Class 11

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 15 Waves Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 15 Waves

PSEB 11th Class Physics Guide Waves Textbook Questions and Answers

Question 1.
A string of mass 50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Solution:
Mass of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, l = 20.0 m
Mass per unit length, µ = \(\frac{M}{l}=\frac{2.50}{20} \) = 0.125 kg m^-1
The velocity (υ) of the transverse wave in the string is given by the relation:
υ = \(\sqrt{\frac{T}{\mu}}=\sqrt{\frac{200}{0.125}}=\sqrt{1600} \) = 40 m/s

∴ Time taken by the disturbance to reach the other end, t = \(\frac{l}{v}=\frac{20}{40}\) = 0.5 s

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340ms^-1?(g=9.8ms^-2)
Solution:
Height of the tower, s = 300 m
The initial velocity of the stone, µ = 0
Acceleration, a = g = 9.8 m/s²
Speed of sound in air = 340 rn/s

The time (t₁) taken by the stone to strike the water in the pond can be calculated using the second equation of motion, as:
s=ut₁+\(\frac{1}{2}\) gt₁²
300 = 0+ \( \frac{1}{2}\) × 9.8 × t₁²
∴ t₁ =\( \sqrt{\frac{300 \times 2}{9.8}}\) = 7.82 s
Time taken by the sound to reach the top of the tower,
t₂ =\(\frac{300}{340}\) = 0.88 s
Therefore, the time after which the splash is heard, t = t₁ +t₂
= 7.82+0.88= 8.7 s .

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 ms^-1.
Solution:
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.10 kg
Velocity of the transverse wave, ν = 343 m/s
Mass per unit length, µ =\(\frac{m}{l}=\frac{2.10}{12}\) = 0.175 kg m^-1
For tension T, velocity of the transverse wave can be obtained using the relation:
υ = \(\sqrt{\frac{T}{\mu}}\)
∴ T = υ²µ = (343)² × 0.175 = 20588.575 ≈ 2.06 ×10⁴ N

Question 4.
Use the formula υ = \(\sqrt{\frac{\gamma \boldsymbol{P}}{\rho}}\) to explain why the speed of sound in
air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Solution:
(a) Given the relation:
υ = \( \sqrt{\frac{\gamma \boldsymbol{P}}{\rho}} \) ……………………………. (i)
Density, ρ = \(\frac{\text { Mass }}{\text { Volume }}=\frac{M}{V}\)
where, M = Molecular weight of the gas; V = Volume of the gas
Hence, equation ‘(i) reduces to:
υ = \(\sqrt{\frac{\gamma P V}{M}}\) …………………………………………. (ii)
Now, from the ideal gas equation for n = 1 :
PV = RT
For constant T, PV = Constant
Since both M and γ are constants, υ = Constant
Hence, at a constant temperature, the speed of sound in a gaseous
medium is independent of the change in the pressure of the gas.

(b) Given the relation:
υ = \( \sqrt{\frac{\gamma \boldsymbol{P}}{\rho}} \) …………………………………. (i)
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
P = \(\frac{R T}{V}\)

Substituting equation (ii) in equation (i), we get:
υ = \(\sqrt{\frac{\gamma R T}{V \rho}}=\sqrt{\frac{\gamma R T}{M}}\) ……………………………… (iii)
where, M = mass = ρV is a constant; γ and R are also constants We conclude from equation (iii) that v ∝ \(\sqrt{T}\) .
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i. e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.

(c) Let υ_m and υ_d be the speeds of sound in moist air and dry air respectively.
Let ρ_m and ρ_d be the densities of moist air and dry air respectively.
Take the relation:
υ = \(\sqrt{\frac{\gamma P}{\rho}}\)
Hence, the speed of sound in moist air is:
υ_m = \(\sqrt{\frac{\gamma P}{\rho_{m}}}\) ………………………….. (i)
And the speed of sound in dry air is:
υ_d = \(\sqrt{\frac{Y P}{P_{d}}}\) ………………………………… (ii)

On dividing equations (i) and (ii), we get:
\(\frac{v_{m}}{v_{d}} \)
On dividing equations (i) and (ii), we get:
\(\frac{v_{m}}{v_{d}}=\sqrt{\frac{\gamma P}{\rho_{m}}} \times \frac{\rho_{d}}{\gamma P}=\sqrt{\frac{\rho_{d}}{\rho_{m}}} \)
However, the presence of water vapour reduces the density of air, i.e.,
ρ_d>ρ_m
∴ υ_m>υ_d
Hence the speed of sound in moist air greater than it is in dry air.
Thus, in a gaseous medium, the speed of sound increase with humidity.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear
in the combination x-υt or x+υt, i.e., y=f(x±υt). Is the converse true? Examine if the following functions for y can
possibly represent a traveIliig wave:
(a) (x—υt)² (b)log \(\left[\frac{x+v t}{x_{0}}\right] \) (c) \(\frac{1}{(x+v t)}\)
Solution:
No, the converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should
remain finite for all values of x and t.

(a) Does not represent a wave
Explanation :
For x = 0 and t = 0, the function (x – υt)² becomes 0.
Hence, for x = 0 and t = 0, the function represents a point and not a wave,

(b) Represents a wave Explanation:
For x = 0 and t = 0, the function log \(\left(\frac{x+v t}{x_{0}}\right)\) = log 0 = ∞
Since the function does not converge to a finite value for x = 0 and t = 0, it represents a travelling wave.

(c) Does not represents a wave
Explanation :
For x = 0 and t = 0, the function
\(\frac{1}{x+v t}\) = log \(\frac{1}{0} \) = ∞
Since the function does not converge to a finite value for x = 0 and t = 0, it does not represent a travelling wave.

Question 6.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 ms^-1 and in water 1486 ms^-1.
Solution:
(a) Frequency of the ultrasonic sound, υ = 1000 kHz = 10⁶ Hz
Speed of sound in water, υ_a = 340 m/s
The wavelength (λ_r)of the transmitted sound is given as:
λ_r = \(\frac{v}{v}=\frac{340}{10^{6}}\) = 3.4 × 10^-4 m

(b) Frequency of the ultrasonic sound, v = 1000 kHz = 10⁶ Hz
Speed of sound in water, υ_w, =1486 m/s
The wavelength of the transmitted sound is given as:
λ_t= \(\frac{1486}{10^{6}}\) = 1.49 × 10^-3 m

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Solution:
Speed of sound in the tissue, υ = 1.7 km/s = 1.7 x 10 3 m/s
Operating frequency of the scanner, v = 4.2 MHz = 4.2 x 10⁶ Hz
The wavelength of sound in the tissue is given as:
λ = \(\frac{v}{v}=\frac{1.7 \times 10^{3}}{4.2 \times 10^{6}}\) = 4.1 x 10^-4m.

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0sin(36t+0.018x+\(\frac{\pi}{4}\))
where x and y are in cm and t in s. The positive direction of x is from left to right.
Is this a travelling wave or a stationary wave?
(a) If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Solution:
(a) Yes.
The equation of a progressive wave travelling from right to left is given by the displacement function:
y(x,t) = a sin(ωt + kx + Φ) ……………………………………. (i)
The given equation is
y(x, t) = 3.0 sin( 36t +0.018x+\(\frac{\pi}{4}\)) …………………………………. (ii)
On comparing both the equations, we find that equation (ii) represents a travelling wave, propagating from right to left.
Now, using equations (i) and (ii), we can write:
ω = 36 rad/s and k = 0.018 cm^-1
We know that
v = \(\frac{\omega}{2 \pi}\) and λ = \(\frac{2 \pi}{k}\)
Also,
υ = vλ
∴ υ = \(\left(\frac{\omega}{2 \pi}\right) \times\left(\frac{2 \pi}{k}\right)\) = \(\frac{\omega}{k}=\frac{36}{0.018} \) = 2000 cm/s = 20 m/s
Hence, the speed of the given travelling wave is 20 m/s.

(b) Amplitude of the given wave, a =3 cm (Given)
Frequency of the given wave:
v = \(\frac{\omega}{2 \pi}=\frac{36}{2 \times 3.14}\) = 5.73 Hz

(c) On comparing çquations (i) and (ii), we find that the initial phase angle, Φ = \(\frac{\pi}{4}\)

(d) The distance between two successive crests or troughs is equal to the
wavelength of the wave.
Wavelength is given by the relation:
k= \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{0.018}\) = 348.89 cm = 3.49 m.

Question 9.
For the wave described in question 8, plot the displacement (y) versus (t) graphs for s =0,2 and 4 çm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Solution:
All the waves have different phases. The given transverse harmonic wave is
y(x,t) = 3.0 sin (36t+0.018x + \(\frac{\pi}{4}\))
For x = 0, the equation reduces to
y(0,t) = 3.0 sin (36t+\(\frac{\pi}{4}\)) ………………………….. (i)
Also,
ω = \(\frac{2 \pi}{T}\) = 36 rad/s
∴ T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{36} \) = \(\frac{\pi}{18}\) s
For different values of t, we calculate y using eq. (i). These values are tabulated below

On plotting y versus t graph, we obtain a sinusoidal curve as shown in figure below.

Similar graphs are obtained for x = 2 cm and x = 4 cm.
The oscillatory motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three cases.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080x +0.35)
where, x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4m
(b) 0.5m
(c) \(\frac{\lambda}{2}\)
(d) \(\frac{3 \lambda}{4}\)
Solution:
Equation for a travelling harmonic wave is given as
y{x,t) = 2.0 cos 2π(10t -0.0080x +0.35)
= = 2.0 cos (20πt – 0.016πx+0.70π)

where, propagation constant, k = 0.0160π
Amplitude, a = 2 cm
Angular frequency, ω = 20 π rad/s
Phase difference is given by the relation:
Φ =kx=\(\frac{2 \pi}{\lambda} \)

(a)
For x=4m=400cm
Φ =0.016 π × 400 =6.4 π rad

(b) For 0.5 m=50cm
Φ = 0.1016 π × 50 = 0.8 π rad

(c) For x= \(\frac{\lambda}{2}\)
Φ = \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) = π rad

(d) For x= \(\frac{3 \lambda}{4}\)
Φ = \(\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4} \) = 1.5π rad

Question 11.
The transverse displacement of a string (clamped at its both ends) is given by y(x,t) = 0.06 sin \(\frac{\mathbf{2} \pi}{\mathbf{3}}\) x cos (120πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^-2 Kg Answer the following
(a) Does the function represents a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength,
frequency, and speed of each wave?
(c) Determine the tension in the string.
Solution:
(a) The general equation representing a stationary wave is given by the displacement function:
y(x,t) = 2asinkxcosωt
This equation is similar to the given equation:
y(x,t)= 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos (120πt)
Hence, the given function represents a stationary wave.

(b) A wave travelling along the positive x -direction is given as
y₁ =asin(ωt -kx)
The wave travelling along the negative x -direction is given as:
y₂ = -asin(ωt +kx)
The superposition of these two waves yields:
y= y₁+y₂ = asin(ωt -kx)-asin(ωt +kr)
= asin(ωt)cos(kx) – asin(kx)cos(ωt)- asin(ωt)cos(kx) – asin(kx)cos(ωt)
= -2asin(kx)cos(ωt)
= – 2asin \(\left(\frac{2 \pi}{\lambda} x\right)\)cos (2πcvt) …………………………….. (i)

The transverse displacement of the string is given as y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)cos (120πt) ………………………………. (ii)
Comparing equations (i) and (ii), we have
\(\frac{2 \pi}{\lambda}=\frac{2 \pi}{3}\)
∴ Wavelength, λ = 3 m
it is given that
120 π =2πv
Frequency, ν =60 Hz
Wave speed, υ = vλ
=60 × 3=180 m/s

(c) The velocity of a transverse wave travelling in a string is given by the relation
υ = \(\sqrt{\frac{T}{\mu}} \) ………………………… (iii)
where, µ = Mass per unit length of the string = \(\frac{m}{l}=\frac{3.0}{1.5} \times 10^{-2}\)
=2 x 10^-2 kgm^-1
T = Tension in the string = T
From equation (iii), tension can be obtained as
T =ν²µ=(180)² x 2 x 10^-2 =648 N

Question 12.
(i) For the wave on a string described in question 11, do all the points on the string oscillate with the same
(a) frequency,
(b) phase, (c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Solution:
(I) (a) Yes, except at the nodes; All the points on the string oscillate with the same frequency, except at
the nodes which have zero frequency.

(b) Yes, except at the nodes;
All the points in any vibrating loop have the same phase, except at the nodes.

(C) No;
All the points in any vibrating loop have different amplitudes of vibration.

(ii) The given equation is .
y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)cos (120πt)
For x = 0.375m and t =0
Amplitude = Displacement 0.06sin \(\left(\frac{2 \pi}{3} x\right) \cos 0^{\circ}\)
= 0.06 sin \(\left(\frac{2 \pi}{3} \times 0.375\right) \times 1\)
= 0.06 sin(0.25π) = 0.06 sin\(\left(\frac{\pi}{4}\right)\)
= 0.06 x \(\frac{1}{\sqrt{2}}\) = 0.042 m

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:
(a) y = 2cos(3x)sin(10t)
(b) y = 2\(\sqrt{x-v t}\)
(c) y = 3sin(5x – 0.5t) + 4cos(5x – 0.5t)
(d) Y = cos x sin t + cos 2x sin 2t
Solution:
(a) The given equation represents a stationary wave because the harmonic terms kx and cot appear separately in the equation.

(b) The given equation does not contain any harmonic term. Therefore, it does not represent either a travelling wave or a stationary wave.’

(c) The given equation represents a travelling wave as the harmonic terms kx and cot are in the combination of kx – cot.

(d) The given equation represents a stationary wave because the harmonic terms kx and cot appear separately in the equation. This equation actually represents the superposition of two stationary waves.

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x10^-2 kg and its linear mass density is 40 x 10^-2 kgm^-1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Solution:
Mass of the wire, m = 3.5×10 ^-2 kg
Linear mass density, μ = \(\frac{m}{l}\) = 4.0 × 10^-2 kg m^-1
Frequency of vibration, μ = 45 Hz
∴ Length of the wire, l = \(\frac{m}{\mu}=\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}\) =0.875 m
The wavelength of the stationary wave (λ,) is related to the length of the wire by the relation:
λ = \(\frac{2 l}{n}\)
where, n = Number of nodes in the wire For fundamental node, n = 1:
λ =2l
λ =2 x 0.875 = 1.75 m
(a) The speed of the transverse wave in the string is given as
υ = vλ = 45 x 1.75 = 78.75 m/s

(b) The tension produced in the string is given by the relation:
T =υ² μ
= (78.75)² x 4.0 x 10^-2 =248.06 N

Question 15.
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz)when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Solution:
Frequency of the turning fork, v = 340 Hz
Since the given pipe is attached with a piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation l₁ = \(\frac{\lambda}{4}\)
where, length of the pipe, = 25.5 cm = 0.255 m
λ = 4l₁ =4 x 0.255 = 1.02 m
The speed of sound is given by the relation:
υ = vλ = 340 x 1.02 = 346.8 m/s

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?
Solution:
Length of the steel rod, l = 100 cm = lm
v Fundamental frequency of vibration, v = 2.53 kHz = 2.53 x 10³ Hz When the rod is plucked at its middle, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is \(\frac{\lambda}{2}\)
l = \( \frac{\lambda}{2}\)
λ=2l=2 x l=2m
The speed of sound in steel is given by the relation:
v =vλ = 2.53 x 10³ x 2
= 5.06 x 10 ³ m/s = 5.06 km/s

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 ms^-1).
Solution:
First (Fundamental); No
Length of the pipe, l = 20 cm = 0.2 m
Source frequency = n^th normal mode of frequency, v_n = 430 Hz Speed of sound, ν = 340 m/s
In a closed pipe, n^th the rth normal mode of frequency is given by the relation
v_n = (2n -1) \(\frac{v}{4 l}\) ; n is an integer = 0,1,2,3 ………………
430 = (2n -1) \(\frac{340}{4 \times 0.2} \)
2n-1 = \(\frac{430 \times 4 \times 0.2}{340}\) = 1.01
2n =1.01+1
2n = 2.01
n ≈ 1

Hence, the first mode of vibration frequency is resonantly excited by the given source.
In a pipe open at both ends, the n^th mode of vibration frequency is given by the relation:
v_n = \(\frac{n v}{2 l}\)
n = \(\frac{2 l v_{n}}{v}=\frac{2 \times 0.2 \times 430}{340}\) = 0.5
Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Solution:
Frequency of string A, f_A = 324 Hz
Frequency of string B = f_B
Beat’s frequency, n = 6 Hz
Beat’s frequency is given as
n = |f_A ±f_B|
6 =324 ±f_B
f_B =330 Hz or 318 Hz

The frequency decreases with a decrease in the tension in a string. This is because the frequency is directly proportional to the square root of the tension. It is given as
v ∝ \(\sqrt{T}\)
Hence, the beat frequency cannot be 330 Hz.
∴ f_B= 318 Hz

Question 19.
Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa,
(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) The shape of a pulse gets distorted during propagation in a dispersive medium.
Solution:
(a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum.
Therefore, a displacement node is nothing but a pressure antinode and vice versa.

(b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

(d) Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse wave is such that it produces shearing *’ stress in a medium. The propagation of such a wave is possible only in solids, and not in gases. ‘
Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids.

(e) A pulse is actually a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air
(i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 ms^-1,
(b) recedes from the platform with a speed of 10 ms^-1?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms ^-1.
Solution:
(i)
(a) Frequency of the whistle, ν = 400 Hz
Speed of the train, υ_T = 10 m/s
Speed of sound, υ = 340 m/s
The apparent frequency (v’) of the whisde as the train approaches the platform is given by the relation
υ’ = \(=\left(\frac{v}{v-v_{T}}\right) \mathrm{v}=\left(\frac{340}{340-10}\right) \times 400\) = 412.12 Hz
(b) The apparent frequency (v”) of the whistle as the train recedes from the platform is given by the relation
v” = \(\left(\frac{v}{v+v_{T}}\right) \mathrm{v}=\left(\frac{340}{340+10}\right) \times 400\) = 388.57 Hz

(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same, i.e.,340 m/s.

Question 21.
A train standing in a station yard blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 ms^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms^-1? The speed of sound in still air can be taken as 340 ms^-1.
Solution:
For the stationary observer:
Frequency of the sound produced by the whistle, v = 400 Hz
Speed of sound = 340 m/s
Velocity of the wind, ν = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard .by the observer will be the same as that produced by the source, i. e., 400 Hz.
The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, υ_e = 340 +10 = 350 m/s

The wavelength (λ) of the sound heard by the observer is given by the relation:
λ = \(\frac{v_{e}}{v}=\frac{350}{400}\) = 0.857 m

For the running observer:
Velocity of the observer, υ₀ = 10 m/s
The observer is moving toward the source. As a result of the
motions of the source and the observer, there is a change in (v’).
This is given by the relation:
υ’ = \(\left(\frac{v+v_{o}}{v}\right) v=\left(\frac{340+10}{340}\right) \times 400\) = 411.76 Hz
Since the air is still, the effective speed of sound = 340 + 0 = The source is at rest. Hence, the wavelength of the sound will i. e., λ remains 0.875 m
Hence, the given two situations are not exactly identical.

Additional Exercises

Question 22.
A travelling harmonic wave on a string is described by
y(x,t) = 7.5 sin (0.0050x + 12t+\(\frac{\pi}{4}\))
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
Solution:
(a) The given harmonic wave is
y(x,t) = 7.5sin(0.0050x + 12t+\(\frac{\pi}{4}\))
For x = 1 cm and t = 1 s,
y(1, 1) = 7.5sin(0.0050x + 12t+\(\frac{\pi}{4}\))
= 7.5 sin (12.0050+\(\frac{\pi}{4}\)) = 7.5sinθ
where, 0 = 12.0050 + \(\frac{\pi}{4}\) = 12.0050 + \( \) = 12.79 rad 4
= \(\frac{180}{3.14} \times 12.79\) = 732.810
∴ y(1,1) = 7.5 sin (732.810) = 7.5sin (90 × 8 +12.81°) = 7.5 sin 12.81°
= 7.5 × 0.2217
= 1.6229 ≈ 1.663 cm
The velocity of the oscillation at a given point and time is given as

At x = 1 cm and t = 1 s
v = y(1, 1) = 90 cos(12.005 +\(\frac{\pi}{4}\))
= 90 cos (732.81 ° ) = 90 cos (90 x 8 +12.81 ° ) = 90cos(12.81°) = 90 x 0.975 = 87.75 cm/s

Now, the equation of a propagating wave is given by
y(x,t) = asin(kx +ωt +Φ)
where, k = \(\frac{2 \pi}{\lambda} \)
∴ λ = \(\frac{2 \pi}{k}\)
And ω = 2πv
∴ v = \(\frac{\omega}{2 \pi}\)
Speed, υ = vλ = \(\frac{\omega}{k}\)
where, ω = 12 rad/s
k = 0.0050 cm-1
∴ v = \(\frac{12}{0.0050}\) = 2400 cm/s
Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:
k = \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{0.0050}\) = 1256 cm = 12.56 cm
Therefore, all the points at distances nλ{n = ±1,±2… and so on), i.e., ±12.56 m, + 25.12m, … and so on for x =1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s and 11s.

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium,
(a) Does the pulse have a definite
(i) frequency,
(ii) wavelength,
(iii) speed of propagation?
(b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note
produced by the whistle equal to \(\frac{1}{20}\) or 0.05 Hz?
Solution:
(a) (i) No;
(ii) No;
(iii) Yes;
Explanation:
The narrow sound pulse does not have a fixed wavelength or frequency. However, the speed of the sound pulse remains the same, which is equal to the speed of sound in that medium.
(b) No;
The short pip produced after every 20 s does not mean that the frequency of the whistle is \(\frac{1}{20}\) or 0.05 Hz. It means that 0.05 Hz is the frequency of the repetition of the pip of the whisde.

Question 24.
One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude.

At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as a function of x and t that describes the wave on the string.
Solution:
The equation of a Gravelling wave propagating along the positive y-direction is given by the displacement equation
y(x, t) = a sin (cot – kx) ………………………. (i)
Linear mass density, μ = 8.0 x 10 -3 kg m-1
Frequency of the tuning fork, v = 256 Hz
The amplitude of the wave, a = 5.0 cm = 0.05 m ……………………………. (ii)
Mass of the pan, m = 90 kg
Tension in the string, T = mg = 90 x 9.8 = 882 N

The velocity of the transverse wave υ, is given by the relation:
υ = \(=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{882}{8.0 \times 10^{-3}}}\) = 332 m/s
Angular Frequency, ω = 2πv
= 2 x 3.14 x 256
= 1607.68 = 16 x 103 rad/s ………………………….. (iii)
Wavelength λ = \(\frac{v}{v}=\frac{332}{256}\)m
∴ propagation constant, k = \(\frac{2 \pi}{\lambda}=\frac{2 \times 3.14}{\frac{332}{256}}\) = 4.84 m-1 ……….. (iv)
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:
y(x,t) = 0.05sin(1.6 x 103t -4.84 x)
where x and y are in and t in s.

Question 25.
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 ms-1.
Solution:
Operating frequency of the SONAR system, v = 40 kHz
Speed of the enemy submarine, ve = 360 km/h = 100 m/s
Speed of sound in water, v = 1450 m/s
The source is at rest and the observer (enemy submarine) is moving toward it.

Hence, the apparent frequency (v’) received and reflected by the submarine is given by the relation:
The frequency (v”) received by the enemy submarine is given by the relation
v’ = \( =\left(\frac{v+v_{e}}{v}\right) v=\left(\frac{1450+100}{1450}\right) \times 40\) = 42.76 kHz
The frequency (V”) received by the enemy submarine is given by the relation
v” = \(\left(\frac{v}{v-v_{s}}\right) v^{\prime}\)
Where vs = 100 m/s
∴ v” = \(\left(\frac{1450}{1450-100}\right) \times 42.76 \) = 45.93 kHz

Question 28.
Earthquakes generate sound waves inside the Earth. Unlike a gas, the Earth can experience both transverse (S) and
longitudinal (P) sound waves. Typically the speed of S wave is about 4.0kms-1 , and that of P wave is 8.0 kms1. A
seismograph records P and S waves from an Earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the Earthquake occur?
Solution:
Let νs and vp, be the velocities of S and P waves respectively.
Let L be the distance between the epicentre and the seismograph.
We have
L = νstsub>s ……………………….. (i)
L = νptsub>p …………………………(ii)

where ts and tp are the respective times taken by the S and P waves to
reach the seismograph from the epicentre
It is given that
νp =8km/s
νs =4km/s

From equations (i) and (ii), we have
υsts = υptp
4ts = 8tp
ts = 2tp …………………………..(iii)
It is also given that
ts – tp =4 min=240s
2tp-tp= 240
tp = 240
and = 2×240 =840 s
From equation (ii), we get

L =8×240=1920 km
Hence, the Earthquake occurs at a distance of 1920 km from the seismograph.

Question 27.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound hi air. What frequency does the bat hear reflected off the wall?
Solution:
Ultrasonic beep frequency emitted by the bat, ν = 40 kHz
The velocity of the bat, νb = 0.03 ν
where, ν = velocity of sound in air
The appartment frequency of the sound strìking the wall is given as
v’ = \(\left(\frac{v}{v-v_{b}}\right) v=\left(\frac{v}{v-0.03 v}\right) \times 40 \) = \(\frac{40}{0.97}\) kHz
This frequency is reflected by the stationary wall (νs = 0) toward the bat.
The frequency (ν”) of the received sound is given by the relation:
ν” = \(\left(\frac{v+v_{b}}{v}\right) \mathrm{v}^{\prime}=\left(\frac{v+0.03 v}{v}\right) \times \frac{40}{0.97}=\frac{1.03 \times 40}{0.97}\) = 42.47 kHz

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 14 Oscillations Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 14 Oscillations

PSEB 11th Class Physics Guide Oscillations Textbook Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Solution:
(b) and (c)
Explanations :
(a) The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.

(c) When a hydrogen molecule rotates about its center of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.

(d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a 17-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Solution:
(b) and (c) are SHMs; (a) and (d) are periodic, but not SHMs
Explanations :
(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.

(b) An oscillating mercury column in a [/-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

Question 3.
Figure depicts four x-t plots for linear motion of a particle. Which of the plots represents periodic motion? What is the period of motion (in case of periodic motion)?
(a)

(b)

(c)

(d)

Solution:
(b) and (d) are periodic
Explanation :
(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.
(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.
(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time. In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.

Question 4.
Which of the following functions of time represent (a) simple ‘ harmonic, (b) periodic but not simple harmonic, and (c) non¬periodic motion? Give period for each case of periodic motion (a is any positive constant):
(a) sin ωt – cos ωt
(b) sin ³ωt
(c) 3cos(\(\pi / 4 \) -2ωt)
(d) cos ωt +cos 3 ωt+cos 5ωt
(e) exp(-ω²t²)
(f) 1+ ωt+ω^2t
Solution:
(a) SHM
The given function is:

This function represents SHM as it can be written in the form: a sin (ωt +Φ)
Its period is: \(\frac{2 \pi}{\omega}\)

(b) Periodic, but not SHM The given function is:
sin³ ωt = \(\frac{1}{4}\) [3sinωt -sin3ωt] (∵ sin3θ = 3sinθ – 4sin³ θ)
The terms sin cot and sin 3 ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

Period of\(\frac{3}{4}\)sin ωt = \(\frac{2 \pi}{\omega}\) = T
Period of\(\frac{1}{4}\)sin3ωt = \(\frac{2 \pi}{3 \omega}\) = T’ = \(\frac{T}{3}\)
Thus, period of the combination
= Minimum time after which the combined function repeats
= LCM of T and \(\frac{T}{3}\) = T
Its period is 2 \(\pi / \omega\)

(c) SHM
The given function is:
3 cos \(\left[\frac{\pi}{4}-2 \omega t\right]\) = 3 cos \(\left[2 \omega t-\frac{\pi}{4}\right]\)
This function represents simple harmonic motion because it can be written in the form:
acos(ωt +Φ)

Its period is :
\(\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}\)

(d) Periodic, but not SHM
The given function is cosωt +cos3ωt +cos5ωt. Each individual cosme function represents SHM. However, the superposition of threc simple harmonic motions is periodic, but not simple harmonic.

cosωt represents SHM with period = \(\frac{2 \pi}{\omega}\) T (say)
cos 3ωt represents SHM with period = \(\frac{2 \pi}{3 \omega}=\frac{T}{3}\)
cos 5ωt represents SHM with period = \(\frac{2 \pi}{5 \omega}=\frac{T}{5}\)
The minimum time after which the combined function repeats its value is T. Hence, the given function represents periodic function but not SHM, with period T.

(e) Non-periodic motion: .
The given function exp(- ω²t²) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) Non-periodic motion
The given function is 1+ ωt + ω²t²
Here no repetition of values. Hence, it represents non-periodic motion.

Question 5.
A particle Is in linear simple harmonic motion between two points, A and B, 10 cm apart Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when It Is
(a) at the end A,
(b) at the end B,
(c) at the midpoint of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Solution:
The given situation is shown in the following figure. Points A and B are the two endpoints, with AB =10cm. 0 is the midpoint of the path.

A particle is in linear simple harmonic motion between the endpoints
(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.
Its acceleration is positive as it is directed along with AO.
Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.
Its acceleration is negative as it is directed along B.
Force is also negative in this case as the particle is directed leftward.

(c)
The particle is extending a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the partide Is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)
The particle is moving towards point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to R. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

(e)

The particle is moving towards point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the value for velocity, acceleration, and force are all positive.

(f)

This case is similar to the one given in (d).

Question 6.
Whi ch of the following relationships between the acceleration a and the displacement x of a particle involves simple harmonic motion?
(a) a=0.7x
(b) a=-200x²
(c) a= – 10 x (d) a=100x³
Solution:
A motion represents simple harmonic motion if it is governed by the force law:
F=-kx
ma’= -kx
∴ a = – \(\frac{k}{m}\) x

where F is the force
m is the mass (a constant for a body)
x is the displacement
a is the acceleration.
k is a constant
Among the given equations, only equation a = -10 x is written in the
above form with \( \frac{k}{m}\) =10. Hence, this relation represents SHM.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function,
x(t) = A cos (ωt+Φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is πs^-1. If instead of the cosine function, we choose the sine function to describe the SHM:x = B sin(ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Solution:
Initially, at t = 0:
Displacement, x = 1 cm Initial velocity, ν = ω cm/s.
Angular frequency, ω = π rad s^-1
It is given that:
x(t) = Acos (ωt+Φ)
1 = Acos(ω x 0 +Φ) = AcosΦ
AcosΦ =1 ……………………………….. (i)

Velocity, ν = \(\frac{d x}{d t}\)
ω = -Aω sin(ωt +Φ)
1 = -Asin(ω x 0 +Φ) = -AsinΦ
Asin Φ = -1 ………………………… (ii)
Squaring and adding equations (i) and (ii), we get
A²(sin²Φ +cos²Φ) = 1+1
A² = 2
∴ A = \(\sqrt{2}\) cm

Dividing equation (ii) by equation (i), we get
tanΦ = -1

SHM is given as
x = Bsin(ωt + α)
Putting the given values in this equation, we get 1 =B sin (ωt + α)
B sin α =1 …………………………… (iii)

Velocity, ν = \(\frac{d x}{d t}\)
ω =(ωB cos (ωt + a)
1 =B cos (ω x 0+α ) = B cos α …………………………………… (iv)
Squaring and adding equations (iii) and (iv), we get
B² [sin²α +cos² α] =1+1
B² =2
B = \(\sqrt{2}\) cm
Dividing equation (iii) by equation (iv), we get

Question 8.
A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Solution:
Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m Time period, T =0.6s
Maximum force exerted on the spring, F = Mg where,
g = acceleration due to gravity = 9.8 m/s²
F = 50 × 9.8 = 490 N
∴ Spring constant , K = \(\frac{F}{l}=\frac{490}{0.2}\) = 2450Nm^-1

Mass m, is suspended from the balance,
Time period, T = \(2 \pi \sqrt{\frac{m}{k}}\)
∴ m = \(\left(\frac{T}{2 \pi}\right)^{2} \times k=\left(\frac{0.6}{2 \times 3.14}\right)^{2} \times 2450 \) = 22.36 kg
∴ Weight of the body = mg = 22.36 x 9.8 = 219.167N
Hence, the weight of the body is about 219 N.

Question 9.
Aspringhavingwith aspiring constant 1200Nm^-1 is mounted on a horizontal table as shown in figure. A mss of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

Determine (i) the frequency of oscillations,
(ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Solution:
Spring constant, k = 1200 Nm^-1
mass,m = 3 Kg
Displacement,A = 2.0 cm = 0.02 m
(i) Frequency of oscillation y, is gyen by the relation
V = \(\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
where, T is the time period
∴ v = \(\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}\) = 3.18 s^-1
Hence, the frequency of oscillations is 3.18 s^-1.

(ii) Maximum acceleration a is given by the relation:
a = ω²A
where,
ω = Angular frequency = \(\sqrt{\frac{k}{m}}\)
A = Maximum displacement
∴ a = \(\frac{k}{m} A=\frac{1200 \times 0.02}{3}\) = 8 ms^-2
Hence, the maximum acceleration of the mass is 8.0 ms²

(iii) Maximum speed, ν_max = Aω
= \(A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}\) = 0.4 m/s
Hence, the maximum speed of the mass is 0.4 m/s.

Question 10.
In question 9, let us take the position of mass when the spring is unstretched as x =0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating massif at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude br the initial phase?
Solution:
(a) The functions have the same frequency and amplitude, but different initial phases.
Distance travelled by the mass sideways, A = 2.0 cm
Force constant of the spring, k =1200 N m^-1
Mass, m =3kg
Angular frequency of oscillation,
ω = \(\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}} \) = \(\sqrt{400}\) = 20 rad s^-1
When the mass is at the mean position, initial phase is 0.
Displacement,
x = A sinωt = 2 sin20 t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is \(\frac{\pi}{2}\)
Displacement, x = Asin \(\left(\omega t+\frac{\pi}{2}\right)\)
=2sin\(\left(20 t+\frac{\pi}{2}\right)\)
= 2 cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is \(\frac{3 \pi}{2}\)
Displacement, x = A sin \(\left(\omega t+\frac{3 \pi}{2}\right)\)
= 2sin \(\left(20 t+\frac{3 \pi}{2}\right)\) = -2cos 20t
The functions have the same frequencyl \(\left(\frac{20}{2 \pi} \mathrm{Hz}\right)\) land amplitude (2cm),
but different initial phases \(\left(0, \frac{\pi}{2}, \frac{3 \pi}{2}\right)\)

Question 11.
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i. e., clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
solution:
Time period,T =2s
Amplitude, A = 3cm
At time, t = O, the radius vector OP makes an angle \(\frac{\pi}{2}\) with the positive x -axis, i.e., phase angle Φ = + \(\frac{\pi}{2}\)
Therefore, the equation of simple harmonic motion for the x —projection of OP, at time t, is given by the displacement equation

(b) Time period, T = 4s
Amplitude, a =2 m
At time t = 0, OP makes an angle ir with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = +π
Therefore, the equation of simple harmonic motion for the x -projection of OP, at time t, is given as

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) x= – 2sin(3t+\(\pi / \mathbf{3}\) )
(b) x=cos (\(\pi / 6\) – t)
(c) x=3 sin (2πt + \(\pi / 4 \) )
(d) x=2 cos πt
Solution:
(a) x = -2 sin \(\left(3 t+\frac{\pi}{3}\right)=+2 \cos \left(3 t+\frac{\pi}{3}+\frac{\pi}{2}\right)=2 \cos \left(3 t+\frac{5 \pi}{6}\right) \)

If this equation is compared with the standard SHM equation,
x =A cos \(\left(\frac{2 \pi}{T} t+\phi\right)\) then we get
Amplitude, A = 2cm
Phase angle, Φ = \(\frac{5 \pi}{6}\) =150°
Angular velocity, ω = \(\frac{2 \pi}{T}\) =3 rad/sec
The motion of the particle can be pokted as shown in the following figure.

(b) x=cos \(\left(\frac{\pi}{6}-t\right)\) =cos \(\left(t-\frac{\pi}{6}\right)\)
If this equation is compared with the standard SHM equation,
x = A cos \(\left(\frac{2 \pi}{T} t+\phi\right)\) then we get
Amplitude, A = 1 cm
Phase angle, Φ = \(-\frac{\pi}{6} \) = – 30°
Angular velocity, ω = \(\frac{2 \pi}{T}\) =1 rad/s
The motion of the particle can be plotted as shown in the following figure.

(c) x =3sin \(\left(2 \pi t+\frac{\pi}{4}\right)\)

If this equation is compared with the standard SHM equation
x = Acos \(\left(\frac{2 \pi}{T} t+\phi\right)\) then we get
Amplitude, A = 3cm
Phase angle, Φ = \(-\frac{\pi}{4}\)
Angular velocity, ω = \(\frac{2 \pi}{T}\) = 2π rad/s
The motion of the particle can be plotted as shown in the following figure.

(d) x=2cosπt
If this equation is compared with the standard SHM equation,
x = A cos \(\left(\frac{2 \pi}{T} t+\phi\right) \) then we get
Amplitude, A = 2cm
Phase angle, Φ = 0
Angular velocity, ω = π rad/s
The motion of the particle can be plotted as shown in the following figure.

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in figure (b) is stretched by the same force F.

(a) What is the minimum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Solution:
(a) For the one block system:
When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as:
F=kl
where k is the spring constant
Hence, the maximum extension produced in the spring, l = \(\frac{F}{k}\)
For the two blocks system:
The displacement (x) produced in this case is:
x = \(\frac{l}{2}\)
Net force, F = +2kx =2k \(\frac{l}{2}\)
∴ l = \(\frac{F}{k}\)

(b) For the one blocks system:
For mass (m) of the block, force is written as
F = ma = m \(\frac{d^{2} x}{d t^{2}}\)
where, x is the displacement of the block in time t
∴ m \(\frac{d^{2} x}{d t^{2}}\) = -kx
It is negative because the direction of elastic force is opposite to the direction of displacement.

where, ω is angular frequency of the oscillation
∴ Time period of the oscillation,
T= \(\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{k}{m}}}=2 \pi \sqrt{\frac{m}{k}}\)

For the two blocks system:
F=m \(\frac{d^{2} x}{d t^{2}}\)
m \(\frac{d^{2} x}{d t^{2}}\) =-2kx

It is negative because the direction of elastic force is opposite to the direction of displacement.
\(\frac{d^{2} x}{d t^{2}}\) = \(-\left[\frac{2 k}{m}\right] x \) = – ω²x
where, Angular frequency, ω = \(\sqrt{\frac{2 k}{m}}\)
∴ Time period T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{2 k}} \)

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 in. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Solution:
Angular frequency of the piston, ω = 200 rad/min.
Stroke =1.0 m
Amplitude, A = \(\frac{1.0}{2}\) = 0.5m
The maximum speed (ν_max) of the piston is given by the relation
ν_max =Aω = 200 x 0.5=100 m/min

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 ms^-2. What is the time period of a simple pendulum on the surface of moon If Its time period on the surface of earth is 3.5 s? (gon the surface of earth is 9.8 ms^-2)
Solution:
Acceleration due to gravity on the surface of moon, g’ = 1.7m s^-2
Acceleration due to gravity on the surface of earth, g = 9.8 ms^-2
Time period of a simple pendulum on earth, T = 3.5 s
T= \(2 \pi \sqrt{\frac{l}{g}}\)

where l is the length of the pendulum
∴ l = \(\frac{T^{2}}{(2 \pi)^{2}} \times g=\frac{(3.5)^{2}}{4 \times(3.14)^{2}} \times 9.8 \mathrm{~m} \)

The length of the pendulum remains constant.
On Moon’s surface, time period,
T’ = \(2 \pi \sqrt{\frac{l}{g^{\prime}}}=2 \pi \sqrt{\frac{(3.5)^{2}}{\frac{4 \times(3.14)^{2}}{1.7}} \times 9.8} \) = 8.4 s
Hence, the time period of the simple pendulum on the surface of Moon is 8.4 s.

Question 16.
Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:
T = \(2 \pi \sqrt{\frac{m}{k}}\) A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation a more involved analysis shows that T is greater than \(2 \pi \sqrt{\frac{l}{g}}\) Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Solution :
(a) The time period of a simple pendulum, T = \(2 \pi \sqrt{\frac{m}{k}}\)
For a simple pendulum, k is expressed in terms of mass m, as
k ∝ m
\(\frac{m}{k}\) = Constant
Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as
F = -mg sinθ
where, F = Restoring force; m = Mass of the bob; g = Acceleration due to
gravity; θ = Angle of displacement
For small θ, sinθ ≈ θ
For large 0,sin0 is greater than 0.
This decreases the effective value of g.
Hence, the time period increases as
T = \(2 \pi \sqrt{\frac{l}{g}}\)
where, l is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.

(d) When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small f oscillations in a radial direction about its equilibrium position, what will be its time period?
Solution:
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.
Acceleration due to gravity = g
Centripetal acceleration = \(\frac{v^{2}}{R}\)
where, v is the uniform speed of the car R is the radius of the track
Effective acceleration (a_eff) is given as
a_eff = \(\sqrt{g^{2}+\left(\frac{v^{2}}{R}\right)^{2}}\)

Time period, T = \( 2 \pi \sqrt{\frac{l}{a_{e f f}}}\)
where, l is the length of the pendulum
∴ Time period, T = \(2 \pi \sqrt{\frac{l}{g^{2}+\frac{v^{4}}{R^{2}}}} \)

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ_l. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
T = \(2 \pi \sqrt{\frac{\boldsymbol{h} \rho}{\rho_{\boldsymbol{l}} \boldsymbol{g}}}\)
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Solution:
Base area of the cork = A
Height of the cork = h
Density of the liquid = ρ_l
Density of the cork = ρ

In equilibrium:
Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
Up-thrust = Restoring force, F = Weight of the extra water displaced
F = -(Volume x Density x g)
Volume = Area x Distance through which the cork is depressed Volume = Ax
∴ F = -Ax ρ_lg ………………………… (i)
According to the force law,
F = kx
k = \(\frac{F}{x}\)

where k is a constant
k = \(\frac{F}{x}\) = -Aρ_lg ………………………………. (ii)
The time period of the oscillations of the cork,
T = \(2 \pi \sqrt{\frac{m}{k}} \) …………………………………… (iii)

where,
m = Mass of the cork
= Volume of the cork x Density
= Base area of the cork x Height of the cork x Density of the cork = Ahρ
Hence, the expression for the time period becomes
T = \(2 \pi \sqrt{\frac{A h \rho}{A \rho_{l} g}}\) = 2\(\pi \sqrt{\frac{h \rho}{\rho_{l} g}} \)

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Solution:
Area of cross-section of the U-tube = A
Density of the mercury column = ρ
Acceleration due to gravity = g
Restoring force, F = Weight of the mercury column of a certain height
F = -(Volume x Density x g)
F = -(A x 2h x ρ x g) = -2Aρgh = -k x Displacement in one of the arms (h)

where, 2h is the height of the mercury column in the two arms
k is a constant, given by k = \(-\frac{F}{h}\) = 2Aρg
Time period = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{2 A \rho g}}\)
where, m is the mass of the mercury column
Let l be the length of the total mercury in the U-tube.
Mass of mercury, m = Volume of mercury x Density of mercury = Alρ
∴ T = \(2 \pi \sqrt{\frac{A l \rho}{2 A \rho g}}=2 \pi \sqrt{\frac{l}{2 g}} \)

Hence the mercury column executes simple harmonic motion with time period \(2 \pi \sqrt{\frac{l}{2 g}} \)

Additional Exercises

Question 20.
An air chamber of volume V has a neck area of cross-section a into which a ball of mass m just fits and can move up and down without any friction (see figure). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Solution:
Volume of the air chamber = V
Area of cross-section of the neck = a
Mass of the ball = m
The pressure inside the chamber is equal to the atmospheric pressure. Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.
Decrease in the volume of the air chamber, ΔV = ax
Volumetric strain =
⇒ \(\frac{\Delta V}{V}=\frac{a x}{V}\)

Bulk Modulus of air, B = \(\frac{\text { Stress }}{\text { Strain }}=\frac{-p}{\frac{a x}{V}}\)
In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.
p = \(\frac{-B a x}{V}\)
The restoring force acting on the ball,
F = p × a = \(\frac{-B a x}{V} \cdot a=\frac{-B a^{2} x}{V}\) ……………………………. (i)
In simple harmonic motion, the equation for restoring force is
F = -kx …………………………………….. (ii)
where, k is the spring constant Comparing equations (i) and (ii), we get
k = \(\frac{B a^{2}}{V}\)
Time period, T = \(2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{V m}{B a^{2}}}\)

Question 21.
You are riding in an automobile of mass 3000kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (6) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Solution:
Mass of the automobile, m = 3000 kg
Displacement in the suspension system, x = 15cm = 0.15 m
There are 4 springs in parallel to the support of the mass of the automobile.
The equation for the restoring force for the system:
F = -4 kx = mg

where, k is the spring constant of the suspension system
Time period, T = \(2 \pi \sqrt{\frac{m}{4 k}}\)
and, k = \(\frac{m g}{4 x}=\frac{3000 \times 10}{4 \times 0.15}\) = 50000 = 5 x 10 ⁴N/m
Spring constant, k = 5 x 10⁴ N/m

Each wheel supports a mass, M = \(\frac{3000}{4}\) = 750 kg
For damping factor b, the equation for displacement is written as:
x = x₀e^-bt/2M

The amplitude of oscillation decreases by 50%

where, Time petiod,t =\(2 \pi \sqrt{\frac{m}{4 k}}=2 \pi \sqrt{\frac{3000}{4 \times 5 \times 10^{4}}}\) =0.7691s
∴ b =\(\frac{2 \times 750 \times 0.693}{0.7691}\) =1351.58kg/s
Therefore, the damping constant of the spring is 1351.58 kg/s.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Solution:
The equation of displacement of a particle executing SHM at an instant t is given as
x = Asinωt
where,
A = Amplitude of oscillation
ω = Angular frequency = \(\sqrt{\frac{k}{M}}\)
The velocity of the particle is
ν = \(\frac{d x}{d t}\) = Aωcosωt

The kinetic energy of the particle is
E_k = \(\frac{1}{2} M v^{2}=\frac{1}{2} M A^{2} \omega^{2} \cos ^{2} \omega t \)
The potential energy of the particle is
E_p = \( \frac{1}{2} k x^{2}=\frac{1}{2} M \omega^{2} A^{2} \sin ^{2} \omega t\)

For time period T, the average kinetic energy over a single cycle is given as


And, average potential energy over one cycle is given as

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -αθ, where J is the restoring couple and 0 the angle of twist).
Solution:
Mass of the circular disc, m = 10 kg
Radius of the disc, r = 15cm = 0.15 m
The torsional oscillations of the disc has a time period, T = 1.5 s The moment of inertia of the disc is
I = \(\frac{1}{2}\) mr² = \(\frac{1}{2} \times(10) \times(0.15)^{2}\) = 0.1125kg-m²

Time period, T = \(2 \pi \sqrt{\frac{I}{\alpha}}\)
where, α is the torsional constant.
An21 4 x(3.14)2x 0.1125 , M
α = \(\frac{4 \pi^{2} I}{T^{2}}=\frac{4 \times(3.14)^{2} \times 0.1125}{(1.5)^{2}} \) = 1.972 N-m/rad
Hence, the torsional spring constant of the wire is 1.972 N-m rad^-1.

Question 24.
A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.
Solution:
Amplitude, A = 5 cm = 0.05m
Time period, T = 0.2 s
(a) For displacement, x = 5 cm = 0.05m
Acceleration is given by
a = -ω²x = \(-\left(\frac{2 \pi}{T}\right)^{2} x=-\left(\frac{2 \pi}{0.2}\right)^{2} \times 0.05\)
Velocity is given by
ν = ω \(\sqrt{A^{2}-x^{2}}=\frac{2 \pi}{T} \sqrt{(0.05)^{2}-(0.05)^{2}}\) = 0
When the displacement of the body is 5 cm, its acceleration is -5π² m/s² and velocity is 0.

(b) For displacement, x =3 cm = 0.03 m
Acceleration is given by
a = – ω²x = – \(\left(\frac{2 \pi}{T}\right)^{2}\)x = \(\left(\frac{2 \pi}{0.2}\right)^{2}\) 0.03 = -3π² m/s²
Velocity is given by

When the displacement of the body is 3 cm, its acceleration is -3π m/s² and velocity is 0.4π m/s.

(c) For displacement, x = 0
Acceleration is given by
a = – ω²x = 0
Velocity is given by

When the displacement of the body is 0, its acceleration is 0, and velocity is0.5π m/s.

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the center with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x_0, and v₀. [Hint: Start with the equation x = a cos(ωt + θ) and note that the initial velocity is negative.]
Solution:
The displacement equation for an oscillating mass is given by
x = Acos(ωt + θ) …………………………… (i)
where A is the amplitude
x is the displacement
θ is the phase constant

Squaring and adding equations (iii) and (iv), we get

Hence, the amplitude of the resulting oscillation is \(\sqrt{x_{0}^{2}+\left(\frac{v_{0}}{\omega}\right)^{2}}\)

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 1 Physical World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 1 Physical World

PSEB 11th Class Physics Guide Physical World Textbook Questions and Answers

Question 1.
Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said: “The most incomprehensible thing about the world is that it is comprehensible”?
Answer:
The physical world around us is full of different complex natural phenomena so the world is incomprehensible. But with the help of study and observations it has been found that all these phenomena are based on some basic physical laws and so it is comprehensible.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science of the validity of this incisive remark.
Answer:
The above statement is true. Validity of this incisive remark can be validated from the example of moment of inertia. It states that the moment of inertia of a body depends on its energy. But according to Einstein’s mass-energy relation (E = mc²), energy depends on the speed of the body.

Question 3.
“Politics is the art of the possible”. Similarly, “Science is the art of the soluble”. Explain this beautiful aphorism on the nature and practice of science.
Answer:
It is well known that to win over votes, politicians would make anything and everything possible even when they are least sure of the same. And in science the various natural phenomena can be explained in terms of some basic laws. So as ‘Politics is the art of possible’ similarly ‘Science is the art of the soluble’.

Question 4.
Though India now has a large base in science and technology, which is fast expanding, it is still a long way from realizing its potential of becoming a world leader in science. Name some important factors, which in your view have hindered the advancement of science in India.
Answer:
Some important factors in our view which have hindered the advancement of science in India are:

• Proper funds are not arranged for the development of research work and laboratories. The labs and scientific instruments are very old and outdated.
• Most of the people in India are uneducated and highly traditional. They don’t understand the importance of science.
• There is no proper employment opportunity for the science educated person in India.
• There are no proper facilities for science education in schools and colleges in India.

Question 5.
No physicist has ever “seen” an electron. Yet, all physicists believe in the existence of electrons. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has “seen” one. How will you refute his argument?
Answer:
No physicist has ever seen an electron but there are practical evidences which prove the presence of electron. Their size is so small, even powerful microscopes find it difficult to measure their sizes. But still its effects could be tested. On the other hand, there is no phenomena which can be explained on the basis of existence of ghosts.

Our senses of sight and hearing are very limited to observe the existence of both.
So, there is no comparison between the two given cases.

Question 6.
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?

(a) A tragic sea accident several centimes ago drowned a young Samurai. As a tribute to his bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.

(b) After the sea tragedy, fishermen in that area, in a gesture of honor to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. ConseQuestionuently, the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.

[Note: This interesting illustration taken from Carl Sagan’s ‘The Cosmos’ highlights the fact that often strange and inexplicable facts which on the first sight appear ‘supernatural’ actually turn out to have simple scientific explanations. Try to think out other examples of this kind].
Answer:
Explanation (b) is correct as it is a scientific explanation of the observed fact.

Question 7.
The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were these advances?
Answer:
More than two centuries ago, England and Western Europe invented steam engine, electricity, theory of gravitation and the explosives. Steam engines helped them in the field of heat and thermodynamics, theory of gravitation in field of motion and making guns and cannons. These progresses brought about industrial revolution in England and Western Europe.
Few of which are given below:

• Steam engine formed on the application of heat and thermodynamics.
• Discovery of electricity helped in designing dynamos and motors.
• Safety lamp which was used safety in mines.
• Invention of powerloom which used steampower was used for spinning and weaving.

Question 8.
It is often said that the world is witnessing now a second industrial revolution, which will transform the society as
radically as did the first. List some key contemporary areas of science and technology, which are responsible for this
revolution.
Answer:
Some of the key contemporary areas of science and technology which may trAnswer:form the society radically are:

• Development of super fast computers.
• Internet and tremendous advancement in information technology.
• Development in Biotechnology.
• Development of super-conducting materials at room temperature.
• Development of robots.

Question 9.
Write in about 1000* words a fiction piece based on your : speculation on the science and technology of the twenty second century.
Answer:
Let us imagine a spaceship moving towards a distant star, 500 light years away. Let it be propelled by current fed into the electric motor consisting of superconducting wires. In space, suppose there is a particular region which has such a high temperature that destroys the superconducting property of electric wires of the motor. At this stage, another spaceship
filled with matter and anti-matter comes to the rescue of the first ship and it (i. e., 1st ship) continues its onward journey.

Another way to put is: Now matter can be changed into energy and energy into matter. A man of 22nd century stands on a plateform of a specially designed machine which energises him and his body disappears in the form of energy. After split of a second, he appears at a place much far away from the previous one just intact.

Question 10.
Attempt to formulate your ‘moral’ views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous conseQuestionuences for the human society. How, if at all, will you resolve your dilemma?
Answer:
In our view a type of discovery which is of great academic interest but harmful for human society should not be made public because science is for the society, society is not for science.

Question 11.
Science, like any knowledge, can be put to good or had use, depending on the user.Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorized:
(a) Mass vaccination against small pox to curb and finally eradicate this disease from the population. (This has already been successfully done in India).
(b) Television for eradication of illiteracy and for mass communication of news and ideas.
(c) Prenatal sex determination
(d) Computers for increase in work efficiency
(e) Putting artificial satellites into orbits around the Earth
(f) Development of nuclear weapons
(g) Development of new and powerful techniQuestionues of chemical and biological warfare.
(h) Purification of water for drinking
(i) Plastic surgery
(j) Cloning
Answer:
(a) Mass vaccination is good as it is used to make the society free from the diseases like Small Pox.

(b) Television for eradication of illiteracy and for mass communication of news and ideas is good as it is a medium which is easily under the reach of common man and also they are very habitual to it.

(c) Prenatal sex determination is bad because people are misusing it. Some of the people after determination of sex of child, think to abort. They do it specially with girl child.

(d) Computers for increase in work efficiency is good as using the computer a man can do much more work with greater efficiency and accuracy as it could do without computers.

(e) Putting artificial satellites into orbits around the Earth is a good development as these satellites serve many purposes like Remote Sensing, Weather Forcasting etc. These informations have very high importance for us as we can plan the things in advance.

(f) Development of nuclear weapons is bad as they can be used in mass destruction.

(g) Development of new and powerful techniQuestionues of chemical and biological warfare are bad as they can also be used for mass destruction.

(h) Purification of water for drinking is good as we can save ourseleves from the diseases which we can have due to drinking the contaminated water.

(i) Plastic surgery is good as with the help of it a man or woman can remove the skin defects occuring due to accidents or some other reasons. It has some bad effects too but they are not very considerable.

(j) Cloning is good as far as animals are concerned with the help of it we can develop some special species which can be used to serve some specific purposes. But it is not good for human beings.

Question 12.
India has had a long and unbroken tradition of great scholarship – in mathematics, astronomy, linguistics, logic and ethics. Yet, in parallel with this, several superstitious and obscurantistic attitudes and practices flourished in our society and unfortunately continue even today – among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?
Answer:
Poverty and illiteracy are the two major factors which make people superstitious in India. So to remove the superstitious and obscurantist attitude we have to first overcome these factors. Everybody should be educated, so that one can have scientific attitude. Knowledge of science can be put to use to prove people’s superstitious wrong by showing them the scientific logic behind everything happening in our world.

Question 13.
Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice give them a secondary status and role. Demolish this view using scientific arguments, and by quoting examples of great women in science and other spheres; and persuade yourself and others that, given equal opportunity, women are on par with men.
Answer:
Some people in our society have the view that women do not have the innate nature, capacity and intelligence.
To demolish this view there are many examples of women who have proven their abilities in science and other fields.
Madam Curie, Mother Teresa, Indira Gandhi, Marget Thatcher, Rani Laxmi Bai, Florence Nightingale are some examples. So in this era women are definitely not behind man in any field.

Question 14.
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P. A. M. Dirac held this view. Criticize this statement.
‘ Look out for some eQuestionuations and results in this book which strike you as beautiful.
Solution:
An equation which agrees with experiment must also be simple and hence beautiful. We have some simple and beautiful equations in physics such as
E = mc² (Energy of light) .
E = hv (Energy of a photon)
KE = 1 / 2 mv ² (Kinetic energy of a moving particle)
PE = mgh (Potential energy of a body at rest)
W = F. d (Work done)
All have the same dimensions. One experiment shows dependency of energy on speed, the other shows dependency on frequency and displacement.
That’s the beauty of equations in physics coming from different experiments.

Question. 15.
Though the statement quoted above may be disputed, most physicists do have a feeling that the great laws of physics are at once simple and beautiful. Some of the notable physicists, besides Dirac, who have articulated this feeling are Einstein, Bohr, Heisenberg, Chandrasekhar and Feynman. You are urged to make special efforts to get access to the general books and writings by these and other great masters of physics (See the Bibliography at the end of this book). Their writings are truly inspiring!
Solution:
There is no doubt that great laws of physics are at once so simple and beautiful and are easy to grasp. For example, let us look at some of these:

• E = mc² is a famous Einstein’s mass energy equivalence relation which has a great impact not only on the various physical phenomena but also on the human lives.
• Plank’s Questionuantum condition i.e., E = hv is also a simple and beautiful eQuestionuation and it is a great law of physics.
• ∆x. ∆p ≥ \(\frac {1}{2}\).\(\frac{h}{2 \pi}\) or ∆ E. A t ≥ \(\frac {1}{2}\).\(\frac{h}{2 \pi}\) is Heisenberg’s.

Uncertainly Principle which is also very simple, beautiful and interesting. It is a direct conseQuestionuence of the dual nature of matter.

• λ = \(\frac{h}{m v}\) is also a famous eQuestionuation in physics known as de-Broglie euation. It is again simple and beautiful.

Question 16.
Textbooks on science may give you a wrong impression that studying science is dry and all too serious and that scientists are absent minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists like any other group of humans have their share of humorists, and many have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books listed in the Bibliography.
Answer:
It is not an exercise as such but is a statement of facts. We can add the names of other scientists who were humorists along with being physicists. They are C.V. Raman, Homi Jahangir Bhabha, Einstein and Bohr. India have several politiciAnswer: like M.M. Joshi, V.P. Singh etc. who are physicists. President A.P.J. Kalafn is also a great nuclear scientist.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 13 Kinetic Theory Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 13 Kinetic Theory

PSEB 11th Class Physics Guide Kinetic Theory Textbook Questions and Answers

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at SIP. Take the diameter of an oxygen molecule to be 3Å.
Solution:
Diameter of an oxygen molecule, d = 3Å
Radius, r = \(\frac{d}{2}=\frac{3}{2}\) =1.5 Å = 1.5 x 10^-8 cm
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm³
Molecular volume of oxygen gas, V = \(\frac{4}{3}\) πr³N

where, N is Avogadro’s number = 6.023 x 10²³ molecules/mole
∴ V = \(\frac{4}{3}\) x 3.14 x (1.5 x 10^-8)3 x 6.023 x 10²³ = 8.51cm³

Ratio of the molecular volume to the actual volume of oxygen = \(\frac{8.51}{22400}\) = 3.8 x 10^-4 ≈ 4 x 10^-4

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Solution:
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as PV = nRT
where, R is the universal gas constant = 8.314 J mol^-1 K^-1
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm = 1.013 x 10⁵ Nm^-2
∴ V = \(\frac{n R T}{P}\)

Hence, the molar volume of a gas at STP is 22.4 litres.

Question 3.
Given figure shows plot of PV IT versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T₁ > T₂ or T₁ <T₂?
(c) What is the value of PVIT where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 x 10³ kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low-pressure high-temperature region of the plot)? (Molecular mass of H₂=2.02u,of O₂ =32.0 u, R= 8.31Jmol^-1K^-1)
Solution:
(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i. e., the ratio \(\frac{\bar{P} V}{T}\) is equal. μR(μ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas.

(b) The dotted plotmn the given graph represents an ideal gas. The curve of the gas at temperatureT₁ is closer to the dotted plot than the curve of the gas at temperature T₂. A real gàs approaches the behaviour of an ideal gas when its temperature increases. Therefore, T₁ > T₂ is true for the given plot.

(c) The value of the ratio PV/T, where the two curves meet, is μ.R. This is because the ideal gas equation is given as:
PV=μRT
\(\frac{P V}{T}\) = μR
where P is the pressure
T is the temperature
V is the volume
μ is the number of moles
R is the unìversal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen =1 x 10^-3 kg = 1 g
R =8.314J mole^-1K^-1
∴ \(\frac{P V}{T}=\frac{1}{32} \times 8.314\) =0.26JK^-1
Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is 0.26JK^-1.

(d) If we obtain similar plots for 1.00 x 10^-3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have \(\frac{P V}{T}\) = 0.26JK^-1
R = 8.314 J mole^-1 K^-1
Molecular mass (M) of H₂ =2.02 u PV
\(\frac{P V}{T}\) = μR at constant temperature
where, μ = \(\frac{m}{M}\) , m = Mass of H₂
∴ m = \(\frac{P V}{T} \times \frac{M}{R}=\frac{0.26 \times 2.02}{8.314}\)
= 6.3 x 10^-2 g = 6.3 x 10^-5 kg
Hence, 6.3 x 10^-5 kg of H₂ will yield the same value of PV/T.

Question 4.
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol-1 K -1, molecular mass of O 2 =32 u).
Solution:
Absolute pressure, p₁ = (15 + 1) atm
[∵ Absolute pressure = Gauge pressure +1 atm] = 16 x 1.013 x 10⁵ Pa
V₁ = 30 L = 30 x 10^-3 m³
T₁ = 273.15 + 27 = 300.15K
Using ideal gas equation,
pV = nRT

Hence, moles removed = 19.48-15.12 = 4.36
Mass removed = 4.36 x 32 g = 139.52 g = 0.1396 kg.
Therefore, 0.14 kg of oxygen is taken out of the cylinder.

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 cm deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Solution:
Volume of the air bubble, = 1.0 cm³ = 1.0 x 10^-6 m³
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T₁ = 12°C = 273 + 12 = 285K
Temperature at the surface of the lake, T₂ = 35°C = 273 + 35 = 308K
The pressure on the surface of the lake,
P₂ =1 atm = 1 x 1.013 x 10⁵Pa
The pressure at the depth of 40 m,
P₁ = 1 atm + dρg
where, ρ is the density of water = 10³ kg/m³
g is the acceleration due to gravity = 9.8 m/s²
∴ P₁ = 1.013 X10⁵+40 X 10³ X 9.8 = 493300 Pa
We have \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
where, V₂ is the volume of the air bubble when it reaches the surface
V₂= \(\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{(493300)\left(1.0 \times 10^{-6}\right) 308}{285 \times 1.013 \times 10^{5}}\)
= 5.263 x 10^-6 m³ or 5.263 cm³
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm³.

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Solution:
Volume of the room, V = 25.0 m³
Temperature of the room, T = 27°C = 273 + 27°C = 300 K
Pressure in the room, P = 1 atm = 1 x 1.013 x 10⁵ Pa
The ideal gas equation relating pressure (?), Volume (V), and absolute temperature (T) can be written as PV = k_BNT
where,
KB is Boltzmann constant = 1.38 x 10 ^-23 m² kg s^-2 K^-1
N is the number of air molecules in the room
∴ N = \(\frac{P V}{k_{B} T}\)
= \(\frac{1.013 \times 10^{5} \times 25}{1.38 \times 10^{-23} \times 300}\) = 6.11 x 10²⁶ molecules
Therefore, the total number of air molecules in the given room is 6.11 x 10²⁶.

Question 7.
Estimate the average thermal energy of a helium atom at (i) room temperature (27°C),
(ii) the temperature on the surface of the Sun (6000K),
(iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).
Solution:
(i) At room temperature, T = 27°C = 273 +27 = 300 K
Average thermal energy, E = \(\frac{3}{2}\)kT
where k is Boltzmann constant = 1.38 x 10^-23 m² kg s^-2 K^-1
∴ E = \(\frac{3}{2}\) x 1.38 x 10^-23 x 300 = 6.21 x 10^-21 J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 x 10^-23 J

(ii) On the surface of the Sun, T = 6000 K
Average thermal energy = \(\frac{3}{2}\) kT = \(\frac{3}{2}\) x 1.38 x 10^-23 x 6000
= 1.241 x 10^-19J
Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.241 x 10^-19J.

(iii) At temperature, T =10⁷ K
Average thermal energy = \(\frac{3}{2}\)kT = \(\frac{3}{2}\) x 1.38 x 10^-23 x 10⁷
= 2.07 x 10^-16 J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 x 10^-16 J.

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is υ_rms the largest?
Solution:
Yes. All contain the same number of the respective molecules.
No. The root means square speed of neon is the largest.
Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, N = 6.023 x 10²³.
The root mean square speed (υ_rms)oi a gas of mass m, and temperature T, is given by the relation: υ_rms = \(\sqrt{\frac{3 k T}{m}} \)
where k is Boltzmann constant
For the given gases, k and T are constants.

Hence, υ_rms depends only on the mass of the atoms, i.e., υ_rms ∝ \(\sqrt{\frac{1}{m}}\)
Therefore, the root mean square speed of the molecules in the three cases is not the same.

Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest.
Hence, neon has the largest root mean square speed among the given gases.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20°C? (atomic mass of Ar = 39.9 u, of He = 4.0u).
Solution:
Temperature of the helium atom, T_He = -20°C = 273 – 20 = 253K
Atomic mass of argon, M_Ar= 39.9 u
Atomic mass of helium, M_He = 4.0 u
Let, (υ_rms)be the rms speed of argon.
Let (υ_rms )He be the rms speed of helium.
The rms speed of argon is given by
(υ_rms)_Ar = \(\sqrt{\frac{3 R T_{\mathrm{Ar}}}{M_{\mathrm{Ar}}}}\) …………………………….. (i)
where, R is the universal gas constant
T_Ar is temperature of argon gas
The rms speed of helium is given by:
(υ_rms)_He = \(\sqrt{\frac{3 R T_{\mathrm{He}}}{M_{\mathrm{He}}}}\) …………………………. (ii)

It is given that: (υ_rms)_Ar = (υ_rms)_He

Therefore, the temperature of the argon atom is 2.52 x 10³ K.

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0u).
Solution:
Pressure inside the cylinder containing nitrogen,
P =2.0atm = 2 x 1.013 x 10⁵ Pa = 2.026 x 10⁵ Pa
Temperature inside the cylinder, T = 17°C = 273 +17 = 290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 x 10,sup>-10 m
Diameter, d = 2 x 1 x 10^-10 = 2 x 10^-10 m
Molecular mass of nitrogen, M = 28.0 g = 28 x 10^-3 kg
The root mean square speed of nitrogen is given by the relation
υ_rms = \(\sqrt{\frac{3 R T}{M}}\)
where, R is the universal gas constant = 8.314 J mole ^-1 K^-1
∴ υ_rms = \(\sqrt{\frac{3 \times 8.314 \times 290}{28 \times 10^{-3}}}\) = 508.26m/s
The mean free path (l) is given by the relation:
l = \(\frac{k T}{\sqrt{2} \times d^{2} \times P}\)
where,
k is the Boltzmann constant = 1.38 x 10^-23 kgm²s^-2 K^-1

Time is taken between successive collisions,

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

Additional Exercises

Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Solution:
Length of the narrow bore, L=1 m = 100 cm
Length of the mercury thread, l = 76 cm
Length of the air column between mercury and the closed-end, l_a = 15cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100-(76+15) = 9 cm

Hence, the total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure.
∴ Length of the air column in the bore = 24 + h cm
and, length of the mercury column = 76 – h cm
Initial pressure, P₁ = 76 cm of mercury
Initial volume,V₁ =15 cm³
Final pressure, P₂ = 76 – (76 – h) = h cm of mercury
Final volume, V₂ = (24 + h) cm³
The temperature remains constant throughout the process.

Height cannot be negative. Hence, 23.8cm of mercury will flow out from the bore and 52.2 cm of mercury will remain in it. The length of the air column will be 24 +23.8 = 47.8 cm.

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas. [Hint: Use Graham’s law of diffusion: R₁/R₂ =(M₂/M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of the kinetic theory.]
Solution:
Rate of diffusion of hydrogen, R₁ = 28.7cm³ s^-1
Rate of diffusion of another gas, R₂ = 7.2 cm³ s^-1
According to Graham’s Law of diffusion, we have
\(\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}\)
where, M₁ is the molecular mass of hydrogen 2.020 g
M₂ is the molecular mass of the unknown gas
∴ M₂ = M₁\(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) = 2.01 \(\left(\frac{28.7}{7.2}\right)^{2}\) = 32.09 g
32g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have a uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n₂ =n₁ exp[-mg(h₂ -h₁) / k_B T]
where n₂,n₁ refer to number density at heights h₂ and h₁ respectively.
Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n₂ = n₁ exp [-mg N_A (ρ -ρ’)(h₂ -h₁) / (ρRT)]
where ρ is the density of the suspended particle, and ρ’ that of surrounding medium. [N_A is Avogadro’s number and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Solution:
According to the law of atmospheres, we have
n₂=n₁ exp [-mg(h₂ – h₁])/k_BT] ………………………………. (i)
where, n₁ is the number density at height h₁, and n₂ is the number density at height h₂
mg is the weight of the particle suspended in the gas column
Density of the medium = ρ’
Density of the suspended particle = ρ
Mass of one suspended particle = m’
Mass of the medium displaced = m
Volume of a suspended particle = V
According to Archimedes’ principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as

Weight of the medium displaced – Weight of the suspended particle
= mg – m’g
= mg – Vρ’g = mg – \(\left(\frac{m}{\rho}\right)\) ρ’g
= mg – \(\left(1-\frac{\rho^{\prime}}{\rho}\right)\) …………………………….. (ii)
Gas constant, R = k_BN
k_B = \(\frac{R}{N}\) …………………………………….. (iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we get

Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:

Substance	Atomic Mass (u)	Density (103 kg m-3)
Carbon (diamond)	12.01	2.22
Gold	197.00	19.32
Nitrogen (liquid)	14.01	1.00
Lithium	6.94	0.53
Fluorine (liquid)	19.00	1.14

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
Atomic mass of a substance = M
Density of the substance = ρ

Avogadro’s number = N = 6.023 x 10²³
Volume of each atom = \(\frac{4}{3} \pi r^{3}\)
Volume of N number of molecules = \(\frac{4}{3} \pi r^{3}\) N …………………………….. (i)
Volume of one mole of a substance = \(\frac{M}{\rho}\) ………………………………….. (ii)

For gold

Hence, the radius of a gold atom is 1.59 Å
For liquid nitrogen

Hence, the radius of a liquid nitrogen atom is 1.77 Å

For lithium

Hence, the radius of a lithium atom is 1.73 Å.

For liquid fluorine

Hence, the radius of liquid fluorine atom is 1.88 Å.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 3 Motion in a Straight Line Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 3 Motion in a Straight Lines

PSEB 11th Class Physics Guide Motion in a Straight Line Textbook Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a), (b)

Explanation
(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
(b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
(c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

Question 2.
The position-time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in below figure. Choose the correct entries in the brackets below;
(a) (A / B) lives closer to the school than (B / A)
(b) (A / B) starts from the school earlier than (B/ A)
(c) (A / B) walks faster than (B / A)
(d) A and B reach home at the (same/different) time
(e) (A / B) overtakes (B / A) on the road (once/twice).

Solution:
(a) Draw normals on graphs from points P and Q. It is clear that OQ > OP. Therefore, child A lives closer to the school than child B.

(b) Child A start from school at time t =0 (become its graph starts from origin) whild child B starts from school at time t = OC. Therefore, child A starts from school earlier than B.

(c) The slope of distance-time graph represents the speed. More the slope of the graph, more will be the speed. As the slope of the x-t graph of B is higher than the slope of the x – t graph of A, therefore child B walks faster than child A.
(d) Corresponding to points P and Q, the value of t from x – t graphs for children A and B is same i. e.,OE. Therefore, children A and B will reach their homes P and Q at the same time.

(e) x – t graphs for children A and B intersect each other at a point D. Child B starts later but reaches home at the same time as that of child A, therefore child B overtake child A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5kmh^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x – t graph of her motion.
Solution:
Speed of the woman = 5 km/h
Distance between her office and home = 2.5 km
Time taken = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac{2.5}{5}\) = 0.5 h = 30 min
Time of arival at office = 9.00 am + 30 min = 9.30 am i. e., at 9.30 am the distance covered will be 2.5 km. This part of journey is represented in graph by OA.
It is given that she covers the same distance in the evening by an auto. Now, speed of the auto = 25 km/h
= \(\frac{2.5}{25}=\frac{1}{10}\) = 0.1 h = 6 mm

She leaves the office at 5.00 pm and take 6 min to reach home. Therefore, she reaches her home at 5.06 pm at this time the distance is zero. This part of journey is represented in graph by BC.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of bis motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Solution:
The x – t graph of the drunkard is shown in figure.

Length of each step =1 m, time taken for each step = 1 s
Time taken to move by 5 steps = 5 s
5 steps forward and 3 steps backward means that the net distance covered by him in first 8 steps i. e., in8s = 5m-3m = 2m
Distance covered by him in first 16 steps orl6s = 2 + 2 = 4m
Distance covered the drunkard in first 24 s i. e., 24 steps = 2 + 2 + 2= 6m
and distance covered in 32 steps i. e. 32 s = 8 m
Distance covered in37 steps = 8 + 5 = 13m
Distance of the pit from the start = 13 m
Total time taken by the drunkard to fall in the pit = 37 s
Since, 1 step requires 1 s of time, so we arrive at the same result from the graph shown.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Speed of the jet airplane, υ_jet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
υ_smoke = -1500 km/h
Speed of its products of combustion with respect to the ground V’_smoke Relative speed of its products of combustion with respect to the airplane,
υ_smoke = υ’_smoke υ _jet
-1500 = υ’_smoke – 500
υ’_smoke = -1000 km/h
The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

Question 6.
A car moving along a straight highway with speed of 126 km h^-1 is brought to a stop within a distance of 200 m, What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Solution:
Initial velocity of the car, u= 126 km/h = 126 × \(\frac{5}{18}\) m/s
= 35 m/s (∵ 1 km/h \(\frac{5}{10}\) m/s)
Final velocity of the car, υ = 0
Distance covered by the car before coming to rest, s = 200 m
From third equation of motion,
υ² – u² = 2as
(0)² – (35)² = 2 × a × 200
a = \(\frac{35 \times 35}{2 \times 200}\) = -3.06 m/s²
From first equation of motion,
v = u +at
t = \(\frac{v-u}{a}=\frac{0-35}{-3.06}=\frac{-35}{-3.06}\) = 11.44s
∴ Car will stop after 11.4 s.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Solution:
For train A:
Initial velocity, u = 72 km/h = 72 × \(\frac{5}{18}\) m/s = 20 m/s
Time, t = 50 s
Acceleration, a_I =0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (S_I) covered by train A can be obtained as :
S_I = ut + \(\frac {1}{2}\)a_It²
= 20 × 50 + 0 = 1000 m

For train B:
Initial velocity, u = 7 2 km/h = 72 × \(\frac{5}{18}\) m/ s = 20 m/ s
Acceleration, a_II = 1 m/s²
Time, t = 50 s
From second equation of motion, distance (S_II) covered by train B can be obtained as :
s_II = ut + \(\frac {1}{2}\) a_IIt²
= 20 × 50 + \(\frac {1}{2}\) × 1 × (50)² = 2250 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 = 1250 m.

Question 8.
On a two lane road, car A is travelling with a speed of 36 kmh^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h 1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Solution:
Velocity of car A, υ_A = 36 km/h = 36 × \(\frac {5}{18}\) m/s = 10 m/s
Velocity of car B, υ_B =54 km/h = 54 × \(\frac {5}{18}\) m/s = 15 m/s
Velocity of car C,υ_C = 54 km/h 54 × \(\frac {5}{18}\) m/s = 15 m/s
Relative velocity of car B with respect to car A,
υ_BA = υ_B – υ_A = 15 – 10 = 5 m/s
Relative velocity of car C with respect to car A,
υ_CA – υ_C – (-υ_A) = 15 + 10 = 25m/s
At a certain instance, both cars B and C are at the same distance from car Ai.e.,
s = 1 km = 1000 m
Time taken (t) by car C to cover 1000 m = \(\frac {1000}{25}\) = 40 s
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
s = ut + \(\frac {1}{2}\)at²
1000 = 5 × 40 + \(\frac {1}{2}\) × a × (40)²
a = \(\frac {1600}{1600}\) = 1ms²

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^-1 in the direction AtoB notices that a bus goes past him every 18 min in the direction of his motion, and eveiy 6 min in opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Solution:
Let V be the speed of the bus running between towns A and B.
Speed of the cyclist, υ = 20 km/h
Relative speed of the bus moving in the direction of the cyclist
= V- υ = (V – 20)km/h
The bus went past the cyclist every 18 min i.e., \(\frac{18}{60}\) h (when he moves in the direction of the bus).
Distance covered by the bus = (V – 20) × \(\frac{18}{60}\) km ……………….. (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to V × \(\frac{T}{60}\) ……………. (ii)
Both equations (i) and (ii) are equal.
V – 20 \(\frac{18}{60}=\frac{V T}{60}\) ……………… (iii)
Relative speed of the bus moving in the opposite direction of the cyclist
= (V + 20)km/h
Time taken by the bus to go past the cyclist = 6 min = \(\frac{6}{60}\)h
∴ (V + 20) \(\frac{6}{60}\) = \(\frac{VT}{60}\) …………………. (iv)
From equations (iii) and (iv), we get
(V + 20) × \(\frac{6}{60}\) = (V – 20) × \(\frac{18}{60}\)
V + 20 = 3 V – 60
2V = 80
V = 4 km/h
Substituting the value of V in equation (iv), we get
(40 + 20) × \(\frac{6}{60}=\frac{40 T}{60}\)
T = \(\frac{360}{40}\) = 9 min

Question 10.
A player throws a hall upwards with an initial speed of 29.4 ms^-1 .
(a) What is the direction of acceleration during the upward motion of the hall?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t₀ = 0 s to be the location and time of
the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s^-2 and neglect air resistance).
Solution:
(a) The ball is moving under the effect of gravity and therefore the direction of acceleration is vertically downward, in the direction of acceleration due to gravity.
(b) At the highest point of its motion velocity is zero and acceleration is equal to the acceleration due to gravity (9.8 m/s) in vertically downward direction.
(c) If we choose the highest point as x = 0 m and t₀ = 0 s and vertically downward direction to be the positive direction of X- axis then,

During upward motion
Sign of position is negative.
Sign of velocity is negative.
Sign of acceleration is positive.

During downward motion Sign of position is positive.
Sign of velocity is positive.
Sign of acceleration is positive.

(d) Let the ball rises upto maximum height h.
Initial velocity of ball (u) = 29.4 m/s
g = 9.8 m/s
Final velocity at maximum height (υ) = 0
Using equation of motion, υ² = u² – 2gh
0 = (29.4)² – 2 × 9.8 × h
or h = \(\frac{29.4 \times 29.4}{2 \times 9.8}\) = 44.1
Again using equation of motion, υ = u – gt
0 = 29.4 -9.8t
or t = \(\frac{29.4}{9.8}\) = 3s
Time of ascent is always equal to the time of descent.
Total time after which the ball returns to the player’s hand = 2t = 2 × 3 = 6s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant’
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True
Explanation: When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.
Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.

(b) False
Explanation: A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.

(c) True
Explanation: This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.

(d) False
Explanation: This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution:
Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s²
Final velocity of the ball = υ
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:
s = ut + \(\frac {1}{2}\)at²
90 = 0 + \(\frac {1}{2}\) × 9.8t²
t = \(\sqrt{18.38}\) = 4.29 s
From first equation of motion, final velocity is given as:
υ = u + at
= 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball, u_r = \(\frac {9}{10}\) υ = \(\frac {9}{10}\) × 42.04 = 37.84 m/s

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
υ =u_r + at’
0 = 37.84 + (-9.8) t’
t’ = \(\frac{-37.84}{-9.8}\) = 3.86s
Total time taken by the ball = t + t’ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor
= \(\frac {9}{10}\) × 37.84 = 34.05m/s
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s The speed-time graph of the ball is represented in the given figure as :

Question 13.
Explain clearly, with examples, the distinction between the following:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval].
Show in both (a) and (h) that the second quantity is either greater than or equal to the first.
When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Solution:
(a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual path length covered by the particle in a given interval of time.
For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.

Whereas, total path length = AB+BC
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.

(b)
For the given particle,
Average velocity = \(\frac{A C}{t}\)

= \(\frac{A B+B C}{t}\)
Since (AB + BC)> AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time? (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero.]
Solution:
(a) A man return his home, therefore total displacement of the man = 0

(b) Speed of man during motion from his home to the market υ₁ = 5 km/h
Speed of man during from market his home υ₂ =7.5 km/h
Distance between his home and market = 2.5 km
(i) Taking time interval 0 to 30 min.
Time taken by the man to reach the market from home,
t₁ = \(\frac{2.5}{5}=\frac{1}{2}\) = h = 30 min
Hence, the man moves from his home to the market in t = 0 to 30 min.

(ii) Taking time interval 0 to 50 min.
Time taken by man in returning to his home from the market

(iii) Taking time interval 0 to 40 min.
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
= 7.5 × \(\frac{10}{60}\) =1.25 km
Net displacement 2.5 -1.25 = 1.25 km
Total distance travelled = 2.5 +1.25 = 3.75 km
Average velocity = \(\frac{1.25}{\left(\frac{40}{60}\right)}\) = \(\frac{1.25 \times 3}{2}\) = 1.875 km/h
Average speed = \(\frac{3.75}{\left(\frac{40}{60}\right)}\) = 5.625 km/h

Question 15.
In questions 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Solution:
Instantaneous velocity is given by the first derivative of distance with respect to time i. e.,
υ_m = \(\frac{d x}{d t}\)
Here, the time interval dt is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.
Therefore, instantaneous speed is always equal to instantaneous velocity.

Question 16.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Solution:
(a) The given x – t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
(b) The given υ – t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
(c) The given υ – t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
(d) The given total path length-time graph, shown in (d), does not represent one dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

Question 17.
Figure shows the x – t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle’ moves in a straight line for t< 0 and on a parabolic path for t > 0 ? If not, suggest a suitable physical context for this graph.

Solution:
No; The x – t graph of a particle moving in a straight line for t < 0 and on a parabolic path for t > 0 cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as t = 0, x = 0. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a theifs car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m s^-1, with what speed does the bullet hit the thief’s car ?
(Note: Obtain that speed which is relevant for damaging the thief’s car).
Solution:
Speed of the police van, υ_p = 30 km/h = 8.33 m/s
Muzzle speed of the bullet, υ_b = 150 m/s
Speed of the thief s car, υ_t =192 km/h = 53.33 m/s
Since the bullet is fired from a moving van, its resultant speed can be obtained as:
= 150 +8.33 = 158.33 m/s
Since, both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief s car can be obtained as:
υ_bt = υ_b – υ_t
= 158.33 – 53.33 = 105 m/s

Question 19.
Suggest a suitable physical situation for each of the following graphs:

Solution:
(a) The given x – t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.

(b) In the given v-t graph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

(c) The given a – t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.

Question 20.
Figure gives the x – t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s,-1.2 s.

Solution:
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a = -ω²x ……. (i)
where, ω → angular frequency
At t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x – t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
At t = 1.2s
In this time interval, x is positive. Thus, the slope of the x – t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
t = -1.2s
In this time interval, x is negative. Thus, the slope of the x – t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

Question 21.
Figure gives the x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Solution:
The average speed of a particle shown in the x – t graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The
sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

Question 22.
Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

Solution:
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the,given interval of time.

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, . average speed of the particle is the greatest in interval 3.

In interval 1: The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2: The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3: The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C and D, acceleration of the particle is zero.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1, 2, 3,…)versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Solution:
Distance covered by a body in n^th second is given by the relation
D_n = u + \(\frac{a}{2}\)(2n – 1) ……………(i)
where, u = Initial velocity, a = Acceleration, n = Time = 1, 2, 3, …. , n
In the given case,
u = 0 and a = 1 m/s²
.-. D_n = \(\frac {1}{2}\)(2n – 1) ……………… (ii)
This relation shows that
D_n ∝ n ………………(iii)
Now, substituting different values of n in equation (ii), we get the following table:

Since the three-wheeler acquires uniform velocity after 10 s, the line , will be parallel to the time-axis after n = 10 s.

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Solution:
Initial velocity of the ball, u = 49 m/s
Acceleration, a = -g = – 9.8 m/s2
Case I:
When the lift was stationary, the boy throws the ball.
Taking upward motion of the ball,
Final velocity, υ of the ball becomes zero at the highest point.
From first equation of motion, time of ascent (t) is given as
υ = u + at
t = \(\frac{v-u}{a}\)
\(\frac{-49}{-9.8}\) = 5s
But, the time of ascent is equal to the time of descent.
Hence, the total time taken by the ball to return to the boy’s hand
= 5 + 5 = 10 s
Motion in a Straight Line 53

Case II:
The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i. e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.

Question 25.
On a long horizontally moving belt (see figure) a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time taken by the child in (a) and (b) ?
Which of the answers alter if motion is viewed by one of the parents?

Solution:
Speed of the belt, υ_B = 4 km/h
Speed of the child, υ_C = 9 km/h

(a) Since the child is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as
υ_CB = υ_C + υ_B = 9 + 4 = 13 km/h

(b) Since the child is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as
υ_CB = υ_C + (-υ_B) = 9 – 4 = 5 km/h

(c) Distance between the child’s parents = 50 m
As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i. e.,
9 km/h = 9 x \(\frac{5}{18}\) m/s = 2.5 m/s.
18
Hence, the time taken by the child in case (a) and (b) is given by
\(\frac{\text { Distance }}{\text { Speed }}=\frac{50}{2.5}\) = 20 s.
If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m/s². Give the equations for the linear and curved parts of the plot.

Solution:
For first stone:
Initial velocity, u₁ =15 m/s
Acceleration, a = -g = -10 m/s²
Using the relation,
x₁ = x₀ + u₁t + \(\frac {1}{2}\)at²
where, x₀ = Height of the cliff = 200 m
x₁ =200 + 15t – 5t² ………………. (i)
When this stone hits the ground, x₁ = 0
-5t² +15t + 200 = 0
t² – 3t – 40 =0
t² – 8t + 5t – 40 = 0
t(t – 8) + 5(t – 8) = 0
(t – 8)(t + 5) = 0
t = 8 s or t = -5s
Since the stone was projected at time t = 0, the negative sign before time is meaningless.
∴ t = 8s

For second stone:
Initial velocity, u₂ = 30 m/s
Acceleration, a = -g = -10 m/s²
Using the relation,
x₂ = x₀ + u₂t + \(\frac {1}{2}\)at²
= 200 + 30t – 5t² ……………. (ii)
At the moment when this stone hits the ground; x₂ = 0
-5t² + 30t + 200 = 0
t² – 6t – 40 = 0
t² -10t + 4t + 40 = 0
t(t – 10) + 4(t – 10) = 0
(t – 10)(t + 4) = 0
t = 10 s or t = -4 s
Here again, the negative sign before time is meaningless.
∴ t = 10 s
Subtracting eq. (i) from eq. (ii), we get
x₂ – x₁ = (200 +30t – 5t²) – (200 + 15t – 5t²)
x₂ – x₁ = 15t …………….. (iii)
Equation (iii) represents the linear path of both stones. Due to this linear relation between
(x₂ – x₁) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t = 8 s.
(x₂ – x₁] )_max = 15 × 8 = 120 m
This is in accordance with the given graph.
After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation :
x₂ – x₁ = 200 + 30t – 5t²
Hence, the equation of linear and curved path is given by
x₂ – x₁ = 15t (Linear path)
x₂ – x₁ = 200 + 30t – 5t² (Curved path)

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in figure given below. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?
Solution:
(a) Distance travelled by the particle = Area under the given graph
= \(\frac{1}{2}\) × (10 – 0) × (12 – 0) = 60 m
Average speed = \(\frac{\text { Distance }}{\text { Time }}\) = \(\frac{60}{10}\) = 6 m/s

(b) Let s₁ and s₂ be the distances covered by the particle between time t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
S = S₁ + s₂ ……………… (i)

For distance S₁:
Let u’ be the velocity of the particle after 2 s and a’ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
υ = u + at
Where, υ = Final velocity of the particle
12. = 0 + a’ × 5
a’ = \(\frac{12}{5}\) = 2.4 m/s² .
Again, from first equation of motion, we have
υ = u + at
= 0 + 2.4 × 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i. e., in 3 s
S₁ = u’t + \(\frac{1}{2}\) a’t²
= 4.8 × 3 + \(\frac{1}{2}\) × 2.4 × (3)²
= 25.2 m ……………… (ii)

For distance S₂:
Let a” be the acceleration of the particle between time t = 5 s and t = 10s.
From first equation of motion,
υ = u + at (where υ = 0 as the particle finally comes to rest)
0 = 12 + a” × 5
a” = \(\frac{-12}{5}\)
= -2.4 m/s²
Distance travelled by the particle in Is (i. e., between t = 5 s and t = 6 s)
S₂ = u”t + \(\frac{1}{2}\) at²
= 12 × a + \(\frac{1}{2}\)(-2.4) × (1)²
= 12 – 1.2 = 10.8 m ……………… (iii)
From equations (i), (ii), and (iii), we get
S = 25.2 + 10.8 = 36 m
∴ Average speed = \(\frac{36}{4}\) = 9 m/s

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in figure.

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ ?
(a) x(t₂) = x(t₁) + υ(t₁)(t₂ – t₁) + \(\frac {1}{2}\) a(t₂ – t₁)2
(b) υ(t₂) = υ(t₁)+a(t₂ – t₁)
(c) _Average = [x(t₂) – x(t₁)] /(t₂ – t₁)
(d) _Average = [(t₂ ) – υ(t₁)] / (t₂ – t₁)
(e) x(t₂) = x(t₁) + υ_Average (t₂ – t₁) + (\(\frac {1}{2}\)) a_Average (t₂ – t₁)²
(f) x (tsub>2) – x (tsub>1) = area under the υ – t curve bounded by the t-axis and the dotted line shown.
Solution:
The slope of the given graph over the time interval tsub>1 to tsub>2 is not constant and is not uniform. It means acceleration is not constant and is not uniform, therefore, relation (a), (b) and (e) are not correct which is for uniform accelerated motion, but relations (c), (d) and (f) are correct, because these relations are true for both uniform or non-uniform accelerated motion.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 9 Mechanical Properties of Solids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

PSEB 11th Class Physics Guide Mechanical Properties of Solids Textbook Questions and Answers

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 x 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10^-5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Given, length of the steel wire, L₁ = 4.7 m
Area of cross-section of the steel wire,A₁ = 3.0 x 10^-5 m²
Length of the copper wire, L₂ = 3.5 m
Area of cross-section of the copper wire, A₂ = 4.0 x 10^-5 m²

Change in length = ΔL₁ = ΔL₂ = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire,
Y₁ = \(\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{\Delta L_{1}} \)
= \(\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L} \) ………………………………. (i)
Young’s modulus of the copper wire,
Y₂ = \(\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{\Delta L_{2}}=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L}\) ………………………………. (ii)
Dividing eq. (i) by eq. (ii), we get:
\(\frac{Y_{1}}{Y_{2}}=\frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}\) = 1.79:1
The ratio of Young’s modulus of steel to that of copper is 1.79: 1.

Question 2.
Figure given below shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strengths for this material?

Solution:
(a) It is clear from the given graph that for stress 150 x 10⁶ N/m², strain is 0.002.
∴ Young s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
= \(\frac{150 \times 10^{6}}{0.002}\) = 7.5 x 10¹⁰ N/m²
Hence, Young’s modulus for the given material is 7.5 x10¹⁰ N/m²

(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit. It is clear from the given graph that the approximate yield strength of this material is 300 x 10⁶ N/m² or 3 x 10⁸ N/m².

Question 3.
The stress-strain graphs for materials A and B are shown in figure given below.

The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Solution:
(a) A, for a given strain, the stress for material A is more than it is for > material B, as shown in the two graphs.
Young’s modulus = \(\frac{\text { Stress }}{\text { Strain }}\)
For a given strain, if the stress for a material is more, then Young’s modulus is also greater for that material. Therefore, Young’s modulus for material A is greater than it is for material B.

(b) A, the amount of stress required for fracturing a material, corresponding to its fracture point, gives the strength of that material. Fracture point is the extreme point in a stress-strain curve. It can be observed that material A can withstand more strain than material B. Hence, material A is stronger than material B.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Solution:
(a) False.
Reason: For a given stress, the strain in rubber is more than it is in steel.
Young’s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }} \)
For a constant stress Y ∝ \(\frac{1}{\text { Strain }}\)
Hence, Young’s modulus for rubber is less than it is for steel.

(b) True.
Reason: Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure given below. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m.
Compute the elongations of the steel and the brass wires.

Solution:
Given, diameter of the wires, d = 0.25 cm
Hence, the radius of the wires, r = \(\frac{d}{2} \) = 0.125cm = 0.125 x 10^-2m
Length of the steel wire, L₁ = 1.5 m
Length of the brass wire, L₂ = 1.0 m
Total force exerted on the steel wire,
F₁ = (4 +6)g= 10×9.8 = 98N .

Young’s modulus for steel
Y₁ = \(\frac{\left(\frac{F_{1}}{A_{1}}\right)}{\left(\frac{\Delta L_{1}}{L_{1}}\right)} \)
where, ΔL₁ = Change in the length of the steel wire
A₁ = Area of cross-section of the steel wire = πr²₁

Young’s modulus of steel, Y₁ = 2.0 x 10¹¹ Pa
∴ ΔL₁ = \(\frac{F_{1} \times L_{1}}{A_{1} \times Y_{1}}=\frac{F_{1} \times L_{1}}{\pi r_{1}^{2} \times Y_{1}}\)
= \(\frac{98 \times 1.5}{3.14\left(0.125 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}}\)
= 1.5 x 10^-4 m

Total force on the brass wire
F₂ =6 x 9.8=58.8N
Young’s modulus for brass
Y₂ = \(\frac{\left(\frac{F_{2}}{A_{2}}\right)}{\left(\frac{\Delta L_{2}}{L_{2}}\right)}\)

where, ΔL₂ = Change in length of the steel wire
A₂ = Area of cross-section of the brass wire
∴ ΔL₂ = \(\frac{F_{2} \times L_{2}}{A_{2} \times Y_{2}}=\frac{F_{2} \times L_{2}}{\pi r_{2}^{2} \times Y_{2}}\)
= \(\frac{58.8 \times 1.0}{3.14 \times\left(0.125 \times 10^{-2}\right)^{2} \times\left(0.91 \times 10^{11}\right)}\)
= 1.3 x 10^-4 m
Hence, elongation of the steel wire =1.49 x 10^-4 m
and elongation of the brass wire = 1.3 x 10^-4 m

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg Is the attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
Given, edge of the aluminium cube, L = 10cm = 0.1 m
The mass attached to the cube, m =100 kg
Shear modulus (ri) of aluminium = 25GPa =25 x 10⁹ Pa
Shear modulus, η = \(\frac{\text { Shear stress }}{\text { Shear strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
where, F = Applied force = mg = 100 x 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 x 0.1 = 0.01 m²

ΔL = Vertical deflection of the cube
ΔL = \(\frac{F L}{A \eta}=\frac{980 \times 0.1}{10^{-2} \times\left(25 \times 10^{9}\right)}\)
= 3.92 x 10^-7 m
The vertical deflection of this face of the cube is 3.92 x 10^-7 m.

Question 7.
Four identical tblloW cylindrical columns of mild steel support a big structure of mass’50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Solution:
Given, mass of the big structure, M = 50000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 x 10¹¹ Pa
Total force exerted, F =Mg = 50000 x 9.8N
Stress = Force exerted on a single column = \(\frac{50000 \times 9.8}{4}\) = 122500 N

Young’s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }} \)
Strain = \(\frac{F}{\frac{A}{Y}} \)
where, Area, A = π(R² – r²) = π[(0.6)² – (0.3)²]
Strain = \(\frac{122500}{3.14\left[(0.6)^{2}-(0.3)^{2}\right] \times 2 \times 10^{11}}\) = 7.22 x 10^-7
Hence, the compressional strain of each column is 7.22 x 10^-7.
∴Compressional strain of all columns is given by
= 7.22 x 10 ^-7 x 4 = 2.88 x 10^-6.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 minxes 19.2 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Given, cross-section area.of copper piece (A) = 15.2 mm x 19.1 mm
= (15.2 x 19.1) x 10 ^-6m²
Force applied (F) = 44500 N
Young’s modulus (Y) =1.1 x 10¹¹ Nm^-2

Young s modulus (Y) = \(=\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
or Longitudinal strain = \(\frac{\text { Longitudinal stress }}{\text { Young’s modulus }}\)
Young’s modulus
= \(\frac{(F / A)}{Y}=\frac{F}{A Y}\)
= \(\frac{44500}{15.2 \times 19.1 \times 10^{-6} \times 1.1 \times 10^{11}}\)
= 0.0013934.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ Nm ^-2, what is the maximum load the cable can support?
Solution:
Radius of the steel cable, r = 1.5cm = 0.015m
Maximum allowable stress = 10⁸ N m^-2
Maximum stress = \(\frac{\text { Maximum force }}{\text { Area of cross – section }} \)
∴ Maximum force = Maximum stress x Area of cross – section
= 10⁸ x π (0.015)²
= 7.065 x 10⁴ N
Hence, the cable can support the maximum load of 7.065 x 10⁴ N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each mid are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
Solution:
The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young’s modulus is given as:
Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\text { Strain }}=\frac{\frac{4 F}{\pi d^{2}}}{\text { Strain }}\) ……………………………. (i)
where, F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that Y ∝ \(\frac{1}{d^{2}}\)
Young’s modulus for iron, Y₁ = 190 x 10⁹ Pa
Diameter of the iron wire = d₁
Young’s modulus for copper, Y₂ = 110 x 10⁹ Pa
Diameter of the copper wire = d₂
Therefore, the ratio of their diameters is given as:
\(\frac{d_{2}}{d_{1}}=\sqrt{\frac{Y_{1}}{Y_{2}}}=\sqrt{\frac{190 \times 10^{9}}{110 \times 10^{9}}}=\sqrt{\frac{19}{11}}\)
= 1.31:11.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle.
The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution:
Given, mass, m = 14.5kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev / s
Cross-sectional area of the wire, a = 0.065cm² = 0.065 x 10^-4 m²
Let δl be died elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is :
F = mg+mlω² ,
= 14.5 x 9.8 +14.5x 1 x (2)² = 200.1 N

Young’s modulus = \(\frac{\text { Stress }}{\text { Strain }} \)
Y = \(\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{F}{A} \cdot \frac{l}{\Delta l}\)
∴ Δl = \(\frac{F l}{A Y}\)

Young’s modulus for steel = 2 x 10¹¹ Pa
∴ Δl = \(\frac{200.1 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}} \) = 1539.23 x 10^-7
= 1.539 x 10^-4
Hence, the elongation of the wire is 1.539 x 10^-4 m.

Question 12.
Compute the bulk modulus of water from the following data: Initial volume =100.0 litre, Pressure increase =100.0atm (1 atm = 1.013 x 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Solution:
Initial volume, V₁ = 100.0 l x 10^-3 m³
Final volume, V₂ = 100.5l = 100.5 x 10^-3 m^-3
Increase in volume, V = V₂ — V₁ = 0.5 x 10^-3 m³
Increase in pressure, Δ_p =100.0 atm = 100 x 1.013 x 10⁵ Pa
Bulk modulus = \( \frac{\Delta p}{\frac{\Delta V}{V_{1}}}=\frac{\Delta p \times V_{1}}{\Delta V}\)
= \(\frac{100 \times 1.013 \times 10^{5} \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}\) = 2.206 x 10⁹ Pa
Bulk modulus of air = 1.0 x 10⁵ Pa
∴ \(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^{9}}{1.0 \times 10^{5}}\)
= 2.026 x 10⁴
This ratio is very high because air is more compressible than water.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 103 x 10³ kgm^-3?
Solution:
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 x 1.01 x 10⁵ Pa
Density of water at the surface, ρ₁ = 1.03 x 10³ kg m^-3
Let ρ₂ be the density of water at the depth h.
Let V₁ be the volume of water of mass m at the surface.
Let V₂ be the volumé of water of mass m at the depth h.
Let ΔV be the change in volume.

ΔV = V₁ – V₂ = \(m\left(\frac{1}{\rho_{1}}-\frac{1}{\rho_{2}}\right)\)
∴ Volumetric strain= \(\frac{\Delta V}{V_{1}}=m\left(\frac{1}{\rho_{1}}-\frac{1}{\rho_{2}}\right) \times \frac{\rho_{1}}{m}\)
∴ \(\frac{\Delta V}{V_{1}}=1-\frac{\rho_{1}}{\rho_{2}}\) ………………………………. (i)

Bulk modulus, B = \(\frac{p V_{1}}{\Delta V}\)
\(\frac{\Delta V}{V_{1}}=\frac{p}{B}\)
Compressibility of water = \(\frac{1}{B}=45.8 \times 10^{-11} \mathrm{~Pa}^{-1}\)
∴ \(\frac{\Delta V}{V_{1}}=80 \times 1.013 \times 10^{5} \times 45.8 \times 10^{-11}\) = 3.71 x 10^-3 ………….(ii)
From equations (i) and (ii), we get

Therefore, the density of water at the given depth (h) is 1.034 x 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 x 1.013 x 10⁵ Pa
Bulk modulus of glass, B = 37 x 10⁹ Nm^-2
Bulk modulus, B = \(\frac{p}{\frac{\Delta V}{V}}\)
where, \(\frac{\Delta V}{V}\) = Fractional change in volume
∴ \(\frac{\Delta V}{V}=\frac{p}{B}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}\)
= 2.73 x 10^-5
Hence, the fractional change in the volume of the glass slab is
2.73 x 10^-5 = 2.73 x 10^-3% = 0.0027%

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0x 10⁶ Pa.
Solution:
Length of an edge of the solid copper cube, l =10 cm = 0.1 m
Hydraulic pressure, p = 7.0 x 10⁶ Pa
Bulk modulus of copper, B = 140 x 10⁹ Pa
Bulk modulus, B = \(\frac{P}{\frac{P}{\Delta V}}\)
where, \(\frac{\Delta V}{V}\) = Volumetric strain
ΔV = Change in volume
V =,Original volume
ΔV = \(\frac{p V}{B}\)
Original volume of the cube, V = l³

Therefore, the volume contraction of the solid copper cube is 0.05 cm³.

Question 16.
How much should the pressure on a litre of water be changed to compress by 0.10%?
Solution:
Volume of water, V =1 L
It is given that water is to be compressed by 0.10%.
∴ Fractional change, \(\frac{\Delta V}{V}=\frac{0.1}{100 \times 1}=10^{-3}\)
Bulk modulus, B = \(\frac{p}{\frac{\Delta V}{V}}\)
p = B x \( \frac{\Delta V}{V}\)

Bulk modulus of water, B = 2.2 x 10⁹ Nm^-2
p = 22 x 10⁹ x 10^-3
=2.2 x 10⁶ Nm^-2
Therefore, the pressure on water should be 2.2 x 10⁶ Nm^-2.

Additional Exercises

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in figure given below, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Solution:
Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 x 10^-3m
radius, r = \( \frac{d}{2}\) = 0.25 x 10^-3 m
Compressional force, F = 50000 N
Pressure at the tip of the anvil,
p = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{\pi r^{2}}\)
= \(\frac{50000}{3.14 \times\left(0.25 \times 10^{-3}\right)^{2}}\)
= 2.55 x 10¹¹ Pa
Therefore, the pressure at the tip of the anvil is 2.55 x 10¹¹ Pa.

Question 18.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure given below. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0mm², respectively. At what point along the rod should a mass m he suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution:
Given, cross-sectional area of wire A, a₁ = 1.0 mm₂ = 1.0 x 10^-6 m²
Cross-sectional area of wire B, a₂ = 2.0 mm² = 2.0 x 10^-6 m²
Young’s modulus for steel, Y₁ = 2 x 10¹¹ Nm^-2
Young’s modulus for aluminium, Y₂ = 7.0 x 10¹⁰ Nm^-2

Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{a}\)
If the two wires have equal stresses, then,
\( \frac{F_{1}}{a_{1}}=\frac{F_{2}}{a_{2}}\)
where, F₁ = Force exerted on the steel wire

F₂ = Force exerted on the aluminium wire
\(\frac{F_{1}}{F_{2}}=\frac{a_{1}}{a_{2}}=\frac{1}{2}\) …………………………. (i)
The situation is shown in the following figure.

Taking torque about the point of suspension, we have
F₁y = F₂(1.05— y)
\(\frac{F_{1}}{F_{2}}=\frac{(1.05-y)}{y}\) ……………………….(ii)
Using equations (i) and (ii), we can write

In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young s modulus = \(\frac{\text { Stress }}{\text { Strain }} \)
Strain = \(\frac{\text { Stress }}{\text { Young’s modulus }}=\frac{\frac{F}{a}}{Y}\)
If the strain in the two wires is equal, then,

Taking torque about the point where mass m, is suspended at a distance y₁ from the side where wire A attached, we get
F₁y₁ =F₂(1.05-y₁)
\(\frac{F_{1}}{F_{2}}=\frac{\left(1.05-y_{1}\right)}{y_{1}}\) ……………………………. (iv)
Using equations (iii) and (iv), we get
\(\frac{\left(1.05-y_{1}\right)}{y_{1}}=\frac{10}{7}\)
7(1.05 – y₁) = 10 y₁
⇒ 17 y₁ = 7.35
y₁ = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Solution:

Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 x 10^-2 cm² = 0.50 x 10^-6 m²
A mass 100 g is suspended from its mid-point.
m = 100 g = 0.1kg
Hence, the wire dips, as shown in the given figure.

Original length = XZ
Depression = l
The length after mass m, is attached to the wire = XO +OZ
Increase in the length of the wire:
Δl = (XO + OZ)-XZ
Where

Expanding and neglecting higher terms, we get:
Δl = \(\frac{l^{2}}{0.5}\)
Strain = \(\frac{\text { Increase in length }}{\text { Original length }}\)
Let T be the tension in the wire.

∴ mg = 2Tcos θ
Using the figure, it can be written as
Cos θ = \(\frac{l}{\left[(0.5)^{2}+l^{2}\right]^{\frac{1}{2}}}=\frac{l}{(0.5)\left[1+\left(\frac{l}{0.5}\right)^{2}\right]^{\frac{1}{2}}}\)
Expanding the expression and eliminating the higher terms, we get
Cos θ = \(\frac{l}{(0.5)\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)}\)
\(\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)\) ≈ 1 for small l

Hence, the depression at the mid-point is 0.0107 m.

Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension’ that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 x 10⁷ Pa?
Assume that each rivet is to carry one-quarter of the load.
Solution:
Diameter of the metal strip, d = 6.0 mm = 6.0 x 10^-3 m
Radius, r = \(\frac{d}{2}\) = 3.0 x 10^-3 m
Maximum shearing stress = 6.9 x 10⁷ Pa
Maximum stress = \(\frac{\text { Maximum load or force }}{\text { Area }}\)
Maximum force = Maximum stress x Area
= 6.9 x 10⁷ x π x (r)²
= 6.9 x 10⁷ x 3.14 x (3 x10^-3)²
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 x 1949.94 = 7799.76 N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 x 10⁸ Pa. A steel ball of initial volume 0.32m is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
Solution:
Water pressure at the bottom, p = 1.1 x 10⁸ Pa
Initial volume of the steel ball, V = 0.32 m³
Bulk modulus of steel, B = 1.6 x 10¹¹ Nm^-2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = \( \frac{p}{\frac{\Delta V}{V}}\)
ΔV = \(\frac{p V}{B}=\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}\)
= 2.2 x 10^-4 m³
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 x 10^-4 m³.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 6 Work, Energy and Power Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

PSEB 11th Class Physics Guide Work, Energy and Power Textbook Questions and Answers

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution:
(a) Positive
In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

(b) Negative
In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.

(c) Negative
Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.

(d) Positive
Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.

(e) Negative
The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

Question 2.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.
Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s, and interpret your results.
Solution:
Mass of the body, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, μ = 0.1
Initial velocity, u = 0
Time, t = 10 s
The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:
a’ = \(\frac{F}{m}\) = \(\frac{7}{2}\) = 3.5 m/s²
Frictional force is given as:
f = μ mg = 0.1 × 2 × 9.8 = -1.96 N
The acceleration produced by the frictional force:
a” = –\(\frac{1.96}{2}\) = -0.98 m/s²
Total acceleration of the body:
a = a’ + a”
= 3.5 + (-0.98) = 2.52 m/s²
The distance travelled by the body is given by the equation of motion:
s = ut + \(\frac{1}{2}\) at²
= 0 + \(\frac{1}{2}\) × 2.52 × (10)² = 126 m
(a) Work done by the applied force, W_a = F × s = 7x 126 = 882 J
(b) Work done by the frictional force, W_f = f × s = -1.96 × 126 = -247 J
(c) Net force = 7 + (-1.96) = 5.04 N
Work done by the net force, W_net = 5.04 × 126 = 635 J
(d) From the first equation of motion, final velocity can be calculated as:
υ = u + at
= 0 + 2.52 × 10 = 25.2 m/s
Change in kinetic energy = \(\frac{1}{2}\) mυ ² – \(\frac{1}{2}\) mu²
= \(\frac{1}{2}\) × 2(υ² – u²) = (25.2)² – 0² = 635 J

Question 3.
Given in figure below are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle muqt have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.


Solution:
Total energy of a system is given by the relation:
E = P.E. + K.E.
> K.E. = E – P.E.
Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative.
(i) In the given case, the potential energy (V₀) of the particle becomes greater than total energy (E) for x > a. Hence, kinetic energy becomes negative in this region. Therefore, the particle will not exist in this region. The minimum total energy of the particle is zero.

(ii) In the given case, the potential energy (V₀) is greater than total energy (E) in all regions. Hence, the particle will not exist in this region.

(iii) In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between x > a and x<b.
The minimum potential energy in this case is -V₁. Therefore, K.E. = E-(-V₁) = E + V₁.
Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than -Vj. So, the minimum total energy the particle must have is -V₁.

(iv) In the given case, the potential energy (V₀) of the particle becomes greater than the total energy (E) for –\(\frac{b}{2}\)< x <\(\frac{b}{2}\) and –\(\frac{a}{2}\)< x <\(\frac{a}{2}\).
Therefore, the particle will not exist in these regions.
The minimum potential energy in this case is -V₁. Therefore, K.E. = E – (-V₁ ) = E + V₁. Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than -V₁. So, the minimum total energy the particle must have is -V₁.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by V (x) = kx² / 2, where k is the force constant of the oscillator. For k = 0.5 Nm^-1, the graph of V (x) versus x is shown in figure below. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ±2m.

Solution:
Total energy of the particle, E = 1 J
Force constant, k = 0.5 Nm^-1
Kinetic energy of the particle, K = \(\frac{1}{2}\)mυ²
According to the conservation law:
E = V + K
1 = \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)mυ ²
At the moment of ‘turn back’, velocity (and hence K) becomes zero.
> 1 = \(\frac{1}{2}\)kx²
\(\frac{1}{2}\) × 0.5 ײ = 1
x² = 4
x = ±2
Hence, the particle turns back when it reaches x = ±2 m.

Question 5.
Answer the following:
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In fig. (i) the man walks 2 m carrying a mass of 15 kg on his hands. In fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Solution:
(a) The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket.
According to the conservation of energy:
Total Energy (T.E.) = Potential energy (P.E.) + Kinetic energy (K.E.)
= mgh + \(\frac{1}{2}\) mυ²
The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

(b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero.

(c) When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.

(d) Case (i)
Mass, m = 15 kg
Displacement, s = 2 m
Work done, W = Fs cosθ
where, θ = Angle between force and displacement
= mgs cosθ = 15 × 2 × 9.8 cos 90°
= 0 ( cos90° = 0)

Case (ii)
Mass, m = 15 kg Displacement, s = 2 m
Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.
Therefore, the angle between them, θ =0°
Since, cos0° = 1
Work done, W = Fs cosθ = mgs
= 15 × 9.8 × 2 = 294J
Hence, more work is done in the second case.

Question 6.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/ remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Solution:
(a) Decreases, A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.

(b) Kinetic energy, The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.

(c) External force, Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many-particle system is proportional to the external forces acting on the system.

(d) Total linear momentum, The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Solution:
(a) False, In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.

(b) False, Although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.

(c) False, The work done in the motion of a body over a closed loop is zero for a conservation force only.

(d) True, In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

Question 8.
Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i. e., when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Solution:
(a) No, In an elastic collision, the total initial kinetic eneigy of the bails will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.

(b) Yes, In an elastic collision, the total linear momentum of the system always remains conserved.

(c) No, In an inelastic collision, there is always a loss of kinetic energy, i. e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision.
Yes, The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.

(d) Elastic, In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) \(t^{\frac{1}{2}}\)
(ii) t
(iii) \(t^{\frac{3}{2}}\)
(iv) t²
Solution:
(ii) t
Let a body of mass m which is initially at rest undergoes one-dimensional motion under a constant force F with a constant acceleration a.
Acceleration (a) = \(\frac{F}{m}\) …………. (i)
Using equation of motion, υ = u + at
⇒ υ = 0 + \(\frac{F}{m}\).t …………. (∵ u = 0)
⇒ υ = \(\frac{F}{m}\)t …………… (ii)
Power delivered (P) = Fυ
Substituting the value from eq. (ii), we get
⇒ P = F × \(\frac{F}{m}\) × t
⇒ P = \(\frac{F^{2}}{m}\)t
Dividing and multiplying by m in R.H.S.
P = \(\frac{F^{2}}{m^{2}}\) × mt= a²mt [Using eq.(i)]
As mass m and acceleration a are constant.
∴ P ∝ t

Question 10.
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(i) \(t^{\frac{1}{2}}\)
(ii) t
(iii) \(t^{\frac{3}{2}}\)
(iv) t²
Solution:
(iii) \(t^{\frac{3}{2}}\)
Power is given by the relation:
P = Fυ
= maυ = mυ \(\frac{d v}{d t}\) = Constant (say,k )
υ dv = \(\frac{k}{m}\) dt
Integrating both sides:
\(\frac{v^{2}}{2}=\frac{k}{m}\)t
υ = \(\sqrt{\frac{2 k t}{m}}\)
For displacement x of the body, we have:
υ = \(\frac{d x}{d t}=\sqrt{\frac{2 k}{m}} t^{\frac{1}{2}}\)
dx = k’\(t^{\frac{1}{2}}\)dt
where, k’ = \(\sqrt{\frac{2 k}{3}}\) = New constant
On integrating both sides, we get:
x = \(\frac{2}{3} k^{\prime} t^{\frac{3}{2}}\)
x ∝ \(t^{\frac{3}{2}}\)

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = -î + 2ĵ + 3k̂N
where î, ĵ, k̂ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Solution:
Force exerted on the body, F = -î + 2ĵ + 3k̂N
Displacement, s = 4 k̂ m
Work done, W = F.s
= (-î + 2ĵ + 3k̂).(4k̂)
= 0 + 0 + 3 × 4 = 12 J
Hence, 12 J of work is done by the force on the body.

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 × 10^-31 kg, proton mass = 1.67 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J).
Solution:
Mass of the electron, m_e = 9.11 × 10^-31 kg
Mass of the proton, m_p =1.67 × 10^-27 kg
Kinetic energy of the electron, E_Ke =10 keV = 10⁴ eV
= 10⁴ × 1.60 × 10^-19
1.60 × 10^-15 J
Kinetic energy of the proton, E_Kp = 100 keV = 10⁵ eV = 1.60 × 10^-14 J
For the velocity of an electron v_e, its kinetic energy is given by the relation:

Question 13.
A rain drop of radius 2 mm falls from a height of500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done hy the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^-1?
Solution:
Radius of the rain drop, r = 2 mm = 2 × 10 ^-3 m
Volume of the rain drop, V = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ m^-3
Density of water, ρ = 10³ kgm^-3
Mass of the rain drop, m = ρV
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³
Gravitational, F = mg
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ 9.8N
The work done by the gravitational force on the drop in the first half of its journey.
W₁ = Fs
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3 )³ × 10³ × 9.8 × 250
= 0.082 J

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i. e., W_II, = 0.082 J.
As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.
∴ Total energy at the top:
E_T = mgh + 0
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × 500
= 0.164 J
Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.
∴ Total energy at the ground:
E_g = \(\frac{1}{2}\) mυ² + 0
= \(\frac{1}{2}\) × \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × (10)²
= 1.675 × 10^-3 J = 0.001675
∴ Resistive force = E_g – E_t = 0.001675 – 0.164 = -0.162 J

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 m s^-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Solution:
Yes; Collision is elastic
The momentum of the gas molecule remains conserved Whether the collision is elastic or inelastic.
The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.
It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Solution:
Volume of the tank, V = 30 m³
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h = 40 m
Efficiency of the pump, η = 30%
Density of water, ρ = 10³ kg/m³
Mass of water, m = ρ V = 30 × 10³ kg
Output power can be obtained as:
P₀ = \(\frac{\text { Work done }}{\text { Time }}=\frac{m g h}{t}\)
= \(\frac{30 \times 10^{3} \times 9.8 \times 40}{900}\) = 13.067 × 10³ W
For input power P_i efficiency η, is given by the relation :
η = \(\frac{P_{o}}{P_{i}}\) = 30%
P_i = \(\frac{13.067}{30}\) × 100 10³
= 0.436 × 10² W = 43.6 kW

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision?

Solution:
It can be observed that the total momentum before and after collision in each case is constant.
For an elastic collision, the total kinetic energy of a system remains conserved before and after collision.
For mass of each ball bearing m, we can write:
Total kinetic energy of the system before collision:
= \(\frac{1}{2}\)mV² + \(\frac{1}{2}\)(2m)0
= \(\frac{1}{2}\)mV²

Case (i)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\)m × 0 + \(\frac{1}{2}\)(2m) (\(\frac{V}{2}\))²
= \(\frac{1}{4}\)mV²
Hence, the kinetic energy of the system is not conserved in case (i).

Case (ii)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\) (2m) × 0 + \(\frac{1}{2}\)mV²
= \(\frac{1}{2}\) mV²
Hence, the kinetic energy of the system is conserved in case (ii).

Case (iii)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\)(3m) (\(\frac{V}{3}\))²
= \(\frac{1}{6}\) mV²
Hence, the kinetic energy of the system is not conserved in case (iii).

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another hob B of the same mass at rest on a table as shown in’ figure given below. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Solution:
Bob A will not rise at all
In an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. In this case, a complete transfer of momentum takes place from the moving mass to the stationary mass. Hence, bob A of mass m, after colliding with bob B of equal mass, will come to rest, while bob B will move with the velocity of bob A at the instant of collision.

Question 18.
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost-point, given that it dissipated 5% of its initial energy against air resistance?
Solution:
Length of the pendulum, l = 1.5 m
Mass of the bob = m
Energy dissipated = 5%
According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:
Potential energy of the bob, E_P = mgl
Kinetic energy of the bob, E_K = 0
Total energy = mgl …………. (i)

At the lowermost point (mean position):
Potential energy of the bob, E_P = 0
Kinetic energy of the bob, E_K = \(\frac{1}{2}\)mυ²
Total energy, E_x = \(\frac{1}{2}\)mυ² ………….. (ii)
As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.
The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i. e.,
\(\frac{1}{2}\)mυ² = \(\frac{95}{100}\) × mgl
υ = \(\sqrt{\frac{2 \times 95 \times 1.5 \times 9.8}{100}}\) = 5.28 m/s

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of ahole on the floor of the trolley at the rate of 0.05 kg s^-1 . What is the speed of the trolley after the entire sand bag is empty?
Solution:
The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sand bag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, die speed of the trolley will remain 27 km/h.

Question 20.
A body of mass 0.5 kg travels in a straight line with velocity υ = ax^3/2 where a = 5 m^-1/2s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Solution:
Mass of the body, m = 0.5 kg
Velocity of the body is governed by the equation, υ = a \(x^{\frac{3}{2}}\) where,
a = 5m^-1/2s^-1
Initial velocity, u (at x = 0) = 0
Final velocity, υ (at x = 2 m) = 5 × (2)^3/2 m/s = 10√2 m/s
Work done, W = Change in kinetic energy
= \(\frac{1}{2}\)m(υ² – u²)
= \(\frac{1}{2}\) × 0.5[(10√2)² – (0)²]
= \(\frac{1}{2}\) × 0.5 × 10 × 10 × 2
= 50 J

Question 21.
The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m², υ = 36km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced?
Solution:
(a) Area of the circle swept by the windmill = A
Velocity of the wind = υ
Density of air = ρ
Volume of the wind flowing through the windmill per sec = Aυ
Mass of the wind flowing through the windmill per sec = ρ Aυ
Mass m, of the wind flowing through the windmill in time t = ρ A υ t

(b) Kinetic energy of air = \(\frac{1}{2}\) mυ²
= \(\frac{1}{2}\) (ρAυt)v² = \(\frac{1}{2}\)ρ Aυ³t

(c) Area of the circle swept by the windmill, A = 30 m²
Velocity of the wind, υ =36 km/h = 36 × \(\frac{5}{18}\) m/s
= 10 m/s [1 km/s = \(\frac{5}{18}\) m/s]
Density of air, ρ = 1.2 kg m^-3
Electric energy produced = 25% of the wind energy
= \(\frac{25}{100}\) ×Kinetic energy of air
= \(\frac{1}{8}\)ρAυ³t

\(\) = \(\frac{1}{8}\)ρAυ³
= \(\frac{1}{8}\) × 1.2 × 30 × (10)³
= 4.5 × 10³ W = 4.5 kW

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution:
(a) Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5
m Number of times the weight is lifted, n = 1000
∴ Work done against gravitational force:
= n(mgh)
=1000 × 10 × 9.8 × 0.5
= 49 × 10³ J = 49 kJ

(b) Energy equivalent of 1 kg of fat = 3.8 × 10⁷ J
Efficiency rate = 20%
Mechanical energy supplied by the person’s body
= \(\frac{20}{100}\) × 3.8 × 10⁷ J
= \(\frac{1}{5}\) × 3.8 × 107 J 5
Equivalent mass of fat lost by the dieter
= \(\frac{1}{\frac{1}{5} \times 3.8 \times 10^{7}}\) × 49 × ³
= \(\frac{245}{3.8}\)× 3.8 × 107
= × 10^-4
= 6.45 × 10^-3 kg

Question 23.
A family uses 8 kW of power, (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
Solution:
(a) Power used by the family, P = 8 kW = 8 × 10³ W
Solar energy received per square meter = 200 W
Efficiency of conversion from solar to electrical energy = 20%
Area required to generate the desired electricity = A
As per the information given in the question, we have
8 × 10³ = 20% × (A × 200)
= \(\frac{20}{100}\) × A × 200
> A = \(\frac{8 \times 10^{3}}{40}\) = 200m²

(b) In order to compare this area to that of the roof of a typical house, let ‘a’ be the side of the roof
∴ area of roof =a × a = a²
Thus a² = 200 m²
or a = \(\sqrt{200 \mathrm{~m}^{2}}\) = 14.14 m
∴ area of roof = 14.14 × 14.14 m²
Thus 200 m² is comparable to the roof of a typical house of dimensions 14.14 m × 14.14 m = 14 m × 14 m.

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Solution:
Mass of the bullet, m = 0.012 kg
Initial speed of the bullet, u_b = 70 m/s
Mass of the wooden block, M = 0.4 kg
Initial speed of the wooden block, u_B = 0
Final speed of the system of the bullet and the block = υ
Applying the law of conservation of momentum,
mu_b +MU_B = (m + M)υ
012 × 70 + 0.4 × 0 = (0.012 + 0.4)υ
∴ υ = \(\frac{0.84}{0.412}\) = 2.04 m/s

For the system of the bullet and the wooden block,
Mass of the system, m’ = 0.412 kg
Velocity of the system = 2.04 m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system,
Potential energy at the highest point
= Kinetic energy at the lowest point
m’ gh = \(\frac{1}{2}\)m’ υ²
h = \(\frac{1}{2}\)(\(\frac{v^{2}}{g}\))
= \(\frac{1}{2}\) (\(\frac{(2.04)^{2}}{9.8}\)) = 0.2123m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet
– Kinetic energy of the system
= \(\frac{1}{2}\) mu_b– \(\frac{1}{2}\)m’ υ²
= \(\frac{1}{2}\) × 0.012 × (70)² – \(\frac{1}{2}\) × 0.412 × (2.04)²
= 29.4 – 0.857 = 28.54 J

Question 25.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (see figure). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60°and h = 10m, what are the speeds and times taken by the two stones?

Solution:
No; the stone moving down the steep plane will reach the bottom first Yes; the stones will reach the bottom with the same speed The given situation can be shown as in the following figure:

Here, the initial height (AD) for both the stones is the same (h).
Hence, both will have the same potential energy at point A.
As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i. e.,
\(\frac{1}{2}\) mυ₁² = \(\frac{1}{2}\)mυ₂²
υ₁ = υ₁ = υ, say
where,
m = Mass of each stone
υ = Speed of each stone at points B and C
Hence, both stones will reach the bottom with the same speed, υ.

For stone I: Net force acting on this stone is given by:
F_net = ma₁ = mgsinθ₁
a₁ = gsinθ₁

For stone II: a₂ = gsinθ₂
∵ θ₂ > θ₁
∴ sinθ₂ > sinθ₁
∴ a₂ > a₁
Using the first equation of motion, the time of slide can be obtained as:
υ = u + at
∴ t = \(\frac{v}{a}\) (∵ u = 0)
For stone I: t₁ = \(\frac{v}{a_{1}}\)
For stone II: t₂ = \(\frac{v}{a_{2}}\)
∵ a₂ > a₁
∴ t₂ < t₁
Hence, the stone moving down the steep plane will reach the bottom first. The speed (υ) of each stone at points B and C is given by the relation obtained from the law of conservation of energy.

Question 26.
A1 kg block situated on a rough incline is connected to a spring of spring constant 100 Nm^-1 as shown in figure given below. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Solution:
Mass of the block, m = 1 kg
Spring constant, k = 100 N m^-1
Displacement in the block, x = 10 cm = 0.1 m
The given situation can be shown as in the following figure.

At equilibrium:
Normal reaction, R = mg cos 37°
Frictional force, f = μR = p mg cos 37°
where, μ is the coefficient of friction
Net force acting on the block = mg sin 3 7° – f
= mgsin37° – μmgcos37°
= mg (sin37° – μcos37°)
At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,
mg (sin37° – μcos37°) x = \(\frac{1}{2}\)kx ²
1 × 9.8 (sin 37° – μcos37°) = \(\frac{1}{2}\) × 100 × 0.1
0.6018 – μ × 0.7986 = 0.5102
0.7986 μ = 0.6018 – 0.5102 = 0.0916
∴ μ = \(\frac{0.0916}{0.7986}\) = 0.115

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m^-1. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Solution:
Mass of the bolt, m = 0.3 kg
Speed of the elevator = 7 m/s
Height, h = 3 m
Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.
Heat produced = Loss of potential energy
= mgh = 0.3 × 9.8 × 3 = 8.82 J
The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

Question 28.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^-1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Solution:
Mass of the trolley, M = 200 kg
Speed of the trolley, υ = 36 km/h = 36 × \(\frac{5}{18}\)m/s = 10 m/s
Mass of the boy, m = 20 kg
Initial momentum of the system of the boy and the trolley
= (M + m)υ
= (200 + 20) × 10
= 2200 kg-m/s
Let v’ be the final velocity of the trolley with respect to the ground.
Final velocity of the boy with respect to the ground = υ’ – 4
Final momentum = Mυ’ + m(υ’ – 4)
= 200 υ’ + 20υ’ – 80
= 220 υ’ – 80
As per the law of conservation of momentum,
Initial momentum = Final momentum
2200 =220 υ’ – 80
∴ υ’ = \(\frac{2280}{220}\) = 10.36 m/s
Length of the trolley, l = 10 m
Speed of the boy, υ” = 4 m/s
Time taken by the boy to run, t = \(\frac{10}{4}\) = 2.5 s
∴ Distance moved by the trolley = υ’ × t= 10.36 × 2.5 = 25.9 m

Question 29.
Which of the following potential energy curves in the given figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Solution:
(i), (ii), (iii), (iv), and (vi)
The potential energy of a system of two masses is inversely proportional to the separation between them. In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero [i.e., V (r) = 0] when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

Question 30.
Consider the decay of a free neutron at rest: n → p + e^–. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (see figure).

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of P-decay. This particle is known as
neutrino. We now know that it is a particle of intrinsic spin \(\frac{1}{2}\) (like e^–, p or n), but is neutral, and either massless or having an extremely small mass
(compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e^– + υ]
Solution:
The decay process of free neutron at rest is given as:
n → p + e^–
From Einstein’s mass-energy relation, we have the energy of electron as
Δ𝜏 mc²
where,
Δ𝜏 m = Mass defect
= Mass of neutron – (Mass of proton + Mass of electron)
c = Speed of light
Δ𝜏 m and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the p-decay of a neutron or a nucleus. The presence of neutrino v on the LHS of the decay correctly explains the continuous energy distribution.
Here, Δ𝜏 m= mass of neutron – (mass of proton + mass of electron)
= 1.6747 × 10^-24 -(1.6724 × 10^-24 + 9.11 × 10^-28)
= (1.6747 – 1.6733) × 10^-24
= 0.0014 × 10^-24 gms ‘
∴ E = 0.0014 × 10^-24 × (3 × 10¹⁰)² ergs
= 0.0126 × 10^-4 ergs
= [Latex]\frac{0.0126 \times 10^{-4}}{1.6 \times 10^{-12}}[/Latex]
( ∵ 1 eV = 1.6 × 10^-19 J = 1.6 × 10^-19 × 10⁷ ergs)
= 0.007875 × 10 ⁸ eV
= 0.7875 × 10⁶ eV
= 0.79 MeV

A positron has the same mass as an electron but an opposite charge of+e. When an electron and a positron come close to each other, they destroy each other. Their masses, are converted into energy according to Einstein’s relation and the energy so obtained is released in the form of y-rays and is given by
E’ = mc² = 2 × 9.1 × 10^-31 kgx (3 × 10⁸ ms^-1)²
= 1.638 × 10^-13 J
= \(\frac{1.638 \times 10^{-13}}{1.6 \times 10^{-13}}\) MeV
= 1.02MeV (∵1 MeV = 1.6 × 10^-13 eV)
= Minimum energy a photon must possess for pair production.

Alternate method
Decay of a free neutron at rest,
n → p + e^–
Let Δ m be the mass defect during this process.
∴ Mass defect (Δ m) = (Mass of neutron) – (Mass of proton and electron) This mass defect is fixed and hence electron of fixed energy should be produced. Therefore, two-body decay of this type cannot explain the observed continuous energy distribution in the (3-decay of a neutron in a nucleus.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 2 Biological Classification Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 2 Biological Classification

PSEB 11th Class Biology Guide Biological Classification Textbook Questions and Answers

Question 1.
Discuss how classification systems have undergone several changes over a period of time?
Answer:
(i) Linnaeus proposed a two kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively. But as this system did not distinguish between the eukaryotes and prokaryotes, unicellular and multicellular organisms and photosynthetic (green algae) and non-photosynthetic (fungi) organisms, so scientists found it an inadequate system of classification. Classification systems for the living organisms have hence, undergone several changes over time.

(ii) The two kingdom system of classification was replaced by three kingdom system, then by four and finally by five kingdom system of classification of RH Whittaker (1969).

(iii) The five kingdoms included Monera, Protista, Fungi, Plantae and Animalia. This is the most accepted system of classification of living organisms.

(iv) But, Whittaker has not described viruses and lichens. Then Stanley described viruses, viroids, etc.
Thus, over a period of time, classification systems have undergone several changes.

Question 2.
State two economically important uses of:
(a) Heterotrophic bacteria
(b) Archaebacteria
Answer:
(a) Heterotrophic Bacteria

• Maintain soil fertility by nitrogen fixation, ammonification and nitrification, e.g., Rhizobium bacteria (in the root nodules of legumes).
• The milk products such as butter, cheese, curd, etc., are obtained by the action of bacteria. The milk contains bacterial forms like Streptococcus lacti, Escherichia coli, Lactobacillus lactis and Clostridium sp., etc.

(b) Archaebacteria

• Methanogens are responsible for the production of methane (biogas) from the dung of animals.
• Archaebacteria help in the degradation of waste materials.

Question 3.
What is the nature of cell walls in diatoms?
Answer:
In case of diatoms, the cell wall forms two thin overlapping cells, which fit together as in a soap box. The cell wall is made up of silica. Due to siliceous nature of cell wall, it is known as diatomite or diatomaceous Earth. Diatomaceous Earth is a whitish, highly porous, chemically inert, highly absorbant and fire proof substance.

Question 4.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify?
Answer:
Sometimes, greqn algae such as Chlorella, Scenedesmus and Spirogyra, etc,, grow in excess in water bodies and impart green colour to the water. These are called algal blooms. Red dianoflagellates (Gonyaulax) grow in abundance in sea and impart red colour to the ocean. This looks like red tides. Both due to algal blooms and ‘red tide’ the animal life declines due to toxins and deficiency of oxygen inside water.

Question 5.
How are viroids different from viruses?
Answer:
Viroids different from viruses as follows:

Virus	Viroids
1. These are smaller than bacteria.	Smaller than viruses.
2. Both RNA and DNA present.	Only RNA is present.
3. Protein coat present.	Protein coat absent.
4. Causes diseases like mumps and AIDS.	Causes plant diseases like spindle tuber diseases – potato.

Question 6.
Describe briefly the four major groups of Protozoa.
Answer:
Protozoans are divided into four phyla on the basis of locomotory organelles – Zooflagellata, Sarcodina, Sporozoa and Ciliates.
(i) Zooflagellates: These protozoans possess one to several flagella for locomotion. Zooflagellates are generally uninucleate, occasionally multinucleate.
The body is covered by a firm pellicle. There is also present cyst formation.
Examples: Giardia, Trypanosoma, Leishmania and Trichonympha, etc.

(ii) Sarcodines: These protozoans possess pseudopodia for locomotion. Pseudopodia are of four types, i.e., lobopodia, filopodia, axopodia and reticulopodia. Pseudopodia are also used for engulfing food particles. Sarcodines are mostly free living, found in freshwater, sea water and on damp soil only a few are parasitic. Nutrition is commonly holozoic. Sarcodines are generally uninucleates. Sarcodines are of four types – Amoeboids (i.e.,Amoeba, etc.), radiolarians (i.e., Acanthometra, etc.), foraminiferans (i.e., Elphidium, etc.) and heliozoans (i.e., Actinophrys, etc.).

(iii) Sporozoans: All of them are endoparasites. Locomotoryorganelles (cilia, flagella, pseudopodia, etc.) are absent. Nutrition is parasitic (absorptive), Phogotrophy is rare. The body is covered with an elastic pellicle or cuticle. Nucleus is single. Contractile vacuoles are absent. Life cyle consists of two distinct asexual and sexual phases. They may be passed in one (monogenetic) or two different hosts (digenetic),e.g., Plasmodium, Monocystis, etc.

(iv) Ciliates: These are aquatic, actively moving organisms because of the presence of thousands of cilia. They have a cavity (gullet) that opens to the outside of the cell surface. The coordinated movements of rows of cilia causes the water laden with food to enter into the gullet, e.g., Paramecium.

Question 7.
Plants are autotrophic. Can you think of some plants that are partially heterotrophic?
Answer:
Plants are autotrophs, i.e., they prepare their own food through the process of photosynthesis. But, in nature there are also some other plants which are partjally heterotrophic, i.e., they partially depend upon another organisms for food requirements, e.g.,
(i) Loranthus and Viscum are partial stem parasites which have leathery leaves. They attack several fruit and forest trees and with the help of their haustoria draw sap from the xylem tissue of the host.

(ii) Insectivorous plants have special leaves to trap insects. The trapped insects are killed and digested by proteolytic enzymes secreted by the epidermis of the leaves, e.g., pitcher plant.

(iii) Parasitic plant, e.g., Cuscutta develops haustoria, which penetrate, the vascular bundles of the host plant to absorb water and solutes.

Question 8.
What do the terms phycobiont and mycobiont signify?
Answer:
In case of lichens (t. e., an association of algae and fungi), the algal partner which is capable of carrying out photosynthesis is known as phycobiont, whereas the fungal partner which is heterotrophic in nature is known as mycobiont.

Question 9.
Give a comparative account of the classes of Kingdom Fungi under the following:
(i) Mode of nutrition
(ii) Mode of reproduction

Fungal Class	Mode of Nutrition	Mode of Reproduction
Myxomycetes	Heterotrophic and mostly saprophytic	Asexual and sexual reproduction
Phycomycetes	Mostly parasites	Asexual and sexual methods
Zygomycetes	Mostly saprophytic	Asexual and sexual reproduction
Ascomycetes	Saprophytes or parasites	Asexual and sexual reproduction
Basidiomycetes	Saprophytes or parasites	Asexual and sexual method
Deuteromycetes	Saprophytes or parasites	Only asexual reproduction

Question 10.
What are the characteristic features of Euglenoids?
Answer:
The characteristic features of euglenoids are as follows :

• They occur in freshwater habitats and damp soils.
• A single long flagella present at the anterior end.
• Creeping movements occur by expansion and expansion of their body known as euglenoid movements.
• Mode of nutrition is holophytic, saprobic or holozoic.
• Reserve food material is paramylum.
• Euglenoids are known as plant and animal.
  Plant characters of them are as follows:
  (a) Presence of chloroplasts with chlorophyll.
  (b) Holophytic nutrition
  Animal characters of them are as follows:
  (a) Presence of pellicle, which is made up of proteins and not a cellulose.
  (b) Presence of stigma.
  (c) Presence of contractile vacuole.
  (d) Presence of longitudinal binary fission.
• Under favourable conditions euglenoids multiply by longitudinal binary fission, e.g., Euglena, Phacus, Paranema, etc.

Question 11.
Give a brief account of viruses with respect to their structure ’ and nature of genetic material. Also name four common viral diseases.
Answer:
Viruses are non-cellular, ultramicroscopic, infectious particles. They are made up of envelope, capsid, nucleoid and occasionally one or two enzymes. Viruses possess an outer thin loose covering called envelope. The central portion of nucleoid is surrounded by capsid that is made up of ( smaller sub-units known as capsomeres.

The nucleic acid present in the viruses is known as nucleoid. It is the r infective part of the virus which utilises the host cell machinery. The
genetic material of viruses is of four types –

(i) Double stranded DNA (dsDNA) as found in pox virus, hepatitis-B virus and herpes virus, etc.
(ii) Single stranded DNA (ssDNA) occur in coliphage Φ, coliphage Φ x 174.
(iii) Double stranded RNA (dsRNA) occur in Reo virus,
(iv) Single stranded RNA (ssRNA) occur in TMV virus, polio virus, etc.
Four common viral diseases are – (i) Polio, (ii) AIDS, (iii) Hepatitis-B (iv) Rabies.

Question 12.
Organise a discussion in your class on the topic—Are viruses living or non-living?
Answer:
Viruses are intermediate between living and non-living objects. They resemble non-living objects in:

• Lacking protoplast. ‘
• Ability to get crystallised.
• High specific gravity which is found only in non-living objects.
• Absence of respiration and energy storing system.
• Absence of growth and division.
• Cannot live independent of a living cell.

They resemble living objects in:

• Presence of genetic material (DNA or RNA).
• Property of mutation.
• Irritability.
• Can grow and multiply inside the host cell.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 8 Gravitation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 8 Gravitation

PSEB 11th Class Physics Guide Gravitation Textbook Questions and Answers

Question 1.
Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?
Answer:
(a) No, Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects.

(b) Yes, If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).

(c) Tindal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

Question 2.
Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula -G Mm(1 /r₂ -1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth.
Solution:
(a) Decreases,
Explanation : Acceleration due to gravity at altitude h is given by the relation:
g_h = \(\left(1-\frac{2 h}{R_{e}}\right) g\)
where, R_e = Radius of the Earth
g = Acceleration due to gravity on the surface of the Earth
It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

(b) Decreases,
Explanation : Acceleration due to gravity at depth d is given by the relation:
g_d = \(\left(1-\frac{d}{R_{e}}\right) g\)
It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

(c) Mass of the body,
Explanation : Acceleration due to gravity of body of mass m is given by the relation:
g = \(\frac{G M_{e}}{R^{2}}\)
where, G = Universal gravitational constant
M_e = Mass of the Earth
R_e = Radius of the Earth
Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More,
Explanation : Gravitational potential energy of two points r₂ and r₁ distance away from the centre of the Earth is respectively given by:
V(r₁) = -GmM
V(r₂) = – \(\frac{G m M}{r_{2}}\)
V = V (r₂) – V(r₁) = -GmM\(\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)\).
Hence, this formula is more accurate than the formula mg (r₂ – r₁).

Question 3.
Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Solution:
Time taken by the Earth to complete one revolution around the Sun,
T_e = 1 year
Orbital radius of the Earth in its orbit, R_e = 1 AU
Time taken by the planet to complete one revolution around the Sun,
T_p = \(\frac{1}{2} T_{e}=\frac{1}{2}\) year
Orbital radius of the planet = R_p
From Kepler’s third law of planetary motion, we can write

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

Question 4.
I₀, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 x 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Solution:
Orbital period of I₀, TI₀ = 1.769 days = 1.769 x 24 x 60 x 60 s
Orbital radius of I_I0, R_I0 = 4.22 x 108 m
Satellite I₀ is revolving around the Jupiter Mass of the latter is given by the relation:
M_j = \(\frac{4 \pi^{2} R_{I_{O}}^{3}}{G T_{I_{O}}^{2}}\) ……………………….. (i)

where M_j = Mass of Jupiter
G = Universal gravitational constant Orbital period of the Earth,
T_e = 365.25 days = 365.25 x 24 x 60 x 60 s Orbital radius of the Earth,
Re =1 AU = 1.496 x 10¹¹ m
Mass of Sun is given as:
M_s = \(\frac{4 \pi^{2} R_{e}^{3}}{G T_{e}^{2}}\) ………………………. (ii)

= 1045.04
∴ \(\frac{M_{s}}{M_{j}} \sim 1000\)
\(M_{s} \sim 1000 \times M_{j}\)
Hence, it can be inferred that the mass of Jupiter is about one thousandth that of the Sun.

Question 5.
Let us assume that our only consists of 2.5 x 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic center take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.
Solution:
Number of stars in our galaxy, = 2.5 x 10¹¹
Mass of each star = one solar mass =2 x 10³⁰ kg
Mass of our galaxy,M =2.5 x 10¹¹ x 2 x 10³⁰ = 5 x 10⁴¹ kg
Diameter of Milky Way, d = 10⁵ ly
Radius of Milky Way,r= 5 x 10⁴ ly
∵ 1 ly=946 x 10¹⁵m
∴ r=5 x 10⁴ x 9.46 x 10¹⁵
=4.73 x 10²⁰m

Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
T = \(\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{1 / 2}\)

1 year = 365 x 324 x 60 x 60 s
∵ 1 s = \(\frac{1}{365 \times 24 \times 60 \times 60} \) years
∴ 1.12 x 10¹⁶ s = \(\frac{1.12 \times 10^{16}}{365 \times 24 \times 60 \times 60}\) = 3.55 x 10 ⁸ years

Question 6.
Choose the correct alternative:
(a) If the zero of potential energy Is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence Is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Solution:
(a) Kinetic energy
Explanation: Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (maybe negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative.

Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.

(b) Less
Explanation: An orbiting satellite acquires certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

Question 7.
Does the escape speed of a body from the Earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Answer:
(a) No, we know that the escape velocity of the body is given by
V_e = \(\sqrt{\frac{2 G M}{R}} \) , where M and R are mass and radius of Earth. Thus clearly, it does not depend on the grass of the body as V_e is independent of it.

(b) Yes, we know that V_e depends upon the gravitational potential at the point from where the body is launched. Since the gravitational potential depends on the latitude and height of the point, therefore the escape velocity depends on the location of the point from where it is projected. It can also be experienced as:
V_e = \(\sqrt{2 g r}\) . As g has different values at different heights. Therefore, V_e depends upon the height of location.

(c) No, it does not depend on the direction of projection as V_e is independent of the direction of projection.

(d) Yes, it depends on the height of location from where the body is launched as explained in (b).

Question 8.
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
(a) No, according to law of conservation of linear momentum L = mvr constant, therefore the comet moves faster when it is close to the sun and moves slower when it is farther away from the sun. Therefore, the speed of the comet does not remain constant.

(b) No, as the linear speed varies, the angular speed also varies. Therefore, angular speed of the comet does not remain constant.

(c) Yes, as no external torque is acting on the comet, therefore, according to law of conservation of angular momentum, the angular momentum of the comet remain constant.

(d) No, kinetic energy of the comet = \(\frac{1}{2}\) mν² As the linear speed of the comet changes as its kinetic energy also changes. Therefore, its KE does not remains constant.

(e) No, potential energy of the comet changes as its kinetic energy changes.

(f) Yes, total energy of a comet remain constant throughout its orbit.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem?
Answer:
(b), (c), and (d)
Explanations:
(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.

(b) A swollen face is caused generally because of apparent weightlessness in space sense organs such as eyes, ears, nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.

(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.

(d) Space has different orientations. Therefore, orientational problems can affect an astronaut in space.

Question 10.
In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the center of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure) (i) a, (ii) b, (iii) c, (iv) 0.

Solution:
(iii) Gravitational potential (V) is constant at all points in a spherical shell.
Hence, the gravitational potential gradient \(\left(\frac{d V}{d r}\right)\) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at center O will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at center O of the given hemispherical shell has the direction as indicated by arrow c.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow
(i) d,
(ii) e,
(iii) f,
(iv) g.
Solution:
(ii) Gravitational potential (V) is constant at all points in a spherical shell.
Hence, the gravitational potential gradient \(\left(\frac{d V}{d r}\right)\) inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. Ibis indicates that gravitational forces acting at a point in a spherical shell are symmetric. If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Question 12.
A rocket Is fired from the Earth towards the Sun. At what distance from the Earth’s centre is the gravitational force on the rocket zero? Mass of the Sun = 2 x 10³⁰kg, mass of the Earth = 6 x 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 10¹¹ m)
Solution:
Mass of the Sun, M_s = 2 x 10³⁰ kg
Mass of the Earth, M_e = 6 x 10 ²⁴ kg
Orbital radius, r = 1.5 x 10¹¹ m
Mass of the rocket = m

Let x be the distance from the center of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:

= 2.59 x 10⁸ m

Question 13.
How will you ‘weigh the Sun’, that is estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 x 10⁸ km.
Solution:
Orbital radius of the Earth around the Sun, r = 1.5 x 10⁸ km = 1.5 x 10¹¹ m
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days = 365.25 x 24 x 60 x 60 s
Universal gravitational constant, G = 6.67 x 10^-11N-m² kg^-2
Thus, mass of the Sun can be calculated using the relation,
M = \(\frac{4 \pi^{2} r^{3}}{G T^{2}}\)
= \(\frac{4 \times(3.14)^{2} \times\left(1.5 \times 10^{11}\right)^{3}}{6.67 \times 10^{-11} \times(365.25 \times 24 \times 60 \times 60)^{2}}\)
= \(\frac{133.24 \times 10^{33}}{6.64 \times 10^{4}}\) = 2.0 x 10³⁰kg
Hence, the mass of the Sun is 2.0 x 10³⁰ kg.

Question 14.
A Saáurn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.5 x 10⁸ km away from the Sun?
Solution:
Distance of the Earth from the Sun, r_e = 1.5x 10⁸ km= 1.5 x 10¹¹ m
Time period of the Earth = T_e
Time period of Saturn, T_s = 29.5 T_e
Distance of Saturn from the Sun = r_s
From Kepler’s third law of planetary motion, we have
T=\(\left(\frac{4 \pi^{2} r^{3}}{G M}\right)^{\frac{1}{2}}\)
For Saturn and Sun, we can write

= 1.43 x 10¹² m

Question 15.
A body weigths 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Solution:
Weight of the body, W =63 N
Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
g’ = \(\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}\)
where, g = Acceleration due to gravity on the Earth’s surface
R_e =Radius of the Earth
For h= \(\frac{R_{e}}{2}\)
g’ = \(\frac{g}{\left(1+\frac{R_{e}}{2 \times R_{e}}\right)^{2}}=\frac{g}{\left(1+\frac{1}{2}\right)^{2}}=\frac{4}{9} g\)

Weight of a body of mass m at height h is given as
W’ = mg’
= m x \(\frac{4}{9}\) g = \(\frac{4}{9}\) x mg
= \(\frac{4}{9}\) W
= \(\frac{4}{9}\) x 63 =28 N

Question 16.
Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the Earth if it weighed 250 N on the surface?
Solution:
Weight of a body of mass m at the Earth’s surface, W = mg = 250 N
Body of mass m is located at depth, d = \(\frac{1}{2}\) R_e
where, R_e = Radius of the Earth
Acceleration due to gravity at depth d is given by the relation
g’ = \(\left(1-\frac{d}{R_{e}}\right) g=\left(1-\frac{R_{e}}{2 \times R_{e}}\right) g=\frac{1}{2} g\)
Weight of the body at depth d,
W’ = mg’
= m x \(\frac{1}{2} g=\frac{1}{2}\) mg = \(\frac{1}{2}\) W
= \(\frac{1}{2}\) x 250 = 125 N

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the Earth’s surface. How far from the Earth does the rocket ‘ go before returning to the earth? Mass of the Earth = 6.0 x 10²⁴ kg; mean radius of the Earth = 6.4 x 10⁶ m;
G = 6.67 x 10 ^-11 N-m² kg^-2.
Solution:
Velocity of the rocket, ν = 5 km/s = 5 x 10³ m/s
Mass of the Earth, M_e = 6.0 x 10²⁴ kg
Radius of the Earth, R_e = 6.4 x 10⁶ m
Height reached by rocket mass, m = h

At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= \(\frac{1}{2} m v^{2}+\left(\frac{-G M_{e} m}{R_{e}}\right)\)
At highest point h,
ν = 0
and, Potential energy = – \(\frac{G M_{e} m}{R_{e}+h}\)
Total energy of the rocket = 0+ \(\left(-\frac{G M_{e} m}{R_{e}+h}\right)=-\frac{G M_{e} m}{R_{e}+h}\)
From the law of conservation of energy, we have
Total energy of the rocket at the Earth’s surface=Total energy at height h


Height achieved by the rocket with respect to the centre of the Earth = R_e +h
= 6.4 x 10⁶ +1.6 x 10⁶ = 8.0 x 10⁶ m

Question 18.
The escape speed of a projectile on the Earth’s surface is 11.2 km s^-1. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.
Solution:
Escape velocity of a projectile from the Earth, υ_esc = 11.2 km/s
Projection velocity of the projectile, v_p = 3 v_esc
Mass of the projectile = m
Velocity of the projectile far away from the Earth = v_f
Total energy of the projectile on the Earth = \(\frac{1}{2} m v_{p}^{2}-\frac{1}{2} m v_{\mathrm{esc}}^{2}\)
The gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth = \(\frac{1}{2}\) mv²_f
From the law of conservation of energy, we have

Question 19.
A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out Of the Earth’s gravitational influence? Mass of the satellite = 200 kg mass of the Earth = 6.0 x 10²⁴ kg; radius of the Earth=6.4x 10⁶ m;G=6.67 x 10^-11 N-m² kg^-2.
Solution:
Mass of the Earth, M_e =6.0 x 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 x 10⁶ m
Universal gravitational constant, G = 6.67 x 10^-11 N-m²kg²
Heightofthesatellite,h =400 km=4 x 10⁵ m=0.4 x 10⁶ m

Total energy of the satellite at height h = \(\frac{1}{2} m v^{2}+\left(\frac{-G M_{e} m}{R_{e}+h}\right)\)
Orbital velocity of the satellite, ν = \(\sqrt{\frac{G M_{e}}{R_{e}+h}}\)
Total energy of height, h = \(\frac{1}{2} m\left(\frac{G M_{e}}{R_{e}+h}\right)-\frac{G M_{e} m}{R_{e}+h}=-\frac{1}{2}\left(\frac{G M_{e} m}{R_{e}+h}\right) \)

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.
Energy required to send the satellite out of its orbit = – (Bound energy)

Question 20.
Two stars each of one solar mass (= 2x 10³⁰ kg) are approaching each other for a head-on collision. When they are a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Mass of each star, M = 2 x 10³⁰ kg
Radius of each star, R = 10⁴ km = 10⁷ m
Distance between the stars, r = 10⁹ km = 10¹² m
For negligible speeds, ν = 0 total energy of two stars separated at distance r
= \(\frac{-G M M}{r}+\frac{1}{2} m v^{2}\)
= \(\frac{-G M M}{r}+0\) = 0 ………………………… (i)

Now, consider the case when the stars are about to collide:
The velocity of the stars = ν
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = \(\frac{1}{2} M v^{2}+\frac{1}{2} M v^{2}\) = Mv²
Total potential energy of both stars = \(\frac{-G M M}{2 R}\)
Total energy of the two stars = Mv² – \(\frac{-G M M}{2 R}\) ………………………. (ii)
Using the law of conservation of energy, we can write

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Solution:
The situation is represented in the following figure :

Mass of each sphere, M = 100 kg
Separation between the spheres, r = 1 m
X is the mid-point between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions. Gravitational potential at point X

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

Additional Exercises

Question 22.
As you have learnt in the text, a geostationary satellite orbits the Earth at a height of nearly 36,000 km from the surface of the Earth. What is the potential due to Earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the Earth = 6.0 x 10²⁴ kg, radius = 6400 km.
Solution:
Mass of the Earth, M = 6.0 x 10²⁴ kg
Radius of the Earth, R = 6400 km = 6.4 x 10⁶ m
Height of a geostationary satellite from the surface of the Earth,
h = 36000 km = 3.6 x 10⁷m
Gravitational potential energy due to Earth’s gravity at height h,

Question 23.
A star 2.5 times the mass of the Sun and collapsed to a size of 12 km rotates with a speed of 1.2 revs. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the Sun = 2 x 10³⁰ kg).
Solution:
Yes, A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.
Gravitational force, fg = \(\frac{G M m}{R^{2}}\)
where,M =Mass of the star=2.5 x 2 x 10³⁰ = 5 x 10³⁰ kg
m = Mass of the body

R=Radiusofthestar=12 km=1.2 x 10⁴ m
∴ f_g = \(\frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{\left(1.2 \times 10^{4}\right)^{2}}\)
= 2.31 x 10¹² mN
Centrifugal force, f_c = mrω²

where, ω = Angular speed = 2 πv
ν = Angular frequency = 1.2 rev s^-1
f _c=mR(2πv)²
= m x (1.2 x 10⁴) x 4 x(3.14)² x (1.2)² = 1.7 x 10⁵ mN
Since f_g > f_c, the body will remain stuck to the surface of the star.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the spaceship = 1000 kg; mass of the Sun = 2 x 10³⁰ kg; mass of mars = 6.4 x 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 x 10⁸ km;
G = 6.67 x 10^-11 N-m²kg^-2.
Solution:
Given, mass of the Sun, M = 2 x 10³⁰ kg

Mass of the mars, m = 6.4 x 10²³ kg
Mass of spaceship, Δm = 1000 kg
Radius of orbit of the mars, r₀ = 2.28 x 10¹¹
Radius of the mars, r = 3.395 x 10⁶ m
If v is the orbital velocity of mars, then

Since the velocity of the spaceship is the same as that of the mars, Kinetic energy of the spaceship,


Total potential energy of the spaceship,
U = Potential energy of the spaceship due to its being in the gravitational
field of the mars + potential energy of the spaceship due to its being in the gravitational field of the Sun.

Total energy of the spaceship E=K +U = 2.925 x 10¹¹ J – 5.977 x 10¹¹J
= – 3.025 x 10¹¹ J = -3.1 x 10¹¹ J
Negative energy denotes that the spaceship is bound to the solar system. Thus, the energy needed by the spaceship to escape from the solar system = 3.1 x 10¹¹ J.

Question 25.
A rocket is fired ‘vertically’ from file surface of mars with a speed of 2 kms^-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it?
Mass of mars = 6.4x 10²³ kg; radius of mars = 3395 km;
G = 6.67 x 10^-11 N-m² kg².
Solution:
Initial velocity of the rocket, ν = 2 km/s = 2 x 10³ m/s
Mass of Mars, M = 6.4 x 10²³ kg .
Radius of Mars, R = 3395 km = 3.395 x 10⁶ m
Universal gravitational constant, G = 6.67 x 10^-11 N-m² kg^-2
Mass of the rocket = m
Initial kinetic energy of the rocket = \(\frac{1}{2} m v^{2}\)
Initial potential energy of the rocket = \(\frac{-G M m}{R}\)
Total initial energy = \(\frac{1}{2} m v^{2}-\frac{G M m}{R}\)

If 20 % of initial kinetic energy is lost due to martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.
Total initial energy available = \(\frac{80}{100} \times \frac{1}{2} m v^{2}-\frac{G M m}{R}=0.4 m v^{2}-\frac{G M m}{R} \)
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h = \(-\frac{G M m}{(R+h)}\)
Applying the law of conservation of energy for the rocket, we can write

= 495 x 10³ m = 495 km

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 1 The Living World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 1 The Living World

PSEB 11th Class Biology Guide The Living World Textbook Questions and Answers

Question 1.
Why are living organisms classified?
Answer:
Living organisms are classified because:

• there are millions of organisms on the earth, which need a proper system of classification for identification.
• a number of new organisms are discovered each year. They require a particular system to be identified and to find out their correct position in a group.

Question 2.
Why are the classification systems changing every now and then?
Answer:
Evolution is the major factor responsible for the change in classification systems. Since, evolution still continues, so many different species of plants and animals are added in the already existed biodiversity. These newly discovered plant and animal specimens are then identified, classified and named according to the already existing classification systems. Due to evolution, animal and plant species keep on changing, so necessary changes in the already existed classification systems are necessary to place every newly discovered plant and animal in their respective ranks.

Question 3.
What different criteria would you choose to classify people that you meet often?
Answer:
The different scientific criteria to classify people that we meet often would be :
(i) Nomenclature: It is the science of providing distinct and proper names to the organisms. It is the determination of correct name as per established universal practices and rules.

(ii) Classification: It is the arrangement of organisms into categories based on systematic planning. In classification various categories used are class, order, family, genus and species.

(iii) Identification: It is the determination of correct name and place of an organism. Identification is used to tell that a particular species is similar to other organism of known identity. This includes assigning an organism to a particular taxonomic group.
The same criteria can be applied to the people we meet daily. We can identify them will their names classify them according to their living areas, profession, etc.

Question 4.
What do we learn from identification of individuals and populations?
Answer:
Identification of individuals and population categorized it into a species. Each species has unique characteristic features. On the basis of these features, it can be distinguished from other closely related species, e. g.,

Question 5.
Given below is the scientific name of Mango. Identify the correctly written name.
(i) Mangifera Indica
(ii) Mangifera indica.
Answer:
(ii) Mangifera indica (the name of species can never begins with a capital letter).

Question 6.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
Taxon is a grouping of organisms of any level in hierarchical classification, which is based on some common characteristics, e.g., insects represent a class of phylum – Arthropoda. All the insects possess common characters of three pairs of jointed legs. The term ‘taxon’ was introduced by ICBN in 1956. Examples of taxa are kingdom, phylum or division, class, order, family, genus and species. These taxa form taxonomic hierarchy, e.g., taxa for human :
Phylum – Chordata
Class – Mammalia
Order – Primata
Family – Hominidae
Genus – Homo
Species – sapiens

Question 7.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
The correct sequence of taxonomical categories is as follows :
Species → Genus → Order → Phylum

Question 8.
Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meanings of species in case of higher plants and animals on one hand, and bacteria on the other hand.
Answer:
A group of individual organisms with fundamental similarities is called species. It can be distinguished from other closely related species on the basis of distinct morphological differences.
In case of higher plants and animals, one genus may have one or more than one species, e.g.,Panthera leo (lion) and Panthera tigris (tiger).
In this example, Panthera is genus, which includes leo (lion) and tigris (tiger) as species.
In case of bacteria, different categories are present on the basis of shape. These are spherical, coccus, rod-shaped, comma and spiral-shaped. Thus, meaning of species in case of higher organisms and bacteria are different.

Question 9.
Define and understand the following terms:
(i) Phylum,
(ii) Class,
(iii) Family,
(iv) Order,
(v) Genus.
Answer:
(i) Phylum: Phylum comes next to Kingdom in the taxonomical hierarchy. All broad characteristics of an animal or plant are defined in a phylum. For example all chordates have a notochord and gill at some stage of life cycle. Similarly all arthropods have jointed legs made of chitin.

(ii) Class: The category class includes related orders. It is higher than order and lower than phylum. For example, class – Mammalia has order – Carnivora, Primata, etc.

(iii) Family: It is the category higher than genus and lower than order, which has one or more related genera having some common features. For example, Felidae, Canidae, etc.

(iv) Order: Order further zeroes down on characteristics and includes related genus. For example humans and monkeys belong to the order primates. Both humans and monkeys can use their hands to manipulate objects and can walk on their hind legs.

(v) Genus: It comprises a group of related species which has more characters in common in comparison to species of other genera. We can say that genera are aggregates of closely related species. For example, j potato, tomato and brinjal are three different species but all belong to the genus Solatium. Lion (Panthera leo), leopard (P. pardus) and tiger (P. tigris) with several common features, are all species of the genus Panthera. This genus differs from another genus Felis which includes cats.

Question 10.
How is a key helpful in the identification and classification of an
organism?
Answer:
Key is a device (scheme) of diagnostic alternate (contrasting) characters, which provide an easy means for the identification of unknown organism. The keys are taxonomic literature based on the contrasting characters generally a pair called couplet. Each statement in the key is called a lead. Separate taxonomic keys are required for each taxonomic category such as family, genus and species for identification purposes. Being analytical ‘ in nature, two types of keys are commonly used-indented key and bracketed key.

(i) Indented key provides sequence of choice between two or more statements of characters of species. The user has to make correct choise for identification.

(ii) Bracketed key (1) are used for contrasting characters like indented key but they are not repeated by intervening sub-dividing character and each character is given a number in brackets.

Question 11.
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Answer:

Taxonomical hierarchy is the system of arrangement of taxonomic categories in a descending order depending upon their relative dimensions. It was introduced by Linnaeus (1751) and is therefore, also called Linnaeus hierarchy. Each category, referred to as a unit of classification commonly called as taxon (pi. taxa), e.g., taxonomic categories and hierarchy can be illustrated by a group of organisms, i.e., insects. The common features of insects is ‘three pair of jointed legs’. It means insects are recognisable objects which can be classified, so given a rank or category.
Category further denotes a rank. Each rank or taxon, represents a unit of classification taxonomic studies of all plants and animals led to the development of common categories such as kingdom, phylum or division (for plants), class, order, family, genus and species. All organisms, including those in the plant and animal, kingdoms have ‘species’ as the lowest category.

To place an organism in various categories is to have the knowledge of characters of an individual or group of organism. This helps to identify similarities and dissimilarities among the individual of the same kind of organisms as well as of other kinds of organism. Some organisms with their taxonomical categories are given in following table: