is given by a = ω²r

viii) The force providing this acceleration should also be along the same direction, hence centripetal.
∴ \(\overrightarrow{\mathrm{F}}\) = m\(\overrightarrow{\mathrm{a}}\) = -mω²\(\overrightarrow{\mathrm{r}}\)
This is the expression for the centripetal force on a particle undergoing uniform circular motion.

ix) Magnitude of F = mω²r = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mωv

[Note: The definition of angular velocity is not mentioned in this chapter but is in Ch.2 Mathematical Methods.]

3. Solve the following problems.

Question 1.
An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 sec. What is the velocity of the aeroplane at the take off ?
Answer:
Given: Length of runway (s) = 500 m, t = 30 s
To find: Velocity (y)
Formulae. i) s = ut + \(\frac{1}{2}\) at²
ii) v = u + at
Calculation: As the plane was initially at rest, u = 0
From formula (1),
500 = 0 + \(\frac{1}{2}\) × a × (30)²
∴ 500 = 450 a
∴ a = \(\frac{10}{9}\) m/s²
From formula (ii),
v = 0 + \(\frac{10}{9}\) × 30
∴ v = \(\frac{100}{3}\) m/s = (\(\frac{100}{3} \times \frac{18}{5}\)) km/hr
∴ v = 120 km/hr
The velocity of the aeroplane at the take off is 120 km/hr.

Question 2.
A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes. The car covers a distance of 100 m before it stops. Calculate
(i) the average retardation of the car
(ii) time taken by the car to come to rest.
Answer:
Given: u = 120 kmh^-1 = 120 × \(\frac{5}{18}\) = \(\frac{100}{3}\) ms^-1
s = 100 m, v = 0
To find: i) Average retardation of the car (a)
ii) Time taken by car (t)

Formulae: i) v² – u² = 2as
ii) v = u + at
Calculation: From formula (i),

i) Average retardation of the car is \(\frac{50}{9}\) ms² (in magnitude).
ii) Time taken by the car to come to rest is 6 s.

Question 3.
A car travels at a speed of 50 km/hr for 30 minutes, at 30 km/hr for next 15 minutes and then 70 km/hr for next 45 minutes. What is the average speed of the car?
Answer:
Given: v₁ = 50 km/hr. t₁ = 30 minutes = 0.5 hr,
v₂ = 30 km/hr, t₂ = 15 minutes = 0.25 hr,
v₃ = 70 km/hr, t₃ = 45 minutes 0.75 hr
To find: Average speed of car (v_av)
Formula v_av = \(\frac{\text { total path length }}{\text { total time interval }}\)
Calculation:
Path length,
x₁ = v₁ × t₁ = 50 × 0.5 = 25km
x₂ = v₂ × t₂ = 30 × 0.25 = 7.5 km
x₃ = v₃ × t₃ = 70 × 0.75 = 52.5 km
From formula,
v_av = \(\frac{x_{1}+x_{2}+x_{3}}{t_{1}+t_{2}+t_{3}}\)
∴ v_av = \(\frac{25+7.5+52.5}{0.5+0.25+0.75}=\frac{85}{1.5}\)
∴ v_av = 56.66 km/hr

Question 4.
A velocity-time graph is shown in the adjoining figure.

Determine:

1. initial speed of the car
2. maximum speed attained by the car
3. part of the graph showing zero acceleration
4. part of the graph showing constant retardation
5. distance travelled by the car in first 6 sec.

Answer:

1. Initial speed is at origin i.e. 0 m/s.
2. Maximum speed attained by car, v_max = speed from A to B = 20 m/s.
3. The part of the graph which shows zero acceleration is between t = 3 s and t = 6 s i.e., AB. This is because, during AB there is no change in velocity.
4. The graph shows constant retardation from t = 6 s to t = 8 s i.e., BC.
5. Distance travelled by car in first 6 s
   = Area of OABDO
   = A(△OAE) + A(rect. ABDE)
   = \(\frac{1}{2}\) × 3 × 20 + 3 × 20
   = 30 + 60
   ∴ Distance travelled by car in first 6 s = 90 m

Question 5.
A man throws a ball to maximum horizontal distance of 80 meters. Calculate the maximum height reached.
Answer:
Given: R = 80m
To find: Maximum height reached (H_max)
Formula: R_max = 4H_max
Calculation: From formula,
∴ H_max = \(\frac{\mathrm{R}_{\max }}{4}=\frac{80}{4}\) = 20 m
The maximum height reached by the ball is 20m.

Question 6.
A particle is projected with speed v₀ at angle θ to the horizontal on an inclined surface making an angle Φ (Φ < θ) to the horizontal. Find the range of the projectile along the inclined surface.
Answer:

i) The equation of trajectory of projectile is given by,
y(tan θ)x – (\(\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\))x² …………..(1)

ii) In this case to find R substitute,
y = R sinΦ ………….. (2)
x = R cosΦ ………….. (3)

iii) From equations (1), (2) and (3),
we have,

Question 7.
A metro train runs from station A to B to C. It takes 4 minutes in travelling from station A to station B. The train halts at station B for 20 s. Then it starts from station B and reaches station C in the next 3 minutes. At the start, the train accelerates for 10 sec to reach the constant speed of 72 km/hr. The train moving at the constant speed is brought to rest in 10 sec. at next station.
(i) Plot the velocity-time graph for the train travelling from station A to B to C.
(ii) Calculate the distance between stations A, B, and C.
Answer:

The metro train travels from station A to station B in 4 minutes = 240 s.
The trains halts at station B for 20 s.
The train travels from station B’ to station C in 3 minutes= 180 s.
∴ Total time taken by the metro train in travelling from station A to B to C
= 240 + 20 + 180 = 440 s.
At start, the train accelerates for 10 seconds to reach a constant speed of 72 km/hr = 20 m/s.
The train moving is brought to rest in 10 s at next station.
The velocity-time graph for the train travelling from station A to B to C is as follows:
Distance travelled by the train from station A to station B
= Area of PQRS
= A ( △PQQ’) A (☐QRR’) + A(SRR’)
= (\(\frac{1}{2}\) × 10 × 20 + (220 × 20) + (\(\frac{1}{2}\) 10 × 20)
= 100 + 4400 + 100
= 4600m = 4.6km

Distance travelled by the train from station B’ to station C
= Area of EFGD
= A(△EFF’) + A(☐F’FGG’) + A(△DGG’)
= (\(\frac{1}{2}\) × 10 × 20) × (160 × 20) + (\(\frac{1}{2}\) × 10 × 20)
= 100 + 3200 + 100
= 3400m = 3.4km

Question 8.
A train is moving eastward at 10 m/sec. A waiter is walking eastward at 1.2m/sec; and a fly is flying toward the north across the waiter’s tray at 2 m/s. What is the velocity of the fly relative to Earth.
Answer:
Given

Question 9.
A car moves in a circle at the constant speed of 50 m/s and completes one revolution in 40 s. Determine the magnitude of the acceleration of the car.
Answer:
Given: v = 50 m/s, t = 40 s, s = 2πr
To find: acceleration (a)
Formulae: i) v = \(\frac{\mathrm{s}}{\mathrm{t}}\)
ii) a = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
Calculation: From formula (i),
50 = \(\frac{2 \pi \mathrm{r}}{40}\)
∴ r = \(\frac{50 \times 40}{2 \pi}\)
∴ r = \(\frac{1000}{\pi}\) cm
From formula (ii),
a = \(\frac{v^{2}}{r}=\frac{50^{2}}{1000 / \pi}\)
∴ a = \(\frac{5 \pi}{2}\) = 7.85 m/s²
The magnitude of acceleration of the car is 7.85 m/s.

Alternate method:
Given: v = 50 m/s, t = 40 s,
To find: acceleration (a)
Formula: a = rω² = vω
Calculation: From formula,
a = vω
= v(\(\frac{2 \pi}{\mathrm{t}}\))
= 50(\(\frac{2 \times 3.142}{40}\))
= \(\frac{5}{2}\) × 3.142
∴ a = 7.85m/s²

Question 10.
A particle moves in a circle with constant speed of 15 m/s. The radius of the circle is 2 m. Determine the centripetal acceleration of the particle.
Answer:
Given: v = 15 m/s, r = 2m
To find: Centripetal acceleration (a)
Formula: a = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
Calculation: From formula,
a = \(\frac{(15)^{2}}{2}=\frac{225}{2}\)
∴ a = 112.5m/s²
The centripetal acceleration of the particle is 112.5 m/s².

Question 11.
A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50 m? Assume g = 10 m s^-2.
Answer:
Given: 40 ≤ R ≤ 50, θ = 300, g = 10 m/s²
To find: Range of initial velocity (u)
Formula: R = \(\frac{\mathrm{u}^{2} \sin (2 \theta)}{\mathrm{g}}\)
Calculation: From formula,
The range of initial velocity,

∴ 21.49m/s ≤ u ≤ 24.03m/s
The range of initial velocity should be between 21.49 m/s ≤ u ≤ 24.03 m/s.

11th Physics Digest Chapter 3 Motion in a Plane Intext Questions and Answers

Can you recall? (Textbook Page No. 30)

Question 1.
What ¡s meant by motion?
Answer:
The change ¡n the position of an object with respect to its surroundings is called motion.

Question 2.
What Is rectilinear motion?
Answer:
Motion in which an object travels along a straight line is called rectilinear motion.

Question 3.
What is the difference between displacement and distance travelled?
Answer:

• Displacement is the shortest distance between the initial and final points of movement.
• Distance is the actual path followed by a body between the points in which it moves.

Question 4.
What is the difference between uniform and non-uniform motion?
Answer:

• A body is said to have uniform motion if it covers equal distances in equal intervals of time.
• A body is said to have non-uniform motion if it covers unequal distances in equal intervals of time.

Internet my friend (Textbook Page No. 44)

i. hyperphysics.phy-astr.gsu.eduJhbase/mot.html#motcon
ii. www .college-physics.comlbook/mechanics
[Students are expected to visit the above mentioned webs ires and collect more information.]

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

1. Choose the most correct option.

Question A.
A sample of pure water, whatever the source always contains …………. by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
Answer:
a. 88.9

Question B.
Which of the following compounds can NOT demonstrate the law of multiple proportions?
a. NO, NO₂
b. CO, CO₂
c. H₂O, H₂O₂
d. Na₂S, NaF
Answer:
d. Na₂S, NaF

Question C.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
a. – 40°
b. + 40°
c. – 80°
d. – 20°
Answer:
a. – 40°

Question D.
SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
Answer:
b. Ampere

Question E.
In the reaction N₂ + 3H₂ → 2NH₃, the ratio by volume of N₂, H₂ and NH₃ is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
Answer:
d. gaseous volumes

Question F.
Which of the following has maximum number of molecules ?
a. 7 g N₂
b. 2 g H₂
c. 8 g O₂
d. 20 g NO₂
Answer:
b. 2 g H₂

Question G.
How many g of H₂O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
Answer:
a. 4.5

Question H.
The number of molecules in 22.4 cm³ of nitrogen gas at STP is
a. 6.022 × 10²⁰
b. 6.022 × 10²³
c. 22.4 × 10²⁰
d. 22.4 × 10²³
Answer:
a. 6.022 × 10²⁰

Question I.
Which of the following has the largest number of atoms ?
a. 1g Au_(s)
b. 1g Na_(s)
c. 1g Li_(s)
d. 1g Cl_2(g)
Answer:
c. 1g Li_(s)

2. Answer the following questions.

Question A.
State and explain Avogadro’s law.
Answer:
i. In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.

ii. Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:

According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:

Question B.
Point out the difference between 12 g of carbon and 12 u of carbon.
Answer:
12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom.

Question C.
How many grams does an atom of hydrogen weigh ?
Answer:
The mass of a hydrogen atom is 1.6736 × 10^-24 g.

Question D.
Calculate the molecular mass of the following in u.
a. NH₃
b. CH₃COOH
c. C₂H₅OH
Answer:
i. Molecular mass of NH₃ = (1 × Average atomic mass of N) + (3 × Average atomic mass of H)
= (1 × 14.0 u) +(3 × 1.0 u)
= 17 u

ii. Molecular mass of CH₃COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u)
= 60 u

iii. Molecular mass of C₂H₅OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u)
= 46 u
Ans: i. The molecular mass of NH₃ = 17 u
ii. The molecular mass of CH₃COOH = 60 u
iii. The molecular mass of C₂H₅OH = 46 u

Question E.
How many particles are present in 1 mole of a substance ?
Answer:
The number of particles in one mole is 6.0221367 × 10²³.

Question F.
What is the SI unit of amount of a substance ?
Answer:
The SI unit for the amount of a substance is mole (mol).

Question G.
What is meant by molar volume of a gas ?
Answer:
The volume occupied by one mole of a gas at standard temperature (0 °C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm³.

Question H.
State and explain the law of conservation of mass.
Answer:
Law of conservation of mass:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• Antoine Lavoisier who is often referred to as the father of modem chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
• Both his experiments resulted in increased weight of products.
• After several experiments, in burning of phosphorus, he found that the weight gained by the phosphoms was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
• When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.

Question I.
State the law of multiple proportions.
Answer:
The law states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.

3. Give one example of each

Question A.
Homogeneous mixture
Answer:
Homogeneous mixture: Solution (An aqueous solution of sugar)

Question B.
Heterogeneous mixture
Answer:
Heterogeneous mixture: Suspension (of sand in water)

Question C.
Element
Answer:
Element: Gold

Question D.
Compound
Answer:
Compound: Distilled water.

4. Solve problems :

Question A.
What is the ratio of molecules in 1 mole of NH₃ and 1 mole of HNO₃.
Answer:
One mole of any substance contains particles equal to 6.022 × 10²³.
1 mole of NH₃ = 6.022 × 10²³ molecules of NH₃
I mole of HNO₃ = 6.022 × 10²³ molecules of HNO₃
∴ Ratio = \(\frac{6.022 \times 10^{23}}{6.022 \times 10^{23}}\) = 1 : 1
Ans: The ratio of molecules is = 1 : 1.

Question B.
Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen

Molar volume of a gas = 22.4 dm³ mol^-1 = 22.4 L at STP

Ans: Number of moles of hydrogen = 0.02 mol

Question C.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
∴ Mass of 18 atoms of hydrogen = 18 × 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u

Question D.
Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ x = \(\frac{254 \mathrm{u}}{127 \mathrm{u}}\) = 2 atoms

b. 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol^-1
Now,

Now,
Number of atoms = Number of moles × Avogadro’s constant
= 2 mol × 6.022 × 10²³ atoms/mol
= 12.044 × 10²³ atoms
= 1.2044 × 10²⁴ atoms
Ans. i.Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = 1.2044 × 10²⁴ atoms

Question E.
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. 5 mg carbon = 5 × 10^-3 g carbon
Atomic mass of carbon = 12 u
∴ Molar mass of carbon 12 g mol^-1

b. 12 mg carbon = 12 × 10^-3 g carbon

Number of atoms = Number of moles × Avogadro’s constant
Number of atoms of carbon = 1 × 10^-3 mol × 6.022 × 10²³ atoms/mol
= 6.022 × 10²⁰ atoms
Ans: Number of moles of carbon in his homework writing = 4.167 × 10^-4 mol
Number of atoms of carbon in 12 mg homework writing = 6.022 × 10²⁰ atoms

Question F.
Arjun purchased 250 g of glucose (C₆H₁₂O₆) for Rs 40. Find the cost of glucose per mole.
Answer:
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C₆H₁₂O₆
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C₆H₁₂O₆).
Molecular mass of glucose
= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)
=180 u
∴ Molar mass of glucose = 180 g mol^-1

Now,
\(\frac {250}{180}\) mol of glucose cost = Rs 40
1 mol glucose cost = x
∴ x = \(\frac{40 \times 180}{250}\) = Rs 28.8/mol of glucose
Ans. The cost of glucose per mole is Rs 28.8.

[ Calculation using log table:
\(\frac{40 \times 180}{250}\)
= Antilog₁₀ [log₁₀(40) + log₁₀(180) + log₁₀(250)]
= Antilog₁₀ [1.6021 + 2.2553 – 2.3979]
= Antilog₁₀ [1.4595] = 28.80 ]

Question G.
The natural isotopic abundance of ¹⁰B is 19.60% and ¹¹B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron(B)

Ans. Average atomic mass of boron = 10.84 u

Question H.
Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 °C
b. 30 °C
Answer:
a. Given: Temperature in degree Celsius =40°C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 40 °C in the formula,
°F = \(\frac {9}{5}\) (°C)+32
= \(\frac {9}{5}\) (40) + 32
= 72 + 32
= 104 °F

b. Given: Temperature in degree Celsius = 30 °C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 30 °C in the formula,
°F = \(\frac {9}{5}\)(°C) + 32
= \(\frac {9}{5}\)(30) + 32
= 54 + 32
= 86 °F
Ans: i. The temperature 40 °C corresponds to 104 °F.
ii. The temperature 30 °C corresponds to 86 °F.

Question I.
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Calculator: Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH₃COOH
= (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12 u) + (4 × 1 u) + (2 × 16 u) = 60 u
∴ Molar mass of acetic acid = 60 g mol^-1

Now,
Number of molecules of acetic acid = Number of moles × Avogadro’s constant
= 0.367 mol × 6.022 × 10²³ molecules/mol
= 2.210 × 10²³ molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 × 10²³ molecules

Question J.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 32 = 64 g) of oxygen to give (2 × 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g

Question K.
Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles × Avogadro’s constant
= 0.4 mol × 6.022 × 10²³ atoms/mol
= 2.4088 × 10²³ atoms of N

b. 1.6 g of Sulphur (S)
Molar mass of sulphur = 32 g mol^-1

Number of atoms of S = Number of moles × Avogadro’s constant
= 0.05 mol × 6.022 × 10²³ atoms/mol
= 0.3011 × 10²³ atoms
= 3.011 × 10²² atoms of S
Ans: a. Number of nitrogen atoms in 0.4 mole = 2.4088 × 10²³ atoms of N
b. Number of sulphur atoms in 1.6 g = 3.011 × 10²² atoms of S

Question L.
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Answer:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
% of oxygen = \(\frac{1.2}{3.2}\) × 100 = 37.5%
% of metal = \(\frac{2.0}{3.2}\) × 100 = 62.5%
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g
% of oxygen = \(\frac{0.85}{2.27}\) × 100 = 37.44 ≈ 37.5%
% of metal = \(\frac{1.42}{2.27}\) × 100 = 62.56 ≈ 62.5%
Therefore, irrespective of the source, the given compound contains same elements in the same proportion. The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight”. Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.

Question M.
In two moles of acetaldehyde (CH₃CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer:
Molecular formula of acetaldehyde: C₂H₄O
Moles of acetaldehyde = 2 mol
a. Number of moles of carbon atoms = Moles of acetaldehyde × Number of carbon atoms
= 2 × 2
= 4 moles of carbon atoms

b. Number of moles of hydrogen atoms = Moles of acetaldehyde × Number of hydrogen atoms
= 2 × 4
= 8 moles of hydrogen atoms

c. Number of moles of oxygen atoms = Moles of acetaldehyde × Number of oxygen atoms
= 2 × 1
= 2 moles of oxygen atoms

d. Number of molecules of acetaldehyde = Moles of acetaldehyde × Avogadro number
= 2 mol × 6.022 × 10²³ molecules/mol
= 12.044 × 10²³ molecules of acetaldehyde
Ans: i. Number of moles of carbon, hydrogen and oxygen are 4, 8, 2 respectively,
ii. Number of molecules of acetaldehyde = 12.044 × 10²³

Question N.
Calculate the number of moles of magnesium oxide, MgO in
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of Mg = 24 and O = 16)
Answer:
Given: i. Mass of MgO = 80 g
ii. Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae: Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
Calculation: i. Molecular mass of MgO = (1 × Average atomic mass of Mg) + (1 × Average atomic mass of O)
= (1 × 24u) + (1 × 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol^-1
Mass of MgO = 80 g
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{80 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 2 mol

ii. Mass of MgO = 10 g, Molar mass of MgO = 40 g mol^-1
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{10 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.25 mol
Ans: i. The number of moles in 80 g of magnesium oxide, MgO = 2 mol
ii. The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol

Question O.
What is volume of carbon dioxide, CO₂ occupying by i. 5 moles and ii. 0.5 mole of CO₂ gas measured at STP.
Answer:
Given: i. Number of moles of CO₂ = 5 mol
ii. Number of moles of CO₂ = 0.5 mol
To find: Volume at STP
Formula: Number of moies of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Calculation: Molar volume of a gas 22.4 dm³ mol^-1 at STP.
Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
∴ i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5mol × 22.4 dm³ mol^-1 = 112 dm³
ii. Volume of the gas at STP Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm³ mol^-1 = 11.2 dm³
Ans: i. Volume of 5 mol of CO₂ = 112 dm³
ii. Volume of 0.5 mol of CO₂ = 11.2 dm³

Question P.
Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1.
Answer:
The molecular formula of potassium chlorate is KClO₃.
Required chemical equation:

2 moles of KClO₃ = 2 × 122.5 = 245 g
3 moles of O₂ at STP occupy = (3 × 22.4 dm³) = 67.2 dm³
Thus, 245 g of potassium chlorate will liberate 67.2 dm³ of oxygen gas.
Let ‘x’ gram of KClO₃ liberate 6.72 dm³ of oxygen gas at S.T.P.
∴ x = \(\frac{245 \times 6.72}{67.2}\) = 24.5 g
Ans: Mass of potassium chlorate required = 24.5 g

Question Q.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH₂)₂CO. Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH₂)₂CO
Molar mass of urea = 60 g mol^-1

∴ Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 2.247 × 10²³ atoms of hydrogen
∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 1.124 × 10²³ atoms of nitrogen
∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of carbon
∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 10²³ atoms of H, 1.124 × 10²³ atoms of N, 0.562 × 10²³ atoms of C and 0.562 × 10²³ atoms of O.

Question R.
Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Answer:
Given: Mass of sulphur (reactant) = 16 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 32 g oxygen to form 64 g of sulphur dioxide as follows:

Hence, (0.5 × 32 = 16 g) of sulphur will combine with (0.5 × 32 = 16 g) of oxygen to give (0.5 × 64 = 32 g) sulphur dioxide.
Ans: Mass of sulphur dioxide produced = 32 g

5. Explain

Question A.
The need of the term average atomic mass.
Answer:

• Several naturally occurring elements exist as a mixture of two or more isotopes.
• Isotopes have different atomic masses.
• The atomic mass of such an element is the average of atomic masses of its isotopes.
• For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.

Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.

Question B.
Molar mass.
Answer:
i. The mass of one mole of a substance (element/compound) in grams is called its molar mass.
ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
e.g.

Element	Atomic mass (u)	Molar mass (g mol-1)
H	1.0	1 0
C	12.0	12.0
O	16.0	16.0

iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
e.g.

Polyatomic substance	Molecular/formula mass (u)	Molar mass (g mol-1)
O2	32.0	32.0
H2O	18.0	18.0
NaCl	58.5	58.5

Question C.
Mole concept.
Answer:

• Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
• Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon -12 isotope.
• One mole is the amount of substance which contains 6.0221367 × 10²³ particles/entities.

Question D.
Formula mass with an example.
Answer:

• The formula mass of a substance is the sum of atomic masses of the atoms present in the formula.
• In substances such as sodium chloride, positive (sodium), and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na⁺) ion is surrounded by six chlorides (Cl^–) ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units.
• Therefore, NaCl is just the formula that is used to represent sodium chloride though it is not a molecule.
• In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass.

e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

Question E.
Molar volume of gas.
Answer:
i. It is more convenient to measure the volume rather than mass of the gas.
ii. It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm³ at standard temperature (0 °C) and pressure (1 atm) (STP).
iii. The volume of 22.4 dm³ at STP is known as molar volume of a gas.
iv. The relationship between number of moles and molar volume can be expressed as follows:

[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol^-1]

Question F.
Types of matter (on the basis of chemical composition).
Answer:
Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.

They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
1. Metals:

• They have a lustre (a shiny appearance).
• They conduct heat and electricity.
• They can be drawn into wire (ductile).
• They can be hammered into thin sheets (malleable).
• e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.

2. Nonmetals:

• They have no lustre, (except diamond, iodine)
• They are poor conductors of heat and electricity, (except graphite)
• They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine

3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.

ii. Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.

Mixtures are of two types:

• Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
  e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
• Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
  e.g. Suspension, which contains insoluble solid in a liquid.

11th Chemistry Digest Chapter 1 Some Basic Concepts of Chemistry Intext Questions and Answers

Can you tell? (Textbook Page No. 1)

Question 1.
Which are mixtures and pure substances from the following?
i. Sea water
ii. Gasoline
iii. Skin
iv. A rusty nail
v. A page of textbook
vi. Diamond
Answer:

No.	Material	Pure substance or mixture
i.	Seawater	Mixture
ii.	Gasoline	Mixture
iii.	Skin	Mixture
iv.	A rusty nail	Mixture
V.	A page of textbook	Mixture
vi.	Diamond	Pure substance

Can you tell? (Textbook Page No. 2)

Question 1.
Classify the following as element and compound.
i. Mercuric oxide
ii. Helium gas
iii. Water
iv. Table salt
v. Iodine
vi. Mercury
vii. Oxygen
viii. Nitrogen
Answer:

No.	Material	Element or compound
i.	Mercuric oxide	Compound
ii.	Helium gas	Element
iii.	Water	Compound
iv.	Table salt	Compound
V.	Iodine	Element
vi.	Mercury	Element
vii.	Oxygen	Element
viii.	Nitrogen	Element

Can you tell? (Textbook Page No. 6)

Question 1.
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:

If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced.

Can you recall? (Textbook Page No. 6)

Question 1.
What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:

• The smallest indivisible particle of an element is called an atom.
• A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
• Every atom of an element has definite mass. The order of magnitude of mass of one atom is 10^-27 kg.
• Isotopes are the atoms of the same element having same atomic number but different mass number.

Try this (Textbook Page No. 8)

Question 1.
Find the formula mass of CaSO₄, if atomic mass of Ca = 40.1 u, S =32.1 u and O = 16.0 u.
Solution:
Formula mass of CaSO₄
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= (40.1) + 32.1 + (4 × 16.0) = 136.2 u
Ans: Formula mass of CaSO₄ = 136.2 u

Can you recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items.

Try this (Textbook Page No. 10)

Question 1.
Calculate the volume in dm³ occupied by 60.0 g of ethane at STP.
Solution:
Given: Mass of ethane at STP = 60.0 g
To find: Volume of ethane
Formulae:

Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP
Molecular mass of ethane = 30 g mol^-1

∴ Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 2 mol × 22.4 dm³ mol^-1 = 44.8 dm³
Ans: Volume of ethane = 44.8 dm³

Activity :

Activity 1.
Collect information from various scientists and prepare charts of their contributions to chemistry.
Answer:

Scientists		Contributions
Joseph Louis Gay-Lussac (1778 – 1850) (French chemist and physicist)	i.	Formulated the gas law.
	ii.	Collected samples of air at different heights and recorded temperatures and moisture contents.
	iii.	Discovered that the composition of the atmosphere does not change with increasing altitude.
Amedeo Avogadro (1776 – 1856) (Italian scholar)	i.	Published article in French journal on determining the relative masses of elementary particles of bodies and proportions by which they enter combinations.
	ii.	Published a research paper titled “New considerations on the theory of proportions and on the determination of the masses of atoms.”

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 14 Semiconductors Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 14 Semiconductors

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:

Intrinsic semiconductors	Extrinsic semiconductors
1. A pure semiconductor is known as intrinsic semiconductors.	The semiconductor, resulting
2. Their conductivity is low	Their conductivity is high even at room temperature.
3. Its electrical conductivity is a function of temperature alone.	Its electrical conductivity depends upon the temperature as well as on the quantity of impurity atoms doped in the structure.
4. The number density of holes (nh) is same as the number density of free electron (ne) (nh = ne).	The number density of free electrons and number density of holes are unequal.

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.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 7 Thermal Properties of Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 7 Thermal Properties of Matter

1. Choose the correct option.

Question 1.
The range of temperature in a clinical thermometer, which measures the temperature of the human body, is
(A) 70 ºC to 100 ºC
(B) 34 ºC to 42 ºC
(C) 0 ºF to 100 ºF
(D) 34 ºF to 80 ºF
Answer:
(B) 34 ºC to 42 ºC

Question 2.
A glass bottle completely filled with water is kept in the freezer. Why does it crack?
(A) Bottle gets contracted
(B) Bottle is expanded
(C) Water expands on freezing
(D) Water contracts on freezing
Answer:
(C) Water expands on freezing

Question 3.
If two temperatures differ by 25 °C on Celsius scale, the difference in temperature on Fahrenheit scale is
(A) 65°
(B) 45°
(C) 38°
(D) 25°
Answer:
(B) 45°

Question 4.
If α, β and γ are coefficients of linear, area l and volume expansion of a solid then
(A) α: β:γ 1:3:2
(B) α:β:γ 1:2:3
(C) α:β:γ 2:3:1
(D) α:β:γ 3:1:2
Answer:
(B) α:β:γ 1:2:3

Question 5.
Consider the following statements-
(I) The coefficient of linear expansion has dimension K^-1
(II) The coefficient of volume expansion has dimension K^-1
(A) I and II are both correct
(B) I is correct but II is wrong
(C) II is correct but I is wrong
(D) I and II are both wrong
Answer:
(A) I and II are both correct

Question 6.
Water falls from a height of 200 m. What is the difference in temperature between the water at the top and bottom of a water fall given that specific heat of water is 4200 J kg^-1 °C^-1?
(A) 0.96 °C
(B) 1.02 °C
(C) 0.46 °C
(D) 1.16 °C
Answer:
(C) 0.46 °C

2. Answer the following questions.

Question 1.
Clearly state the difference between heat and temperature?
Answer:

	Heat	Temperature
i.	Heat is energy in transit. When two bodies at different temperatures are brought in contact, they exchange heat. OR Heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of their temperature difference.	Temperature is a physical quantity that defines the thermodynamic state of a system. OR Heat transfer takes place between the body and the surrounding medium until the body and the surrounding medium are at the same temperature.
ii.	Heat exchange can be measured with the help of a calorimeter.	Temperature is measured with the help of a thermometer.
iii.	Heat (being a form of energy) is a derived quantity.	Temperature is a fundamental quantity.

Question 2.
How a thermometer is calibrated?
Answer:

1. For the calibration of a thermometer, a standard temperature interval is selected between two easily reproducible fixed temperatures.
2. The fact that substances change state from solid to liquid to gas at fixed temperatures is used to define reference temperature called fixed point.
3. The two fixed temperatures selected for this purpose are the melting point of ice or freezing point of water and the boiling point of water.
4. This standard temperature interval is divided into sub-intervals by utilizing some physical property that changes with temperature.
5. Each sub-interval is called as a degree of temperature. Thus, an empirical scale for temperature is set up.

Question 3.
What are different scales of temperature? What is the relation between them?
Answer:

1. Celsius scale:
   • The ice point (melting point of pure ice) is marked as O °C (lower point) and steam point (boiling point of water) is marked as 100 °C (higher point).
   • Both are taken at one atmospheric pressure.
   • The interval between these points is divided into two equal parts. Each of these parts is called as one degree celsius and it is ‘written as 1 °C.
2. Fahrenheit scale:
   • The ice point (melting point of pure ice) is marked as 32 °F and steam point (boiling point of water) is marked as 212 °F.
   • The interval between these two reference points is divided into 180 equal parts. Each part is called as degree fahrenheit and is written as 1 °F.
3. Kelvin scale:
   • The temperature scale that has its zero at -273.15 °C and temperature intervals are same as that on the Celsius scale is called as kelvin scale or absolute scale.
   • The absolute temperature, T and celsius temperature, T_C are related as, T = T_C + 273.15
     eg.: when T_C = 27 °C,
     T = 27+273.15 K = 300.15 K

Relation between different scales of temperature:
\(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}=\frac{\mathrm{T}_{\mathrm{C}}-0}{100}=\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
where,
T_F = temperature in fahrenheit scale,
T_C = temperature in celsius scale,
T_K = temperature in kelvin scale,
[Note: At zero of the kelvin scale, every substance in nature has the least possible activity.]

Question 4.
What is absolute zero?
Answer:

1. When the graph of pressure (P) against temperature T (°C) at constant volume for three ideal gases A, B and C is plotted, in each case, P -T graph is straight line indicating direct proportion between them. The slopes of these graphs are different.
2. The individual straight lines intersect the pressure axis at different values of pressure at O °C. but each line intersects the temperature axis at the same point, i.e., at absolute temperature (-273.15 °C).
3. Similarly graph at constant pressure for three different ideal gases A, B and C extrapolate to the same temperature intercept -273.15 °C i.e., absolute zero temperature.
4. It is seen that all the lines for different gases Cut the temperature axis at the same point at -273.15 °C.
5. This point is termed as the absolute zero of temperature.
6. It is not possible to attain a temperature lower than this value. Even to achieve absolute zero temperature is not possible in practice.
   [Note: The point of zero pressure or zero volume does not depend on am specific gas.]

Question 5.
Derive the relation between three coefficients of thermal expansion.
Answer:
Consider a square plate of side l₀ at 0 °C and h at T °C.

1. l_T = l₀ (1 + αT)
   If area of plate at 0 °C is A₀, A₀ = \(l_{0}^{2}\)
   If area of plate at T °C is A_T,
   A_T = \(l_{\mathrm{T}}^{2}=l_{0}^{2}\) (1 + αT)²
   or A_T = A₀ (1 + αT)² …………… (1)
   Also,
   A_T = A₀(1 + βT)² …………… (2)
   ……………. [∵ β = \(\frac{\mathrm{A}_{\mathrm{T}}-\mathrm{A}_{0}}{\mathrm{~A}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)]
2. Using Equations (1) and (2),
   A₀ (1 + αT)² = A₀(1 + βT)
   ∴ 1 + 2αT + α²T² = 1 + βT
3. Since the values of a are very small, the term α²T² is very small and may be neglected,
   ∴ β = 2a
4. The result is general because any solid can be regarded as a collection of small squares.

Relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).

1. Consider a cube of side l₀ at 0 °C and l_T at T °C.
   ∴ l_T = l₀(1 + αT)
   If volume of the cube at 0 °C is V₀, V₀ = \(l_{0}^{3}\)
   If volume of the cube at T °C is
   V_T, V_T = \(l_{\mathrm{T}}^{3}=l_{0}^{3}\) (1 + αT)³
   V_T = V₀ (1 + αT)³ ………. (1)
   Also,
   V_T = V₀(1 + γT) …………. (2)
   …………. [∵ γ = \(\frac{\mathrm{V}_{\mathrm{T}}-\mathrm{V}_{0}}{\mathrm{~V}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)]
2. Using Equations (1) and (2),
   V₀(1 + αT)³ = V₀(1 + γT)
   ∴ 1 + 3αT + 3α²T² + α³T³ = 1 + γT
3. Since the values of a are very small, the terms with higher powers of a may be neglected,
   ∴ γ = 3α
4. The result is general because any solid can be regarded as a collection of small cubes.

Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}=\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion,
γ = coefficient of cubical expansion.

Question 6.
State applications of thermal expansion.
Answer:
Applications of thermal expansion:

• The steel wheel is heated to expand. This expanded wheel can easily fit over axle. The wheel is then cooled quickly. Upon cooling, wheel contracts and fits tightly upon the axle.
• An electric light bulb gets hot quickly when in use. The wire leads to the filament are sealed into the glass. 1f the glass of the bulbs has significantly different thermal expansivity from the wire leads, the glass and the wire would separate, breaking down the vacuum. To prevent this, wires are made of platinum or some suitable alloy with the same expansivity as ordinary glass.

Question 7.
Why do we generally consider two specific heats for a gas?
Answer:

• A slight change in temperature causes considerable change in pressure as well as volume of the gas.
• Therefore, two principal specific heats are defined for a gas viz., specific heat capacity at constant volume (S_V) and specific heat capacity at constant pressure (S_p).

Question 8.
Are freezing point and melting point same with respect to change of state ? Comment.
Answer:
Though freezing point and melting point mark same temperature (0°C or 32° F), state of change is different for the two points. At freezing point liquid gets converted into solid, whereas at melting point solid gets converted into its liquid state.

Question 9.
Define
(i) Sublimation
(ii) Triple point.
Answer:

1. The change from solid state to vapour stale without passing through the liquid state is called sublimation and the substance is said to sublime.
   Examples: Dry ice (solid CO₂) and iodine.
2. The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure.

Question 10.
Explain the term ‘steady state’.
Answer:

1. When one end of a metal rod is heated, the heat flows by conduction from hot end to the cold end.
   
2. As a result, the temperature of every section of the rod starts increasing.
3. Under this condition, the rod is said to be in a variable temperature state.
4. After some time, the temperature at each section of the rod becomes steady i.e., does not change.
5. Temperature of each cross-section of the rod now becomes constant though not the same. This is called steady state condition.

Question 11.
Define coefficient of thermal conductivity. Derive its expression.
Answer:
Coefficient of thermal conductivity of a material is defined as the quantity of heat that flows in one second between the opposite faces of a cube of side 1 m, the faces being kept at a temperature difference of 1°C (or 1 K).

Expression for coefficient of thermal conductivity:

1. Under steady state condition, the quantity of heat ‘Q’ that flows from the hot face at temperature T₁ to the cold face at temperature T₂ of a cube with side x and area of cross-section A is
   • directly proportional to the cross-sectional area A of the face. i.e.. Q ∝ A
   • directly proportional to the temperature difference between the two faces i.e., Q ∝ (T₁ – T₂)
   • directly proportional to time t (in seconds) for which heat flows i.e.. Q ∝ t
   • inversely proportional to the perpendicular distance x between hot and cold faces i.e., Q ∝ 1/x
2. Combining the above four factors, we have the quantity of heat
   Q ∝ \(\frac{\mathrm{A}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{t}}{\mathrm{x}}\)
   ∴ Q = \(\)
   where k is a constant of proportionality and is called coefficient of thermal conductivity. Its value depends upon the nature of the material.

Question 12.
Give any four applications of thermal conductivity in every day life.
Answer:
Answer: Applications of thermal conductivity:

1. • Thick walls are used in the construction of cold storage rooms.
   • Brick being a bad conductor of heat is used to reduce the flow of heat from the surroundings to the rooms.
   • Better heat insulation is obtained by using hollow bricks.
   • Air being a poorer conductor than a brick, it further avoids the conduction of heat from outside.
2. Street vendors keep ice blocks packed in saw dust to prevent them from melting rapidly.
3. The handle of a cooking utensil is made of a bad conductor of heat, such as ebonite, to protect our hand from the hot utensil.
4. Two bedsheets used together to cover the body help retain body heat better than a single bedsheet of double the thickness. Trapped air being a bad conductor of heat, the layer of air between the two sheets reduces thermal conduction better than a sheet of double the thickness. Similarly, a blanket coupled with a bedsheet is a cheaper alternative to using two blankets.

Question 13.
Explain the term thermal resistance. State its SI unit and dimensions.
Answer:

1. Consider expression for conduction rate,
   P_cond = kA \(\frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{x}}\)
   ⇒ \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\text {cond }}}=\frac{\mathrm{x}}{\mathrm{kA}}\) ……………. (1)
2. Ratio \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\text {cond }}}\) is called as thermal resistance (R_T) of material.

The SI unit of thermal resistance is °C s/kcal or °C s/J and its dimensional formula is [L^-2M^-1T³K¹].

Question 14.
How heat transfer occurs through radiation in absence of a medium?
Answer:

1. All objects possess thermal energy due to their temperature T(T > 0 K).
2. The rapidly moving molecules of a hot body emit EM waves travelling with the velocity of light. These are called thermal radiations.
3. These carry energy with them and transfer it to the low-speed molecules of a cold body on which they fall.
4. This results in an increase in the molecular motion of the cold body and its temperature rises.
5. Thus transfer of heat by radiation is a two fold process-the conversion of thermal energy into waves and reconversion of waves into thermal energy by the body on which they fall.

Question 15.
State Newton’s law of cooling and explain how it can be experimentally verified.
Answer:
The rate of loss of heat dT/dt of the both’ is directly proportional to the difference of temperature (T – T₀) of the body and the surroundings provided the difference in temperatures is small.

Mathematically, Newton’s law of cooling can be expressed as:
\(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ (T – T₀)
∴ \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ∝ C(T – T₀)
where, C is constant of proportionality. Experimental verification of Newton’s law of cooling:

1. Fill a calorimeter upto \(\frac{2}{3}\) of its capacity with a boiling water. Cover it with lid with a hole for passing the thermometer.
2. Insert the thermometer through the hole and adjust it so that the bulb of the thermometer is fully immersed in hot water.
3. Keep calorimeter vessel in constant temperature enclosure or just in open air since room temperature will not change much during the experiment.
4. Note down the temperature (T) on the thermometer at every one minute interval until the temperature of water decreases by about 25 °C.
5. Plot a graph of temperature (T) on Y-axis against time (t) on X-axis. This graph is called cooling curve as shown in figure (a).
   
6. Draw tangents to the curve at suitable points on the curve. The slope of each tangent is \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{T}}{\Delta \mathrm{t}}\) and gives the rate of fall of temperature at that temperature (T).
7. Now the graph of \(\left|\frac{\mathrm{dT}}{\mathrm{dt}}\right|\) on Y-axis against (T – T₀) on X-axis is plotted with (0, 0) origin. The graph is straight line and passes through origin as shown in figure (b), which verities Newton’s law of cooling.
   
   (b) Graphical verification of Newton’s law of cooling

Question 16.
What is thermal stress? Give an example of disadvantages of thermal stress in practical use?
Answer:

1. Consider a metallic rod of length l₀ fixed between two rigid supports at T °C.
   If the temperature of rod is increased by ∆T, length of rod would become,
   l = l₀(1 + α∆T)
   Where, α is the coefficient of linear expansion of material of the rod.
   But the supports prevent expansion of rod. As a result, rod exerts stress on the supports. Such stress is termed as thermal stress.
2. Disadvantage: Thermal stress can lead to fracture or deformation in substance under certain conditions.
3. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps.

Question 17.
Which materials can be used as thermal insulators and why?
Answer:

1. Substances such as glass, wood, rubber, plastic, etc. can be used as thermal insulators.
2. These substances do not have free electrons to conduct heat freely throughout the body. Hence, they arc poor conductors of heat.

3. Solve the following problems.

Question 1.
A glass flask has volume 1 × 10^-4 m³. It is filled with a liquid at 30 ºC. If the temperature of the system is raised to 100 ºC, how much of the liquid will overflow. (Coefficient of volume expansion of glass is 1.2 × 10^-5 (ºC)^-1 while that of the liquid is 75 × 10^-5 ºC^-1)
Solution:
Given: V₁ = 1 × 10^-4 m³ = 10^-4 m³, T₁ = 30°C,
T₂ = 100 °C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
Increase is volume = V₂ – V₁
= γV₁(T₂ – T₁)
increase in volume of beaker
= γ_glass × V₁ (T₂ – T₁)
= 1.2 × 10^-5 × 10^-4 × (100 – 30)
= 1.2 × 70 × 10^-9
= 4 × 10^-9 m³
∴ Increase in volume of beaker
= 84 × 10^-9 m³
Increase in volume of liquid
= γ_liquid × V₁ (T₂ – T₁)
= 75 × 10^-5 × 10 × (100 – 30)
= 75 × 70 × 10
= 5250 × 10^-9 m³
∴ Increase in volume of liquid = 5250 × 10^-9 m³
∴ Volume of liquid which overflows
= (5250 – 84) × 10^-9 m³
= 5166 × 10^-9 m³
= 0.5166 × 10^-7 m³
Volume of liquid that overflows is 0.5166 × 10^-7 m³.
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]

Question 2.
Which will require more energy, heating a 2.0 kg block of lead by 30 K or heating a 4.0 kg block of copper by 5 K? (s_lead = 128 J kg^-1 K^-1, s_copper = 387 J kg^-1 K^-1)
Solution:
Given: m_lead = 2 kg, ∆T_lead = 30 K,
s_lead = 128 J/kg K,
m_Cu =4 kg, ∆T_Cu = 5 K,
s_Cu = 387 J/kg K
To find: Substance requiring more heat energy.
Formula: Q = ms ∆T
Calculation: From formula,
For lead, Q_lead = 2 × 128 × 30 = 7680J
For Copper, Q_Cu = 4 × 387 × 5 = 7740 J
Q_Cu > Q_lead, copper will require more heat energy.
Copper will require more heat energy.

Question 3.
Specific latent heat of vaporization of water is 2.26 × 10⁶ J/kg. Calculate the energy needed to change 5.0 g of water into steam at 100 ºC.
Solution:
Given: L_vap = 2.26 × 10⁶ J/kg
m = 5g = 5 × 10^-3 kg
In this case, no temperature change takes place only change of state occurs.
To find: Heat required to convert water into steam.
Formula: Heat required = mL_vap
Calculation: From formula,
Heat required = 5 × 10^-3 × 2.26 × 10⁶
= 11300J
= 1.13 × 10⁴ J
Heat required to convert water into steam is 1.13 × 10⁴ J
[Note: The answer given above is presented considering standard conventions of writing number with its correct order of magnitude.]

Question 4.
A metal sphere cools at the rate of 0.05 ºC/s when its temperature is 70ºC and at the rate of 0.025 ºC/s when its temperature is 50 ºC. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 ºC.
Solution:

∴ 2(50 – T₀) = 70 – T₀
∴ T₀ = 30 πC
Substituting value of T₀.
0.05 = C (70 – 30)
∴ C = \(\frac{0.05}{40}\) = 0.00125/s.
For T₃ = 40 °C
\(\left(\frac{\mathrm{d} \mathrm{T}}{\mathrm{dt}}\right)_{3}\) = C(T₃ – T₀)
= 0.00125 (40 – 30)
= 0.00125 × 10
= 0.0125°C/s.
i) Temperature of surrounding is 30 °C.
ii) Rate of cooling at 40 °C is 0.0125 °C/s.

Question 5.
The volume of a gas varied linearly with absolute temperature if its pressure is held constant. Suppose the gas does not liquefy even at very low temperatures, at what temperature the volume of the gas will be ideally zero?
Answer:
At temperature of -273.15 °C, the volume of the gas will be ideally zero.

Question 6.
In olden days, while laying the rails for trains, small gaps used to be left between the rail sections to allow for thermal expansion. Suppose the rails are laid at room temperature 27 ºC. If maximum temperature in the region is 45 ºC and the length of each rail section is 10 m, what should be the gap left given that α = 1.2 × 10^-5 K^-1 for the material of the rail section?
Solution:
Given. T₁ = 27 °C, T₂ = 45 °C,
L₁ = 10m.
α = 1.2 × 10^-5 K^-1
To find: Gap that should be left (L₂ – L₁)
Formula: L₂ – L₁ = L₁ α(T₂ – T₁)
Calculation: From formula,
L₂ – L₁ = 10 × 1.2 × 10^-5 × (45 – 27)
= 2.16 × 10^-3 m
= 2.16 mm
The gap that should be left between rail sections is 2.16 mm.

Question 7.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the wooden rim and the iron ring are 1.5 m and 1.47 m respectively at room temperature of 27 ºC. To what temperature the iron ring should be heated so that it can fit the rim of the wheel (α_iron = 1.2 × 10^-5 K^-1).
Solution:
Given: d_w = 1.5 m, d = 1.47 m, T₁ = 27 °C.
α_i = 1.2 × 10^-5/ K
To find: Temperature (T₂)
Formula. α = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
T₂ = \(\frac{\mathrm{d}_{\mathrm{w}}-\mathrm{d}_{\mathrm{i}}}{\mathrm{d}_{\mathrm{i}} \alpha}\) + T₁
= \(\frac{1.5-1.47}{1.47 \times 1.2 \times 10^{-5}}\) + 27
= 1700.7 + 27
= 1727.7 °C
Iron ring should be heated to temperature of 1727.7 °C.

Question 8.
In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C.
Solution:
Here thermometric property P is temperature at some random scale X.
Using equation,
T = \(\frac{100\left(P_{T}-P_{1}\right)}{\left(P_{2}-P_{1}\right)}\)
For P₁ = 20 °X,
P₂ = 200 °X,
T = 62°C
∴ 62 = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-20\right)}{(200-20)}\)
∴ P_T = \(\frac{62 \times(200-20)}{100}\) + 20 = 111.6 + 20
= 131.6 °X
The boiling point of a liquid in this scale is 131.6 °X.

Question 9.
A gas at 900°C is cooled until both its pressure and volume are halved. Calculate its final temperature.
Solution:
Given: T₁ = 900 °C = 900 + 273.15 = 1173.15 K
V₂ = \(\frac{\mathrm{V}_{1}}{2}\), P₂ = \(\frac{\mathrm{P}_{1}}{2}\)
To find: Final temperature (T₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{\mathrm{I}}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula.
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{1173.15}=\frac{\mathrm{P}_{1} \mathrm{~V}_{\mathrm{l}}}{4 \mathrm{~T}_{2}}\)
∴ T₂ = \(\frac{1173.15}{4}\) = 293.29 K
Final temperature of gas is 293.29 K.

Question 10.
An aluminium rod and iron rod show 1.5 m difference in their lengths when heated at all temperature. What are their lengths at 0 °C if coefficient of linear expansion for aluminium is 24.5 × 10^-6 /°C and for iron is 11.9 × 10^-6 /°C
Solution:
Given: (L_T)_i – (L_T)_al = 1.5 m, T₀ = 0 °C
α_al = 24.5 × 10^-6/°C
α_i = 11.9 × 10^-6 /°C
To find: Lengths of aluminium and iron rod (L₀)_al and (L₀)_i
Formula: L_T = L₀[(1 + α(T – T₀)]
Calculation: For T₀ = 0 °C
From formula,
L_T = L₀(1 + αT)
For aluminium,
(L₀)_al = (L₀)_al(1 + α_alT) ……………. (1)
For iron,
(L_T)_i = (L₀)_i (1 + α_iT) ………….. (2)
Subtracting equation (2) by (1),
(L_T)_i – (L_T)_al = [(L₀)_i + (L₀)_i α_iT] – [(L₀)_al + (L₀)_alα_alT]
= (L₀)_i – (L₀)_al + [(L₀)_i α_i – (L₀)_al α_al]T
∴ 1.5 = 1.5 + [(L₀)_i α_i – (L₀)_al α_al)]T
⇒ [(L₀)_iα_i – (L₀)_alα_al] T = 0
∴ (L₀)_alα_al = (L₀)_iα_i

Length of aluminium rod at 0 °C is 1.417 m and that of iron rod is 2.917 m.

Question 11.
What is the specific heat of a metal if 50 cal of heat is needed to raise 6 kg of the metal from 20°C to 62 °C ?
Solution:
Given: Q = 50 cal, m =6 kg,
∆T = 62 – 20 = 42 °C
To find: Specific heat (s)
Formula: Q = ms ∆T
Calculation: From formula,
s = \(\frac{\mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\frac{50}{6 \times 42}\) = 0.198 cal/kg °C
Specific heat of metal is copper 0.198 cal/kg °C.

Question 12.
The rate of flow of heat through a copper rod with temperature difference 30 °C is 1500 cal/s. Find the thermal resistance of copper rod.
Solution:
Given: ∆T = 30 °C, P_cond = 1500 cal/s
To find: Thermal resistance (R_T)
Formula: R_T = \(\frac{\Delta \mathrm{T}}{\mathrm{P}_{\text {cond }}}\)
Calculation: From formula,
R_T = \(\frac{30}{1500}\)
= 0.02 °C s/cal.
Thermal resistance of copper rod is 0.02 °C s/cal.

Question 13.
An electric kettle takes 20 minutes to heat a certain quantity of water from 0°C to its boiling point. It requires 90 minutes to turn all the water at 100°C into steam. Find the latent heat of vaporisation. (Specific heat of water = 1cal/g°C)
Solution:
Let heat supplied by kettle in 20 minutes be Q₁ and that in 90 min. be Q₂.
Using heat temperature of water is raised from O °C to 100 °C.
If mass of water in the kettle is ‘m’ then.
Q₁ = ms_water∆T m × 1 × (100 – 0)
= 100 m ………….. (i)
…………. (∵ S_water = 1 cal/g °C)
Similarly using heat Q₂ water is converted from liquid to gas,
∴ Q₂ = mL_vap ……………. (ii)
Given that heat Q₁, Q₂ are supplied to water in 20 min. (t₁) and 90 min (t₂) respectively.
Kettle being same its conduction rate (P_cond) is same.
Using P_cond = \(\frac{\mathrm{Q}_{1}}{\mathrm{t}_{1}}=\frac{\mathrm{Q}_{2}}{\mathrm{t}_{2}}\) …………… (iii)
From (i), (ii) and (iii),
\(\frac{100 \mathrm{~m}}{20}=\frac{\mathrm{mL}_{\text {vap }}}{90}\),
∴ L_vap = 5 × 90 = 450 cal/g
Latent heat of vaporisation for water is 450 cal/g.

Question 14.
Find the temperature difference between two sides of a steel plate 4 cm thick, when heat is transmitted through the plate at the rate of 400 k cal per minute per square metre at steady state. Thermal conductivity of steel is 0.026 kcal/m s K.
Solution:

Temperature difference between two sides is 10.26 K.
[Note: Above answer is expressed in K (‘kelvin considering that thermal conductivity is expressed in units of kcal / ms K, and not as kcal / m s °C. As 1 °C equivalent to 1 K. conceptually temperature difference of 10.26 K will correspond to 10.26 t]

Question 15.
A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30°C?
Solution:
Given: T₁ = 80 °C, T₂ = 60 °C, T₃ = 40 °C, T₀ = 30 °C, (dt)₁ = 6 min.
To find: Time taken in cooling (dt)₂

Time taken in cooling is 10 min.

11th Physics Digest Chapter 7 Thermal Properties of Matter Intext Questions and Answers

Can you tell? (Textbook Page No. 125)

Question 1.
i) Why the metal wires for electrical transmission lines sag?
ii) Why a railway track is not a continuous piece but is made up of segments separated by gaps?
iii) How a steel wheel is mounted on an axle to fit exactly?
Answer:

1. In hot weather, metal wires get heated due to increased temperature of surrounding. As a result, they expand increasing the slack between transmission line structure, causing them to sag.
2. Railway tracks are made up of metals which expand upon heating. If no gap is kept between tracks, in hot weather, expansion of metal tracks may exert thermal stress on track. This may lead to bending of tracks which would be dangerous. Hence, railway track is not a continuous piece but is made up of segments separated by gaps.
3. The steel wheel is heated to expand. This expanded wheel can easily fit over axle. The wheel is then cooled quickly. Upon cooling, wheel contracts and fits tightly upon the axle.

Intext question. (Textbook Page No 124)

Question 1.
Can you now tell why the balloon bursts sometimes when you try to fill air in it?
Answer:

1. When balloon is blown, air that is blown inside makes the balloon expand.
2. A given size of balloon can expand upto certain limit.
3. Once that limit is reached and air is still blown inside the balloon, balloon cannot expand further.
4. As a result, air causes additional pressure on inner surface of balloon.
5. Since, pressure inside balloon is now greater than pressure outside balloon, balloon bursts equalizing the two pressures.

Can you tell? (Textbook Page No. 125)

Question 1.
Why lakes freeze first at the surface?
Answer:

1. In cold climate, temperature of water in ponds and lakes starts falling.
2. On getting colder, water contracts. As a result, density of water increases and it goes down. To replace it, warmer water from below rises up. This process continues till temperature of water at the bottom of pond becomes 4 °C.
3. Water, due to its anomalous behaviour possesses maximum density at 4 °C.
4. If the temperature lowers further, ice is formed at the surface of pond with water below it.
5. Ice being poor conductor of heat blocks the further heat exchange between atmosphere and water in the pond and maintains water below surface in liquid state.

Activity (Textbook Page No. 129)

Question 1.
To understand the process of change of state:
Take some cubes of ice in a beaker. Note the temperature of ice (0 °C). Start heating it slowly on a constant heat source. Note the temperature after every minute. Continuously stir the mixture of water and ice. Observe the change in temperature. Continue heating even after the whole of ice gets converted into water. Observe the change in temperature as before till vapours start coming out. Plot the graph of temperature (along Y-axis) versus time (along X-axis). Obtain a graph of temperature versus time.
Answer:
[Students are expected to attempt the activity on their own.]

Can you tell? (Textbook Page No. 130)

Question 1.
What is observed after point D in graph? Can steam be hotter than 100 °C?
Answer:
Beyond point D, thermometer again shows rise in temperature. This means, steam can be hotter than 100 °C and is termed as superheated steam.

Question 2.
Why steam at 100 °C causes more harm to our skin than water at 100 °C?
Answer:

1. Though steam and boiling water have same temperature, the heat contained in steam is more than that in boiling water.
2. Steam is formed when boiling water absorbs specific latent heat of vaporisation i.e.. 22.6 × 10⁵ J/kg.
3. As a result, when steam comes in contact with the skin of a person, it gives off additional 22.6 × 10⁵ joule per kilogram causing severe (more serious) burns.
   Hence, burns caused from steam are more serious than those caused from boiling water at same temperature.

Activity (Textbook Page No. 130)

Activity to understand the dependence of boiling point on pressure:
Take a round bottom flask, more than half filled with water. Keep it over a burner and fix a thermometer and steam outlet through the cork of the flask as shown in figure. As water in the flask gets heated, note that first the air, which was dissolved in the water comes out as small bubbles. Later bubbles of steam form at the bottom but as they rise to the cooler water near the top, they condense and disappear. Finally, as the temperature of the entire mass of the water reaches 100 oc, bubbles of steam reach the surface and boiling is said to occur. The steam in the flask may not be visible hut as it comes out of the flask, It condenses as tiny droplets of water giving a foggy appearance.

If now the steam outlet is closed for a few seconds to increase the pressure in the flask, you will notice that boiling stops. More heat would be required to raise the temperature (depending on the increase in pressure) before boiling starts again. Thus, boiling point increases with increase in pressure. Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometers and steam outlet. Close the flask with a air tight cork. Keep the flask turned upside down on a stand. Pour icecold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure and increases with increase in pressure.
Answer:
[Students are expected to attempt the activity an their own.]

Can you tell? (Textbook Page No. 131)

Question 1.
i) Why is cooking difficult at high altitude?
ii) Why is cooking faster in pressure cooker?
Answer:

1. • At high altitude density of air is low which causes reduction in atmospheric pressure.
   • As pressure is less, boiling point of water lowers.
   • Water, at high altitude, starts boiling below 100 OC.
   • As food is cooked mostly through the water boiling, cooking of food becomes difficult.
2. • Pressure cooker operates by expelling air within the cooker and trapping steam produced from the liquid. (mostly water) boiling inside.
   • Due to high internal pressure, boiling point of liquid increases and liquid boils at temperature higher than its boiling point.
   • The increased boiling point allows more absorption of heat by liquid and steam formed is superheated.
   • As a result, food gets cooked quickly.

Internet my friend (Textbook Page No. 139)

i) https ://hyperphysics. phy-astr.gsu.edul/base/hframe.html
ii) https://youtu.be/7ZKHc5J6R5Q
iii) https://physics. info/expansion
Answer:
[Students are expected to visit the above mentioned webs it es and collect more information about the thermal properties of matter.]

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 2 Mathematical Methods Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 2 Mathematical Methods

1. Choose the correct option.

Question 1.
The resultant of two forces 10 N and 15 N acting along + x and – x-axes respectively, is
(A) 25 N along + x-axis
(B) 25 N along – x-axis
(C) 5 N along + x-axis
(D) 5 N along – x-axis
Answer:
(D) 5 N along – x-axis

Question 2.
For two vectors to be equal, they should have the
(A) same magnitude
(B) same direction
(C) same magnitude and direction
(D) same magnitude but opposite direction
Answer:
(C) same magnitude and direction

Question 3.
The magnitude of scalar product of two unit vectors perpendicular to each other is
(A) zero
(B) 1
(C) -1
(D) 2
Answer:
(A) zero

Question 4.
The magnitude of vector product of two unit vectors making an angle of 60° with each other is
(A) 1
(B) 2
(C) \(\frac{3}{2}\)
(D) \(\frac{\sqrt{3}}{2}\)
Answer:
(D) \(\frac{\sqrt{3}}{2}\)

Question 5.
If \(\overrightarrow{\mathrm{A}}\), \(\overrightarrow{\mathrm{B}}\), and \(\overrightarrow{\mathrm{C}}\) are three vectors, then which of the following is not correct?
(A) \(\overrightarrow{\mathrm{A}}\) . (\(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{C}}\)) = \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{C}}\)
(B) \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{B}}\) . \(\overrightarrow{\mathrm{A}}\)
(C) \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{A}}\)
(D) \(\overrightarrow{\mathrm{A}}\) × (\(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{C}}\)) = \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) + \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{C}}\)
Answer:
(C) \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{B}}\) × \(\overrightarrow{\mathrm{A}}\)

2. Answer the following questions.

Question 1.
Show that \(\overrightarrow{\mathrm{A}}\) = \(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\) is a unit vector.
Solution:

Question 2.
If \(\overrightarrow{\mathbf{v}_{1}}\) = 3\(\hat{i}\) + 4\(\hat{j}\) + \(\hat{k}\) and \(\overrightarrow{\mathbf{v}_{2}}\) = \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\), determine the magnitude of \(\overrightarrow{\mathbf{v}_{1}}\) + \(\overrightarrow{\mathbf{v}_{2}}\).
Solution:
\(\overrightarrow{\mathbf{v}_{1}}\) + \(\overrightarrow{\mathbf{v}_{2}}\) = (3\(\hat{i}\) + 4\(\hat{j}\) + \(\hat{k}\)) + (\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\))
= 3\(\hat{i}\) + 3\(\hat{i}\) + 4\(\hat{j}\) – \(\hat{j}\) + \(\hat{k}\) – \(\hat{k}\)
= 4\(\hat{i}\) + 3\(\hat{j}\)
∴ Magnitude of (\(\overrightarrow{\mathbf{v}_{1}}\) + \(\overrightarrow{\mathbf{v}_{2}}\)),
|\(\overrightarrow{\mathbf{v}_{1}}\) + \(\overrightarrow{\mathbf{v}_{2}}\)| = \(\sqrt{4^{2}+3^{2}}\) = \(\sqrt{25}\) = 5 units.
Answer:
Magnitude of \(\overrightarrow{\mathbf{v}_{1}}\) + \(\overrightarrow{\mathbf{v}_{2}}\) = 5 units.

Question 3.
For \(\overline{\mathrm{v}_{1}}\) = 2\(\hat{i}\) – 3\(\hat{j}\) and \(\overline{\mathrm{v}_{2}}\) = -6\(\hat{i}\) + 5\(\hat{j}\), determine the magnitude and direction of \(\overline{\mathrm{v}_{1}}\) + \(\overline{\mathrm{v}_{2}}\).
Answer:
\(\overline{\mathrm{v}_{1}}\) + \(\overline{\mathrm{v}_{2}}\) = (2\(\hat{i}\) – 3\(\hat{j}\)) + (-6\(\hat{i}\) + 5\(\hat{j}\))
= (2\(\hat{i}\) – 6\(\hat{i}\)) + (-3\(\hat{j}\) + 5\(\hat{j}\))
= -4\(\hat{i}\) + 2\(\hat{j}\)
∴ |\(\overline{\mathrm{v}_{1}}\) + \(\overline{\mathrm{v}_{2}}\)| = \(\sqrt{(-4)^{2}+2^{2}}\) = \(\sqrt{20}\) = \(\sqrt{4 \times 5}\) = 2\(\sqrt{5}\)
Comparing \(\overline{\mathrm{v}_{1}}\) + \(\overline{\mathrm{v}_{2}}\), with \(\overrightarrow{\mathrm{R}}\) = R_x\(\hat{i}\) + R_y\(\hat{j}\)
⇒ R_x = -4 and R_y = 2
Taking θ to be angle made by \(\overrightarrow{\mathrm{R}}\) with X-axis,

Answer:
Magnitude and direction of \(\overline{\mathrm{v}_{1}}\) + \(\overline{\mathrm{v}_{2}}\), is
respectively 2\(\sqrt{5}\) and and tan^-1\(\left(-\frac{1}{2}\right)\) with X – axis.

Question 4.
Find a vector which is parallel to \(\overrightarrow{\mathrm{v}}\) = \(\hat{i}\) – 2\(\hat{j}\) and has a magnitude 10.
Answer:

Substituting for w_x in (i) using equation (ii),

Using equation (ii),

Answer:
Required vector is \(\frac{10}{\sqrt{5}} \hat{\mathbf{i}}\) – \(\frac{20}{\sqrt{5}} \hat{\mathbf{j}}\)

Alternate method:

When two vectors are parallel, one vector is scalar multiple of another,
i.e., if \(\overrightarrow{\mathrm{v}}\) and \(\overrightarrow{\mathrm{w}}\) are parallel then, \(\overrightarrow{\mathrm{w}}\) = n\(\overrightarrow{\mathrm{v}}\) where, n is scalar.

Question 5.
Show that vectors \(\vec{a}\) = 2\(\hat{\mathbf{i}}\) + 5\(\hat{\mathbf{j}}\) – 6\(\hat{\mathbf{k}}\) and \(\vec{b}\) = \(\hat{\mathbf{i}}\) + \(\frac{5}{2}\)\(\hat{\mathbf{j}}\) – 3\(\hat{\mathbf{k}}\) are parallel.
Answer:
Let angle between two vectors be θ.

⇒ Two vectors are parallel.

Alternate method:

\(\vec{a}\) = 2(\(\hat{\mathbf{i}}\) + \(\frac{5}{2}\)\(\hat{\mathbf{j}}\) + \(\hat{\mathbf{k}}\)) = 2\(\vec{b}\)
Since \(\vec{a}\) is a scalar multiple of \(\vec{b}\), the vectors are parallel.

3. Solve the following problems.

Question 1.
Determine \(\vec{a}\) × \(\vec{b}\), given \(\vec{a}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) and \(\vec{b}\) = 3\(\hat{\mathbf{i}}\) + 5\(\hat{\mathbf{j}}\).
Answer:
Using determinant for vectors in two dimensions,

Answer:
\(\vec{a}\) × \(\vec{b}\) gives \(\hat{\mathbf{k}}\)

Question 2.
Show that vectors \(\overrightarrow{\mathbf{a}}\) = 2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) + 6\(\hat{\mathbf{k}}\), \(\overrightarrow{\mathbf{b}}\) = 3\(\hat{\mathbf{i}}\) – 6\(\hat{\mathbf{j}}\) + 2\(\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{c}}\) = 6\(\hat{\mathbf{i}}\) + 2\(\hat{\mathbf{j}}\) – 3\(\hat{\mathbf{k}}\) are mutually perpendicular.
Solution:
As dot product of two perpendicular vectors is zero. Taking dot product of \(\vec{a}\) and \(\vec{b}\)


Combining two results, we can say that given three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular to each other.

Question 3.
Determine the vector product of \(\overrightarrow{\mathrm{v}_{1}}\) = 2\(\hat{i}\) + 3\(\hat{j}\) – \(\hat{k}\) and \(\overrightarrow{\mathrm{v}_{2}}\) = \(\hat{i}\) + 2\(\hat{j}\) – 3\(\hat{k}\) are perpendicular to each other, determine the value of a.
Solution:

Answer:
Required vector product is -7\(\hat{i}\) + 5\(\hat{j}\) + \(\hat{k}\)

Question 4.
Given \(\bar{v}_{1}\) = 5\(\hat{i}\) + 2\(\hat{j}\) and \(\bar{v}_{2}\) = a\(\hat{i}\) – 6\(\hat{j}\) are perpendicular to each other, determine the value of a.
Solution:
As \(\bar{v}_{1}\) and \(\bar{v}_{2}\) are perpendicular to each other, θ = 90°

Answer:
Value of a is \(\frac{12}{5}\).

Question 5.
Obtain derivatives of the following functions:
(i) x sin x
(ii) x⁴ + cos x
(iii) x/sin x
Answer:
(i) x sin x
Solution:
\(\frac{d}{d x}\)[f₁(x) × f₂(x)] = f₁(x)\(\frac{\mathrm{df}_{2}(\mathrm{x})}{\mathrm{dx}}\) + \(\frac{\mathrm{df}_{1}(\mathrm{x})}{\mathrm{dx}}\)f₂(x)
For f₁(x) = x and f₂(x) = sin x
\(\frac{d}{d x}\)(x sin x) = x\(\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}\) + \(\frac{d(x)}{d x}\) sin x
= x cos x + 1 × sin x
= sin x + x cos x

(ii) x⁴ + cos x
Solution:

(iii) \(\frac{\mathbf{x}}{\sin x}\)
Solution:

[Note: As derivative of (sin x) is cos x, negative sign that occurs in rule for differentiation for quotient of two functions gets retained in final answer]

Question 6.
Using the rule for differentiation for quotient of two functions, prove that \(\frac{d}{d x}\left(\frac{\sin x}{\cos x}\right)\) = sec²x
Solution:

Question 7.
Evaluate the following integral:
(i) \(\int_{0}^{\pi / 2} \sin x d x\)
(ii) \(\int_{1}^{5} x d x\)
Answer:
(i) \(\int_{0}^{\pi / 2} \sin x d x\)
Solution:

(ii) \(\int_{1}^{5} x d x\)
Solution:

11th Physics Digest Chapter 2 Mathematical Methods Intext Questions and Answers

Can you recall? (Textbook Page No. 16)

Question 1.
Define scalars and vectors.
Answer:

1. Physical quantities which can be completely described b their magnitude (a number and unit) are called scalars.
2. Physical quantities which need magnitude, as well as direction for their complete description, are called vectors.

Question 2.
Which of the following are scalars or vectors?
Displacements, distance travelled, velocity, speed, force, work done, energy
Answer:

1. Scalars: Distance travelled, speed, work done, energy.
2. Vectors: Displacement, velocity, force.

Question 3.
What is the difference between a scalar and a vector?
Answer:

No.	Scalars	Vectors
i.	It has magnitude only	It has magnitude as well as direction.
ii.	Scalars can be added or subtracted according to the rules of the algebra.	Vectors are added or subtracted by the geometrical (graphical) method or vector algebra.
iii.	It has no specific representation.	It is represented by the symbol (→) arrow.
iv.	The division of a scalar by another scalar is valid.	The division of a vector by another vector is not valid.
	Example: Length, mass, time, volume, etc.	Example: Displacement, velocity, acceleration, force, etc.

Internet my friend (Textbook page no. 28)

1. 1. hyperphysics.phy-astr.gsu.edu/hbase/vect. html#veccon
   2. hyperphysics.phy-astr.gsu.edu/hbase/ hframe.html

Answer:
[Students can use links given above as a reference and collect information about mathematical methods]

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

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

1. Choose the correct option.

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

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

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

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

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

2. Answer the following questions.

Question 1.
Star A is farther than star B. Which star will have a large parallax angle?
Answer:

i). ‘b’ is constant for the two stars
∴ θ ∝ \(\frac{1}{D}\)

ii) As star A is farther i.e., D_A > D_B
⇒ θ_A < θ_B
Hence, star B will have larger parallax angle than star A.

Question 2.
What are the dimensions of the quantity l \(\sqrt{l / g}\), l being the length and g the acceleration due to gravity?
Answer:

[Note: When power of symbol expressing fundamental quantity appearing in the dimensional formula is not given, ills taken as 1.]

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

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

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

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

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

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

Rules for determining significant figures:

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

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

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

Question 5.
If the measured values of two quantities are A ± ∆A and B ± ∆B, ∆A and ∆B being the mean absolute errors. What is the maximum possible error in A ± B? Show that if Z = \(\frac{A}{B}\)
\(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)
Answer:
Maximum possible error in (A ± B) is (∆A + ∆B).
Errors in divisions:
i) A
Suppose, Z = \(\frac{A}{B}\) and measured values of A and B are (A ± ∆A) and (B ± ∆B) then,


∴ Maximum relative error of \(\frac{\Delta Z}{Z}=\pm\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}\right)\)

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

Question 6.
Derive the formula for kinetic energy of a particle having mass m and velocity v using dimensional analysis
Answer:
Kinetic energy of a body depends upon mass (m) and velocity (v) of the body.
Let K.E. ∝ m^x v^y
∴ K.E. = km^x v^y …….. (1)
where,
k = dimensionless constant of proportionality. Taking dimensions on both sides of equation (1),
[L²M¹T^-2] – [L⁰M¹T⁰]^x [L¹M⁰T^-1]^y
= [L⁰M^xT⁰] [L^yM⁰T^-y]
= [L^0+yM^x+0T^0-y]
[L²M¹T²] = [L^yM^xT^-y] …………. (2)
Equating dimensions of L, M, T on both sides of equation (2),
x = 1 and y = 2 ,
Substituting x, y in equation (1), we have
K.E. = kmv²

3. Solve numerical examples.

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

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

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

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

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

Question 3.
An electron with charge e enters a uniform. magnetic field \(\vec{B}\) with a velocity \(\vec{v}\). The velocity is perpendicular to the magnetic field. The force on the charge e is given by
|\(\vec{F}\)| = Bev Obtain the dimensions of \(\vec{B}\).
Answer:
Given: |\(\vec{F}\)| = B e v
Considering only magnitude, given equation is simplified to,
F = B e v

∴ B = [L⁰M¹T^-2I^-1]
[Note: The answer given above is calculated in accordance with textual method considering the given data.]

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

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

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

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

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

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

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

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

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

To find:

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

Formulae:

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

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

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

Question 9.
An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. If v depends upon gravitational acceleration g and distance, prove with dimensional analysis that v = k\(\sqrt{g h}\) where k is a constant.
Answer:
Given = v = k\(\sqrt{g h}\)

k being constant is assumed to be dimensionless.
Dimensions of L.H.S. = [v] = [L¹T^-1]
Dimension of R.H.S. = [latex]\sqrt{g h}[/latex]
= [L¹T^-2]^{\(\frac{1}{2}\)} × [L¹]^{\(\frac{1}{2}\)}
= [L²T^-2]^{\(\frac{1}{2}\)}
= [L¹T^-1]
As, [L.H.S.] = [R.H.S.],
=> v = k\(\sqrt{g h}\)is dimensionally correct equation.

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

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

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

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

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

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

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

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

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

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

ii) Releative error \(\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{\Delta V}{V}\)

iii) Percentage error= Relative error × 100%

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

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

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

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

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

Question 15.
Write down the number of significant figures in the following: 0.003 m², 0.1250 gm cm^-2, 6.4 × 10⁶ m, 1.6 × 10^-19 C, 9.1 × 10^-31 kg.
Answer:

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

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

Can you recall (Textbook Page No. 1)

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

1. The standard measure of any quantity is called the unit of that quantity.
2. 
3. MKS or SI system is used mostly. At times. even CGS system is used.

Can you tell? (Textbook Page No. 8)

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

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

Activity (Textbook Page No. 10)

Perform an experiment using a Vernier callipers of least count 0.01cm to measure the external diameter of a hollow cylinder. Take 3 readings at different positions on the cylinder and find (i) the mean diameter (ii) the absolute mean error and (iii) the percentage error in the measurement of diameter.
Answer:
Given: L.C. = 0.01 cm
To measure external diameter of hollow cylinder readings are taken as follows:

[Note: The above table is made assuming zero error in Vernier calipers. If caliper has positive or negative zero error, the zero error correction needs to be introduced into observed reading.]

Internet my friend (Textbook Page No. 12)

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

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

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

Question 1.
Write the equation of the line:
i. parallel to the X-axis and at a distance of 5 units from it and above it.
ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above X-axis, k = 5
∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to Y-axis is x = h. Since the line is at a distance of 5 units to the left of Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2.
Obtain the equation of the line:
i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.
ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.
Here, y-intercept = 3
∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

Question 3.
Obtain the equation of the line containing the point:
i. A(2, – 3) and parallel to the Y-axis.
ii. B(4, – 3) and parallel to the X-axis.
Solution:
i. Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through A(2, – 3), h = 2
∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4.
Find the equation of the line:
i. passing through the points A(2, 0) and B(3,4)
ii. passing through the points P(2, 1) and Q(2,-1)
Solution:
i. The required line passes through the points A(2, 0) and B(3,4).
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁y₁) = (2,0) and (x₁,y₂) = (3,4)
∴ The equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5.
Find the equation of the line:
i. containing the origin and having inclination 60°.
ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).
iii. having slope 1/2 and containing the point (3, -2)
iv. containing the point A(3, 5) and having slope 2/3
v. containing the point A(4, 3) and having inclination 120°.
vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution:
i. Given, Inclination of line = θ = 60°
Slope of the line (m) = tan θ = tan 60°
= \(\sqrt{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
.‘. The equation of the required line is y = \(\sqrt{3}\) x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = \(\frac{7-4}{1-2}\) = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = \(=\frac{1}{2}\) and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y ₁= m(x-x₁)
∴ The equation of the required line is
[y-(- 2)]=\(\frac{1}{2}\)(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = \(\frac{2}{3}\) and the line passes through (3, 5).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is y – 5 = \(\frac{2}{3}\)(x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= – \(\sqrt{3}\)
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is
y- 3 = –\(\sqrt{3}\)(x-4)
∴ y – 3 = –\(\sqrt{3}\) x + 4\(\sqrt{3}\)
∴ \(\sqrt{3}\)x + y – 3 -4\(\sqrt{3}\) = 0

vi.

Given equation of the line is 3x +y = 6.
∴ \(\frac{x}{2}+\frac{y}{6}=1\)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}\) = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( \(\frac{2+0}{2}, \frac{0+6}{2}\) ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{3-0}=\frac{x-0}{1-0}\)
\(\frac{y}{3}=\frac{x}{1}\)
∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = \(\left(\frac{2+0}{2}, \frac{0+6}{2}\right)\) = (1,3)
Slope of OM (m) = \(\frac{3-0}{1-0}\) = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6.
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution:
Given, A(2, 1) and B(3,2)
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7.
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution:
Given, Inclination of line = 0 = 135°
∴ Slope of the line (m) = tan 0 = tan 135°
= tan (90° + 45°)
= – cot 45° = – 1 x-intercept of the required line is 7.
∴ The line passes through (7, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ The equation of the required line is y — 0 = – 1 (x – 7)
∴ y = -x + 7
∴ x + y – 7 = 0

Question 8.
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing
i. side BC
ii. the median AD
iii. the midpoints of sides AB and BC.
Solution:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).
i. Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\)
\(\frac{y}{6}=\frac{x-2}{-3}\)
∴ – 3y = 6x – 12
∴ 6x + 3y – 12 = 0
∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points
A(3,4) and D( \(\frac{1}{2}\) , 3)

∴ The equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
\(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
\(\frac{5}{2}\)(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9.
Find the x and y-intercepts of the following lines:
i. \(\frac{x}{3}+\frac{y}{2}=1\)
ii. \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
iii. 2x – 3y + 12 = 0
Solution:
i. Given equation of the line is latex]\frac{x}{3}+\frac{y}{2}=1[/latex]
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}\) = 1
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}\) = 1
This is of the form = \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10.
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) ………..(i)
Since, the sum of the intercepts of the line is zero.
∴ a + b = 0
∴ b = – a
Substituting b = – a in (i), we get
\(\frac{x}{a}+\frac{y}{(-a)}=1\)
x – y = a .. .(ii)
Since, the line passes through A(1, 3).
∴ 1 – 3 = a
∴ a = – 2
Substituting the value of a in (ii), equation of the required line is
∴ x – y = – 2,
∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = \(\frac{3-0}{1-0}\) = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11.
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …………(i)
This line passes through A(3, 4).
∴ \(\frac{3}{a}+\frac{4}{b}=1\)……………..(ii)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b …(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{4}{a}=1\)
∴ \(\frac{7}{a}=1\)
∴ a = 7
∴ b = 7 …[From (iii)]
Substituting the values of a and b in (i), equation of the required line is
\(\frac{x}{7}+\frac{y}{7}=1\) = 1
∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m = \(=\frac{4-0}{3-0}=\frac{4}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0

Question 12.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution:

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
Slope of BE = \(\frac{-3}{4}\)
…[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope \(\frac{-3}{4}\),
equation of the altitude BE is
y-(-1) = \(\frac{-3}{4}\) (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ Slope of CF = \({2}{3}\) ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , \(\frac{2}{3}\)
equation of the altitude CF is
y-(-3) = \(\frac{2}{3}\)[x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0

Question 13.
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution:

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.
A is the midpoint of side PQ.

Slope of perpendicular bisector of PQ is \(\frac{1}{2}\) and it passes through (\(\frac{3}{2}\)), 3).
Equation of the perpendicular bisector of side PQ is
y – 3 = \(\frac{1}{2}\)(x – \(\frac{3}{2}\))
y – 3 = (\(\frac{1}{2}\left(\frac{2 x-3}{2}\right)\))
∴ 4(y – 3) = 2x – 3
∴ 4y – 12 = 2x – 3
∴ 2x – 4y + 9 = 0
B is the midpoint of side QR
∴ B = \(\left(\frac{4-5}{2}, \frac{-2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-5}{2}\right)\)
Slope of side QR = \(\frac{-3-(-2)}{-5-4}=\frac{-1}{-9}=\frac{1}{9}\)
∴ Slope of perpendicular bisector of QR is -9 and it passes through \(\left(-\frac{1}{2},-\frac{5}{2}\right)\)
∴ Equation of the perpendicular bisector of side QR is

∴ 2y + 5 = -18x – 9
∴ 18x + 2y + 14 = 0
∴ 9x + y + 7 = 0
C is the midpoint of side PR.

Equation of the perpendicular bisector of PR is \(y-\frac{5}{2}=-\frac{4}{11}(x+3)\)
∴ \(11\left(\frac{2 y-5}{2}\right)\) =-4(x + 3)
∴ 11(2y – 5) = – 8 (x + 3)
∴ 22y – 55 = – 8x – 24
∴ 8x + 22y -31 = 0

Question 14.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution:
Let O be the orthocentre of AABC.
Let AM and BN be the altitudes of sides BC and AC respectively.
Now, slope of BC = \(\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\)
Slope of AM = -2 ,..[∵ AM ⊥ BC]
Since AM passes through (2, – 2) and has slope -2,
equation of the altitude AM is y – (- 2) = – 2 (x – 2)
∴ y + 2 = -2x + 4
∴ 2x + y – 2 = 0 …(i)

Also, slope of AC = \(\frac{0-(-2)}{-1-2}=\frac{2}{-3}\)
∴ Slope of BN = \(\frac{3}{2}\) …[∵ BN ⊥ AC]
Since BN passes through (1,1) and has slope \(\frac{3}{2}\), equation of the altitude BN is
y – 1 = \(\frac{3}{2}\)(x-1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0 …(ii)
To find co-ordinates of orthocentre, we have to solve equations (i) and (ii).
By (i) x 2 + (ii), we get
7x – 5 = 0
∴ x = \(\frac{5}{7}\)
substituting x = \(\frac{5}{7}\) in eq (i), we get
2(\(\frac{5}{7}\)) + y – 2 = 0
∴ y = -2(\(\frac{5}{7}\)) + 2
∴ y = \(\frac{-10+14}{7}=\frac{4}{7}\)
∴ Coordinates of orthocentre O = \(\left(\frac{5}{7}, \frac{4}{7}\right)\)

Question 15.
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution:

Slope of ON = \(\frac{-4-0}{3-0}=\frac{-4}{3}\)
Since line L ⊥ ON,
slope of the line L is \(\frac{3}{4}\) and it passes through point N(3, -4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line L is
y-(-4) = \(\frac{3}{4}\)(x-3)
∴ 4(y + 4) = 3(x – 3)
∴ 4y + 16 = 3x – 9
∴ 3x – 4y – 25 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.2

Question 1.
Find the slope of each of the lines which passes through the following points:
i. A(2, -1), B(4,3)
ii. C(- 2,3), D(5, 7)
iii. E(2,3), F(2, – 1)
iv. G(7,1), H(- 3,1)
Solution:
i. Here, A = (2, -1) andB = (4, 3)
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-(-1)}{4-2}=\frac{4}{2}\) = 2

ii. Here, C = (-2, 3) and D = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-3}{5-(-2)}=\frac{4}{7}\)

iii. Here, E s (2, 3) and F = (2, -1)
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{2-2}=\frac{-4}{0}\), which ix not defined.

Alternate Method:
Points E and F have the same x co-ordinates i.e. 2.
Points E and F lie on a line parallel to Y-axis.
∴ The slope of EF is not defined.

iv. Here, G = (7, 1) and H = (-3, 1)
Slope of line GH = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-1}{-3-7}\) = o

Alternate Method:
Points G and H have the same y co-ordinate i.e. 1.
∴ Points G and H lie on a line parallel to the
X-axis.
∴ The slope of GH is 0.

Question 2.
If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, the x-intercept of line L is 2 and the y-intercept of line L is 3
∴ The line L intersects X-axis at (2, 0) and Y-axis at (0,3).
∴ The line L passes through (2, 0) and (0, 3).
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-2}=\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
∴ Slope of the line = tanθ = tan30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is \(\frac{\pi}{4}\)
Solution:
Given, inclination (0) = \(\frac{\pi}{4}\)
∴ Slope of the line = tan θ = tan\(\frac{\pi}{4}\) = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the inclination of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
∴ The line passes through (3, 0) and (0,3).
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-3}\) = -1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan \(\frac{\pi}{4}\)
= tan(π-\(\frac{\pi}{4}\) ) …[v tan(π – θ) = -tan θ]
tan θ = tan \(\frac{3\pi}{4}\)
θ = \(\frac{3\pi}{4}\)
The inclination of the line is \(\frac{3\pi}{4}\).
[Note: Answer given in the textbook is ‘-1 However, as per our calculation it is \(\frac{3\pi}{4}\)]

Question 6.
Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.
Solution:
Given, A(4,4), B(3, 5), C (-1, -1).
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}\) = – 1
Slope of BC = \(\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{-1-4}{-1-4}\) = 1
Slope of AB x slope of AC = – 1 x 1 = – 1
∴ side AB ⊥ side AC
∴ ∆ABC is a right angled triangle right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured anticlockwise.
Solution:

Since the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction,
Inclination of the line (0) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= – cot 45°
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2,1) and R(4,5) are collinear.
Solution:
Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR .
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 4 = 4 (2 – k)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Question 9.
Find the acute angle between the X-axis and the line joining the points A(3, -1) and B(4, – 2).
Solution:
Given, A (3, – 1) and B (4, – 2)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-(-1)}{4-3}\) = – 1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan 45°
= tan (180° -45°)
… [∵ tan (180° – θ) = -tan θ]
= tan 135°
∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis.
Then, α + 0 = 180°
α = 180°- 135° = 45°
∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

Question 10.
A line passes through points A(xi, y0 and B(h, k). If the slope of the line is m, then show that k – y₁ = m (h – x₁).
Solution:
Given, A(x₁, y₁), B(h, k) and
slope of line AB = m
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
∴ m = \(\frac{\mathrm{k}-y_{1}}{\mathrm{~h}-x_{1}}\)
∴ k – y₁ = m (h – x₁)

Question 11.
If the points A(h, 0), B(0, k) and C(a, b) lie on a line, then show that \(\frac{a}{h}+\frac{b}{k}\) = 1. ‘
Solution:
Given, A(h, 0), B(0, k) and C(a, b)
Since the points A, B and C lie on a line, they are collinear.
∴ Slope of AB = slope of BC