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Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 5 Oscillations Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 5 Oscillations

Question 1.
Define :
(1) periodic motion
(2) oscillatory motion. Give two examples.
Answer:
(1) Periodic motion : A motion that repeats itself at definite intervals of time is said to be a periodic motion.
Examples : The motion of the hands of a clock, the motion of the Earth around the Sun.

(2) Oscillatory motion : A periodic motion in which a body moves back and forth over the same path, straight or curved, between alternate extremes is said to be an oscillatory motion.
Examples : The motion of a taut string when plucked, the vibrations of the atoms in a molecule, the oscillations of a simple pendulum.
[Note : The oscillatory motion of a particle is also called a harmonic motion when its position, velocity and acceleration can be expressed in terms of a periodic, sinusoidal functions-sine or cosine, of time.

Question 2.
With a neat diagram, describe a spring-and-block oscillator.
Answer:
Consider a spring-and-block oscillator as shown in below figure in which the block slides on a frictionless horizontal surface. The spring has a relaxed length when the block is at rest at the position O.

The block is then displaced to P by an amount x measured from the equilibrium position O. Upon releasing, the unbalanced force \(\vec{F}\) = –\(k \vec{x}\) toward left accelerates the block and its speed increases. As x gets smaller, |\(\vec{F}\)| and the acceleration decrease proportionately.

k is the elastic constant of the spring called the force constant or spring constant.

At the instant the block passes through the point O, | \(\vec{F}\) | = 0 because x = 0; although there is no acceleration, the speed is maximum.

As soon as the block passes O going to the left, the force on the block and its acceleration increases to the right, because the spring is now compressed. Eventually, the block is brought to rest momentarily at the point Question Then on, the subsequent motion is the same as the motion from P to Q, with all directions reversed.

The acceleration of the block is \(\vec{a}\) = \(\frac{\vec{F}}{m}\) = –\(\frac{k}{m} \vec{x}\) where m is the mass of the block. This shows that the acceleration is also proportional to the displacement and its direction is opposite to that of the displacement, i.e., the force and acceleration are both directed towards the mean or equilibrium position. The motion repeats causing the block to oscillate about equilibrium or mean position O. This oscillatory motion along a straight path is called linear simple harmonic motion (SHM).

The points P and Q are called the extreme positions or the turning points of the motion. One oscillation is a complete to-and-fro motion of the oscillating body (block, in this case) along its path (the motion from O to P, P to Q and Q to O), i.e., two consecutive passages of the body through the point O in the same direction.

Question 3.
In linear SHM, what can you say about the restoring force when the speed of the particle is

1. zero
2. maximum ?

Answer:
The restoring force is

1. maximum
2. zero.

Question 4.
Define period or periodic time, frequency, amplitude and path length of simple harmonic motion (SHM).
Answer:

1. Period or periodic time of SHM : The time taken by a particle performing simple harmonic motion to complete one oscillation is called the period or periodic time of SHM.
2. Frequency of SHM : The number of oscillations performed per unit time by a particle executing SHM is called the frequency of SHM.
3. Amplitude of SHM : The magnitude of the maximum displacement of a particle performing SHM from its mean position is called the amplitude of SHM.
4. Path length of SHM : The length of the path over which a particle performs SHM is twice the amplitude of the motion and is called the path length or range of the SHM.
   [Note : The frequency of SHM is equal to the reciprocal of the period of SHM.]

Question 5.
Obtain the differential equation of linear simple harmonic motion.
Answer:
When a particle performs linear SHM, the force acting on the particle is always directed towards the mean position. The magnitude of the force is directly proportional to the magnitude of the displacement of the particle from the mean position. Thus, if \(\vec{F}\) is the force acting on the particle when its displacement from the mean position is \(\vec{x}\), \(\vec{F}\) = -k\(\vec{x}\) … (1)
where the constant k, the force per unit displacement, is called the force constant. The minus sign indicates that the force and the displacement are oppositely directed.
The velocity of the particle is \(\frac{d \vec{x}}{d t}\) and its acceleration is \(\frac{d^{2} \vec{x}}{d t^{2}}\).
Let m be the mass of the particle.
Force = mass × acceleration
∴ \(\vec{F}\) = m\(\frac{d^{2} \vec{x}}{d t^{2}}\)
Hence, from Eq. (1),
m\(\frac{d^{2} \vec{x}}{d t^{2}}\) = -k\(\vec{x}\)
∴ \(\frac{d^{2} \vec{x}}{d t^{2}}\) + \(\frac{k}{m} \vec{x}\) = 0 … (2)
This is the differential equation of linear SHM.

Question 6.
Obtain the dimensions of force constant in SHM.
Answer:

[Note : The SI unit of force constant is the newton per metre (N/m) while the cgs unit is the dyne per centimetre (dyn/cm).]

Question 7.
State the differential equation of linear SHM. Hence, obtain the expressions for the acceleration, velocity and displacement of a particle performing linear SHM.
Answer:
The differential equation of linear SHM is
\(\frac{d^{2} \vec{x}}{d t^{2}}\) + \(\frac{k}{m} \vec{x}\) = 0
where m = mass of the particle performing SHM, \(\frac{d^{2} \vec{x}}{d t^{2}}\) = acceleration of the particle when its displacement from the mean position is \(\vec{x}\) and k = force constant. For linear motion, we can write the differential equation in scalar form :
\(\frac{d^{2} x}{d t^{2}}\) + \(\frac{k}{m}\)x = 0
Let \(\frac{k}{m}\) = ω², a constant
∴ \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0
∴ Acceleration, a = \(\frac{d^{2} x}{d t^{2}}\) = ω²
The minus sign shows that the acceleration and the displacement have opposite directions. Writing v = \(\frac{d x}{d t}\) as the velocity of the particle.

Hence, EQ. (1) can be written as
v\(\frac{d v}{d x}\) = -ω²x dx
∴ vdv = -ω²x dx
Integrating this expression, we get,
\(\frac{v^{2}}{2}\) = –\(\frac{-\omega^{2} x^{2}}{2}\) + C
where the constant of integration C is found from a boundary condition.

At an extreme position (a turning point of the motion), the velocity of the particle is zero. Thus, v = 0 when x = ± A, where A is the amplitude.

This equation gives the velocity of the particle in terms of the displacement, x. The velocity towards right is taken to be positive and that towards left as negative.
Since, v = dx/dt, we can write Eq. (2) as follows :

where the constant of integration, α, is found from the initial conditions, i.e., the displacement and the velocity of the particle at time t = 0.
From Eq. (3), we have
\(\frac{x}{A}\) = sin (ωt + α)
∴ Displacement as a function of time is,
x = A sin (ωt + α)

Question 8.
From the definition of linear SHM, derive an expression for the angular frequency of a body performing linear SHM.
Answer:
When a body of mass m performs linear SHM, the restoring force on it is always directed towards the mean position and its magnitude is directly proportional to the magnitude of the displacement of the body from the mean position. Thus, if \(\vec{F}\) is the force acting on the body when its displacement from the mean position is \(\vec{x}\),
\(\vec{F}\) = m\(\) = – k\(\vec{x}\)
where the constant k, the force per unit displacement, is the force constant.
Let \(\frac{k}{m}\) = ω², a constant. m
∴ Acceleration, a = –\(\frac{k}{m}\)x = -ω²x
∴ The angular frequency
ω = \(\sqrt{\frac{k}{m}}\) = \(\sqrt{\left|\frac{a}{x}\right|}\)
= \(\sqrt{\text { acceleration per unit displacement }}\)

Question 9.
What is the displacement of a particle at any position, performing linear SHM ?
Answer:
The displacement of a particle performing linear SHM is a specified distance of the particle from the mean position in a specified direction along its path. The general expression for the displacement is x = A sin (ωt + α), where A and ω are respectively the amplitude or maximum displacement and the angular frequency of the motion, and α is the initial phase.

Question 10.
Assuming the general expression for displacement of a particle in SHM, obtain the expressions for the displacement when the particle starts from
(i) the mean position
(ii) an extreme position.
Answer:
The general expression for the displacement of a particle in SHM at time t is x = A sin (ωt + α) … (1) where A is the amplitude and re is a constant in a particular case.
∴ ωt + α = sin^-1\(\frac{x}{A}\) …. (2)
(i) When the particle starts from the mean position, x = 0 at t = 0. Then, from Eq. (2),
α = sin^-1 0 = 0 or π … (3)
Substituting for α into Eq. (1),
x = A sin ωt for α = 0 and x = – A sin ωt for α = π
∴ x = ±A sin ωt … (4)
where the plus sign is taken if the particle’s initial velocity is to the right, while the minus sign is taken when the initial velocity is to the left.

(ii) x = ±A at t = 0 when the particle starts from the right or left extreme position, respectively. Then, from Eq. (2),

where the plus sign is taken when the particle starts from the positive extreme, while the minus sign is taken when the particle starts from the negative extreme.

Question 11.
At what position is the acceleration of a particle in SHM maximum? What is its magnitude? At what position is the acceleration minimum ? What is its magnitude ?
Answer:
The magnitude of the acceleration of a particle performing SHM is
a = ω²x … (1)
where ω is a constant related to the system.
From Eq. (1), the acceleration has a maximum value a_max when displacement x is maximum, |x| = A, i.e., the particle is at the extreme positions.
∴ a_max = ω²A
Also from EQ. (1), the acceleration has a minimum value when x is minimum, x = 0, i.e., the particle is at the mean position.
∴ a_min = 0

Question 12.
At what position is the velocity of a particle in SHM maximum ? What is its magnitude ? At what position is the velocity minimum? What is its magnitude?
Answer:
The velocity of a particle in SHM is
v = ω\(\sqrt{A^{2}-x^{2}}\) … (1)
where ω is a constant related to the system and A is the amplitude of SHM.
From EQ. (1) it is clear that the velocity is maximum when A² – x² is maximum, that is when displacement x = 0, i.e., the particle is at the mean position.
∴ v_max = ωA
Also from Eq. (1), the velocity is minimum when A² – x² is minimum, equal to zero. This occurs when x is maximum, x = ± A, i.e., the particle is at the extreme positions.
∴ v_min = 0

Question 13.
For a particle performing linear SHM, show that its average speed over one oscillation is \(\frac{2 \omega A}{\pi}\), where A is the amplitude of SHM.
OR
Show that the average speed of a particle performing SHM in one oscillation is \(\frac{2}{\pi}\) × maximum speed.
Answer:
During one oscillation, a particle performing SHM covers a total distance equal to 4A, where A is the amplitude of SHM. The time taken to cover this distance is the period (T) of SHM.

Question 14.
A body of mass 200 g performs linear SHM with period 2πs. What is the force constant ?
Answer:
Force constant, k = mω² = m\(\left(\frac{2 \pi}{T}\right)^{2}\)
= 0.2kg × \(\left(\frac{2 \pi}{2 \pi \mathrm{s}}\right)^{2}\) = 0.2N/m.

Question 15.
Derive expressions for the period of SHM in terms of
(1) angular frequency
(2) force constant
(3) acceleration.
Answer:
The general expression for the displacement (x) of a particle performing SHM is x = A sin (ωt + α)
(1) Let T be the period of the SHM and x₁ the displacement after a further time interval T. Then
x₁ = A sin [ω(t + T) + α]
= A sin (ωt + ωT + α)
= A sin (ωt + α + ωT)
Since T ≠ 0, for x₁ to be equal to x, we must have (ωT)_min = 2π.
Hence, the period (T) of SHM is T = 2π/ω
This is the expression for the period in terms of the constant co, the angular frequency.

(2) If m is the mass of the particle and k is the force constant, ω = \(\sqrt{k / m}\).
∴T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi}{\sqrt{k / m}}\) = 2π \(\sqrt{\frac{m}{k}}\)

(3) The acceleration of a particle performing SHM has a magnitude a = ω²x
∴ ω = \(\sqrt{a / x}\)

Question 16.
A small uniform cylinder floats upright to a depth d in a liquid. If it is depressed slightly and released, find its period of oscillations.
Answer:
Consider a cylinder, of length L, area of cross section A and density ρ, floating in a liquid of density σ. If the cylinder floats up to depth d in the liquid, then by the law of floatation, the weight of the cylinder equals the weight of the liquid displaced, i.e.,
ALρg = Adσg
∴ L = dσ/p … (1)
Let the cylinder be pushed down by a distance y. Then, the weight of the liquid displaced by the cylinder of length y will exert a net upward force on the cylinder :
F = Ayσg,
which produces an acceleration,

Question 17.
How does the frequency of an SHM vary with

1. the force constant k
2. the mass of the particle performing SHM ?

Answer:
The frequency of a particle of mass m performing
SHM is f = \(\frac{1}{T}\) = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\).

1. ∴ f ∝\(\sqrt{k}\)
   Thus, the frequency of an SHM is directly proportional to the square root of the force constant of the motion.
2. ∴ f ∝ \(\frac{1}{\sqrt{m}}\)
   Thus, the frequency of an SHM is inversely proportional to the square root of the mass of the particle performing SHM.

Question 18.
In linear SHM, at what position of the particle is the acceleration of the particle half the maximum acceleration?
Answer:
In linear SHM, | a | ∝ | x | ∴ a = \(\frac{a_{\max }}{2}\) when | x | = \(\frac{A}{2}\), where A is the amplitude of SHM.

Question 19.
If the displacement of a particle in SHM is given by x = 0.1 sin (6πrt) metre, what is the frequency of motion ?
Answer:
Comparison of the given equation with
x = A sin (2πft) gives 2πf = 6π rad/s.
∴ Frequency of motion,/= 3 Hz

Question 20.
If the displacement of a particle in SHM is given by x = 0.1 cos (100t) metre, what is the maximum speed of the particle ?
Answer:
Comparison of the given equation with
x = A cos (ωt) gives A = 0.1 m and ω = 100 rad/s.
∴ Maximum speed of the particle = ωA
= 1000 × 0.1 = 10 m/s

Question 21.
A body of mass m tied to a spring performs SHM with period 2 seconds. If the mass is increased by 3m, what will be the period of SHM ?
Answer:

∴ T₂ = 2T₁ = 2 × 2 = 4 seconds gives the required period of SHM.

Question 22.
A particle executing SHM has velocities v₁ and v₂ when at distances x₁ and x₂ respectively from the mean position. Show that its period is T = 2π\(\sqrt{\frac{x_{1}^{2}-x_{2}^{2}}{v_{2}^{2}-v_{1}^{2}}}\) and the amplitude of SHM is A = \(\sqrt{\frac{v_{2}^{2} x_{1}^{2}-v_{1}^{2} x_{2}^{2}}{v_{2}^{2}-v_{1}^{2}}}\)
Answer:
If A is the amplitude and co is the angular frequency, V₁ = ω\(\sqrt{A^{2}-x_{1}^{2}}\) … (1)
and v₂ = ω\(\sqrt{A^{2}-x_{2}^{2}}\) … (2)

Question 23.
Explain
(i) a series combination
(ii) a parallel combination of springs. Obtain the spring constant in each case.
Answer:
(i) Series combination of springs : When two light springs obeying Hooke’s law are connected as shown in below figure and both the springs experience the same force applied to the free end of the combination, they are said to be connected in series.

Consider two springs, 1 and 2, with respective spring constants k₁ and k₂ connected in series and supporting a load F = mg so that the springs are extended. Since the same force acts on each spring, by Hooke’s law,
F = k₁x₁ (for spring 1) and F = k₂x₂ (for spring 2) The system of two springs in series is equivalent to a single spring, of spring constant k_S such that F = k_Sx, where the total extension x of the combination is the sum x₁ + x₂ of their elongations.
x = x₁ + x₂
∴ \(\frac{F}{k_{\mathrm{S}}}\) = \(\frac{F}{k_{1}}\) + \(\frac{F}{k_{2}}\) ∴ \(\frac{1}{k_{\mathrm{S}}}\) = \(\frac{1}{k_{1}}\) + \(\frac{1}{k_{2}}\)
For a series combination of N such springs, of spring constants, k₁, k₂, k₃, … k_N

(ii) Parallel combination of springs : When two light springs obeying Hooke’s law are connected via a thin vertical rod as shown, they are said to be connected in parallel. If a constant force \(\vec{F}\) is exerted on the rod such that the rod remains perpendicular to the direction of the force, the springs undergo the same extension.

Consider two springs, 1 and 2, with respective spring constants k₁ and k₂ connected in series and supporting a load F = mg so that the springs are extended. The two springs stretch by the same amount x but share the load.
F = F₁ + F₂
The system of two springs in parallel is equivalent to a single spring, of spring constant kF such that F = k_PX,
∴ k_Px = k₁x + k₂x ∴ k_P = k₁ + k₂
For a parallel combination of N such springs, of spring constants k₁, k₂, k₃, … k_N
k_P = k₁ + k₂ + k₃ + … + k_N = \(\Sigma_{i=1}^{N} k_{i}\)
Therefore, for a parallel combination of N identical light springs, each of spring constant k, k_P = Nk

Question 24.
Solve the following :

Question 1.
A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate
(i) the angular frequency
(ii) the frequency of oscillation.
Solution :
Data : m = 1 kg, k = 16 N/m

Question 2.
Calculate the time taken by a body performing SHM of period 2 seconds to cover half the amplitude starting from an extreme position.
Solution :
Data : T = 2 s, x₀ = + A (initially at positive extremity), x = \(\frac{A}{2}\)

∴ Starting from the positive extremity, the particle takes \(\frac{1}{3}\) s to cover a distance equal to half the amplitude.

Question 3.
A 3 kg block, attached to a spring, performs linear SHM with the displacement given by x = 2 cos (50t) m. Find the spring constant of the spring.
Solution :
Data : m = 3 kg, x = 2 cos (50t) m
Comparing the given equation with x = A cos ωt,
ω = 50 rad/s
ω² = k/m
∴ The spring constant,
k = mω² = (3)(50)²
= 3 × 2500 = 7500 N/m

Question 4.
A body oscillates in SHM according to the equation x = 5 cos (2πt + \(\frac{\pi}{4}\)), where x and t are
in SI units. Calculate the
(i) displacement and
(ii) speed of the body at t = 1.5 s.
Solution:
Data: x = cos \(\left(2 \pi t+\frac{\pi}{4}\right)\), t = 1.5 s
(i) The displacement at t = 1.5 s is

= 5(1.414)(3.142) = 22.21 m/s

Question 5.
The equation of motion of a particle executing SHM is x = a sin \(\left(\frac{\pi}{6} t\right)\) + b cos \(\left(\frac{\pi}{6} t\right)\), where a = 3 cm and b = 4 cm. Express this equation in the form x = A sin \(\left(\frac{\pi}{6} t+\phi\right)\). Hence, find A and φ.
Solution:
Let a = A cos φ and b = A sin φ, so that

Question 6.
A particle performs SHM of amplitude 10 cm. Its maximum velocity during oscillations is 100 cm/s. What is its displacement, when the velocity is 60 cm/s?
Solution :
Data : A = 10 cm, v_max = 100 cm/s, v = 60 cm/s
v_max = ωA = 100 cm/s

Question 7.
A body of mass M attached to a spring oscillates with a period of 2 seconds. If the mass is increased by 2 kg, the period increases by 1 second. Find the initial mass, assuming that Hooke’s law is obeyed.
Answer:
Data : m₁ = M, T₁ = 2 s, m₂ = M + 2 kg, T₂ = 2s + 1 s = 3s

Question 8.
A load of 100 g increases the length of a light spring by 10 cm. Find the period of its linear SHM if it is allowed to oscillate freely in the vertical direction. What will be the period if the load is increased to 400 g? [g = 9.8 m/s²]
Solution :
Data :m = 100 g = 100 × 10^-3 kg, x = 10 cm = 0.1 m g = 9.8 m/s², m₁ = 400 g = 400 × 10^-3 kg
(1) Stretching force F = mg
Now F = kx (numerically), where k is the force constant.
∴ mg = kx

Question 9.
A particle in SHM has a period of 2 seconds and an amplitude of 10 cm. Calculate its acceleration when it is at 4 cm from its positive extreme position.
Solution :
Data : T = 2s, A = 10 cm, A – x = 4 cm
∴ x = 10 cm – 4 cm = 6 cm

Question 10.
A particle executes SHM with amplitude 5 cm and period 2 s. Find the speed of the particle at a point where its acceleration is half the maximum acceleration.
Solution :
Data: A = 5 cm = 5 × 10^-2 m, T = 2s,

Question 11.
The periodic time of a linear harmonic oscillator is 2π seconds, with maximum displacement of 1 cm. If the particle starts from an extreme position, find the displacement of the particle after π/3 seconds.
Solution :
Data : T = 2π s, A = 1 cm, t = π/3
ω = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{2 \pi}\) = 1 rad/s
x = A cos ωt (∵ particle starts from extreme position)
= (1) cos\(\left(1 \times \frac{\pi}{3}\right)\) = cos \(\left(1 \times \frac{\pi}{3}\right)\) = cos \(\frac{\pi}{3}\) = \(\frac{1}{2}\) cm

Question 12.
A particle performs SHM of period 12 seconds and amplitude 8 cm. If initially the particle is at the positive extremity, how much time will it take to cover a distance of 6 cm from that position?
Solution :
Data : T = 12 s, A = 8 cm
∴ ω = 2π/T = π/6 rad/s
When the particle covers a distance of 6 cm from the positive extremity, its displacement measured from the mean position is x = 8 – 6 = 2 cm.
As the particle starts from the positive extreme position, its displacement is

Question 13.
A particle executes SHM with amplitude 10 cm and period 10 s. Find the velocity and acceleration of the particle at a distance of 5 cm from the equilibrium position.
Solution :
Data : A = 10 cm = 0.1 m, T = 10 s, x = ± 5 cm = ± 0.05 m

Question 14.
A body performs SHM on a path 0.12 m long. Its velocity at the centre of the path is 0.12 m/s. Find the period of SHM. Also find the magnitude of the velocity of the body at \(\sqrt{3}\) × 10^-2 m from the centre of the path.
Solution :
The path length of the SHM is the range 2 A, and the velocity at the centre of the path, i.e., at the equilibrium position, is the maximum velocity v_max.
Data : 2A = 0.12 m, v_max = 0.12 m/s,
x = ± \(\sqrt{3}\) × 10^-2 m

Question 15.
A particle of mass 2 g executes SHM with a period of 12 s and amplitude 10 cm. Find the acceleration of the particle and the restoring force on the particle when it is 2 cm from its mean position. Also find the maximum velocity of the particle.
Solution :
Data : m = 2g = 2 × 10^-3 kg, T = 12 s,
A = 10 cm = 0.1 m, x = ±2 cm = ±2 × 10^-2 m

The acceleration of the particle, a = ω² = (0.5237)² (± 2 × 10^-2)
= ± 0.2743 × 2 × 10^-2 = ± 5.486 × 10^-3 m/s²
The restoring force on the particle at that position, F = ma = ± (2 × 10^-2) (5.486 × 10^-3)
= ±1.097 × 10^-5 N
The maximum velocity of the particle, v_max = ωA = 0.5237 × 0.1 5.237 × 10^-2 m/s

Question 16.
The maximum velocity of a particle performing linear SHM is 0.16 m/s. If its maximum acceleration is 0.64 m/s², calculate its period.
Solution :
Data : v_max = 0.16 m/s, a_max = 0.64 m/s²
v_max = ωA and a_max = ω²A

Question 17.
A particle performing linear SHM has maximum velocity of 25 cm/s and maximum acceleration of 100 cm/s². Find the amplitude and period of oscillation, [π = 3.142]
Solution :
Data : v_max = 25 cm/s, a_max = 100 cm/s²

Question 18.
A particle performing linear SHM has a period of 6.28 seconds and path length of 20 cm. What is the velocity when its displacement is 6 cm from the mean position?
Solution :
Data : T = 6.28 s, 2A = 20 cm ∴ A = 10 cm, x = 6 cm

Question 19.
A uniform wooden rod floats vertically in water with 14 cm of its length immersed in the water. If it is depressed slightly and released, find its period of oscillations.
Solution :
Data : d = 14 cm = 0.14 m

Question 20.
A particle performs UCM. The diameter of the circle is 4 cm. What is the amplitude of linear SHM that is the projection of the UCM on a diameter?
Answer:
Amplitude of linear SHM = radius of the circle = 2 cm.

Question 21.
A particle performs UCM with period 2n seconds along a circle of diameter 10 cm. What is the maximum speed of its shadow on a diameter of the circle ?
Answer:
Maximum speed, v_max = ωA = \(\frac{2 \pi}{T}\)A
= \(\frac{2 \pi}{2 \pi}\) × 5 × 10^-2 = 5 × 10^-2 m/s.

Question 22.
See Question 20 above. What is the maximum acceleration of the shadow ?
Answer:
Maximum acceleration, a_max = ω²A = \(\left(\frac{2 \pi}{T}\right)^{2}\) A
= \(\left(\frac{2 \pi}{2 \pi}\right)^{2}\) × 5 × 10^-2 = 5 × 10^-2 m/s².

Question 23.
What do you understand by the phase and epoch of an SHM ?
Answer:
(1) Phase of simple harmonic motion (SHM) represents the state of oscillation of the particle performing SHM, i.e., it gives the displacement of the particle, its direction of motion from its equilibrium position and the number of oscillations completed.

The displacement of a particle in SHM is given by x = A sin (ωt + α). The angle (ωt + α) is called the phase angle or simply the phase of SHM. The SI unit of phase angle is the radian (symbol, rad).

(2) Epoch of simple harmonic motion (SHM) represents the initial phase of the particle performing SHM, i.e., it gives the displacement of the particle and its direction of motion at time t = 0.

If x₀ is the initial position of the particle, i.e., the position at time t = 0, x₀ = A sin α or α = sin^-1 (x₀/A). The angle α, therefore, determines the initial state of the particle. Hence, the angle α is the epoch or initial phase or phase constant of SHM.
[Note : The symbol for the unit radian is rad, not superscripted c.]

Question 25.
Solve the following.

Question 1.
The differential equation for a particle performing linear SHM is \(\frac{d^{2} x}{d t^{2}}\) = – 4x. If the amplitude is 0.5 m and the initial phase is π/6 radian, obtain the expression for the displacement and find the velocity of the particle at x = 0.3 m.
Solution:
Data : A = 0.5 m, α = π/6 rad
(1) \(\frac{d^{2} x}{d t^{2}}\) = -4x
Comparing this equation with the general equation \(\frac{d^{2} x}{d t^{2}}\) = – ω²x, we get,
ω² = 4 or ω = 2 rad/s
Now, x = A sin (ωt + α)
Substituting the values of A, ω and α, the expression for the displacement for the given SHM is
x = 0.5 sin (2t + π/6) m

(2) The velocity of the particle at x = 0.3 m is v = ± ω \(\sqrt{A^{2}-x^{2}}\)
= ± 2 \(\sqrt{(0.5)^{2}-(0.3)^{2}}\) = ± 0.8 m/s

Question 2.
The displacement of a particle performing linear SHM is given by x = 6 sin (3πt + \(\frac{5 \pi}{6}\)) metre. Find
the amplitude, frequency and the phase constant of the motion.
Solution :
Data : x = 6 sin (3πt + \(\frac{5 \pi}{6}\)) metre
Comparing this equation with x = A sin (ωt + α), we get:

1. Amplitude, A = 6 m
2. ω = 3π rad / s
   ∴ Frequency, f = \(\frac{\omega}{2 \pi}\) = \(\frac{3 \pi}{2 \pi}\) = 1.5 Hz 5%
3. Phase constant, α = \(\frac{5 \pi}{6}\) rad

Question 3.
The equation of linear SHM is a: = 10 sin (4πt + \(\frac{1}{24}\)) cm. Find the amplitude, period and phase constant of the motion. Also, find the phase angle \(\frac{1}{24}\) second after the start.
Solution:
Data : x = 10 sin\(\left(4 \pi t+\frac{\pi}{6}\right)\) + cm, f = \(\frac{1}{24}\) s

(1) Comparing the given equation with x = A sin (ωt +α), we get,
A = 10 cm, ω = 4π rad/s, α = \(\frac{\pi}{6}\) rad
∴

1. Amplitude, A = 10 cm
2. Period, T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi}{4 \pi}\) = 0.5 s
3. Phase constant, α = \(\frac{\pi}{6}\) rad

(2) Phase angle = (ωt + α) = 4πt + \(\frac{\pi}{6}\)
The phase angle \(\frac{1}{24}\) second after the start is obtained by substituting t = \(\frac{1}{24}\) in the above expression.
∴ Phase angle = 4πt + \(\frac{\pi}{6}\) = (4π × \(\frac{1}{24}\)) + \(\frac{\pi}{6}\)
= \(\frac{\pi}{6}\) + \(\frac{\pi}{6}\) = \(\frac{\pi}{3}\) rad

Question 4.
Describe the state of oscillation of a particle if the phase angle of SHM is rad.
Solution :
Data : θ = \(\frac{25 \pi}{4}\) rad
θ = \(\frac{25 \pi}{4}\) = 6π + \(\frac{\pi}{4}\) = 3(2π) rad + \(\frac{\pi}{4}\) rad
The first term indicates that the particle has completed 3 oscillations. The second term indicates that the displacement of the particle in the 4th oscillation is A sin \(\frac{\pi}{4}\) = + \(\frac{1}{\sqrt{2}}\)A, where A is the amplitude of the SHM, and moving towards the positive extreme.

Question 5.
A particle in linear SHM is in its 5th oscillation. If its displacement at that instant is –\(\frac{1}{2}\) A and
is moving toward the mean position, determine its phase at that instant.
Solution :
Data : x = –\(\frac{1}{2}\) A, 5th oscillation
A sin θ₁ = –\(\frac{1}{2}\)A ∴ θ₁ = sin^-1\(\left(-\frac{1}{2}\right)\) = π – \(\frac{\pi}{6}\) rad
As the particle is in its 5th oscillation, its phase is
θ = 2 × 2π + θ₁ = 4π + (π – \(\frac{\pi}{6}\)) = 5π – \(\frac{\pi}{6}\) = \(\frac{29 \pi}{6}\) rad

Question 6.
The amplitude and periodic time of SHM are 5 cm and 6 s, respectively. What is the phase at a distance of 2.5 cm from the mean position?
Solution :
Data : A = 5 cm, T = 6 s, x = 2.5 cm
Since the particle starts from the mean position, its epoch, α = 0.
∴ The equation of motion is x = A sin ωt
∴ The required phase of the particle,
ω = sin^-1\(\frac{x}{A}\)
= sin^-1\(\frac{2.5}{5}\) = sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\) rad

Question 26.
State the expressions for the displacement, velocity and acceleration of a particle performing linear SHM, starting from the mean position towards the positive extreme position. Hence, draw their graphs with respect to time. Draw your conclusions from the graphs.
OR
Represents graphically the displacement, velocity and acceleration against time for a particle performing linear SHM when it starts from the mean position.
Answer:
Consider a particle performing SHM, with amplitude A and period T = 2π/ω starting from the mean position towards the positive extreme position where co is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant are

as the initial phase α = 0

Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.

Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.

Conclusions :

1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graph are sine curves. The v-t graph is a cosine curve.
2. There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
3. There is a phase difference of n radians between x and a.

Question 27.
State the expressions for the displacement, velocity and acceleration of a particle performing linear SHM, starting from the positive extreme position. Hence, draw their graphs with respect to time. Draw your conclusions from the graphs.
OR
A particle performs linear SHM starting from the positive extreme position. Plot the graphs of its displacement, velocity and acceleration against time.
Answer:
Consider a particle performing linear SHM with amplitude A and period T = 2π/ω, starting from the positive extreme position, where ω is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant (t) are
x = A cos ωt = A cos\(\left(\frac{2 \pi}{T} t\right)\) (∵ ω = \(\frac{2 \pi}{T}\))
v = – ωA sin ωt = -ωA sin \(\left(\frac{2 \pi}{T} t\right)\)
a = – ω²A sin ωt = -ω²A cos \(\left(\frac{2 \pi}{T} t\right)\)

Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.

Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.

Conclusions :

1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.
2. There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
3. There is a phase difference of π radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π/2 rad between a and v.

v is minimum (equal to zero) at the extreme positions (i.e., at x = ± A) and v is maximum ( = ± ωA) at the mean position (x = 0).

(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π rad between v and a. a is minimum (equal to zero) at the mean position (x = 0) and a is maximum ( = \(\mp\)ω²A) at the extreme positions (x = ± A).

Question 28.
Discuss analytically the composition of two SHMs of the same period and parallel to each other (along the same path). Find the resultant amplitude when the phase difference is
(1) zero
(2) \(\frac{\pi}{3}\) rad
(3) \(\frac{\pi}{2}\) rad
(4) π rad.
Answer:
Let a particle be subjected to two parallel linear SHMs of the same period along the same path and the same mean position, represented by
x₁ = A₁ sin (ωt + α) and x₂ = A₂ sin (ωt + β),
where A₁ and A₂ are the amplitudes, and α and β are the initial phases of the two SHMs.

According to the principle of superposition, the displacement of the particle at any instant t is the algebraic sum x = x₁ + x₂.
∴ x = A₁ sin (ωt + α) + A₂ sin (ωt + β)
= A₁ sin ωt cos α + A₁ cos ωt sin α + A₂ cos ωt sin β
= (A₁ cos α + A₂ cos β) sin ωt + (A₁ sin α + A₂sin β) cos ωt
Let A₁ cos α + A₂ cos β = R cos δ …. (1)
and A₁ sin α + A₂ sin β = R sin δ …. (2)
∴ x = R cos δ sin ωt + R sin δ cos ωt
∴ x = R cos(ωt + δ) ….. (3)

Equation (3), which gives the displacement of the particle, shows that the resultant motion is also simple harmonic, along the same path as the SHMs superposed, with the same mean position, and amplitude R and initial phase δ but having the same period as the individual SHMs.

Amplitude R of the resultant motion : The resultant amplitude R is found by squaring and adding Eqs. (1) and (2).

Initial phase S of the resultant motion : The initial phase of the resultant motion is found by dividing Eq. (2) by Eq. (1).

Notes :

1. Since the displacements due to the super-posed linear SHMs are along the same path, their vector sum can be replaced by the algebraic sum.
2. To determine δ uniquely, we need to know both sin δ and cos δ.

Question 29.
Solve the following :

Question 1.
Two parallel SHMs are given by x₁ = 20 sin (8πt) cm and x₂ = 10 sin (8πt + π/2) cm. Find the amplitude and the epoch of the resultant SHM.
Solution :
Data : x₁ = 20 sin (8πt) cm = A₁ sin (ωt + α), x₂ = 10 sin (8πt + π/2) cm = A₂ sin (ωt + β)
∴ A₁ = 20 cm, A₂ = 10 cm, α = 0, β = π/2
(1) Resultant amplitude,

(2) Initial phase of resultant SHM,

Question 2.
The displacement of a particle performing SHM is given by x = [5 sin πt + 12 sin \(\left(\pi t+\frac{\pi}{2}\right)\) cm. Determine the amplitude, period and initial phase of the motion.
Solution :
Data : x = [5 sin πt + 12 sin \(\left(\pi t+\frac{\pi}{2}\right)\)] cm
The given expression for displacement may be written as the superposition of two parallel SHMs of the same period as x = x₁ + x₂, where x₁ = 5 sin πt cm = A₁ sin (ωt + α) and
x₂ = 12 sin \(\left(\pi t+\frac{\pi}{2}\right)\) cm = A₂ sin (ωt + β)
∴ A₁ = 5 cm, A₂ = 12 cm, ω = π rad/s, α = 0, β = \(\frac{\pi}{2}\) rad.

Question 3.
An SHM is given by the equation x = [8 sin (4πt) + 6 cos (4πt)] cm. Find its
(1) amplitude
(2) initial phase
(3) period
(4) frequency.
Solution:
Data : x = [8 sin (4πt) + 6 cos (4πt)] cm
x = 8 sin (4πt) + 6 cos (4πt)
= 8 sin (4πt) + 6 sin \(\left(4 \pi t+\frac{\pi}{2}\right)\)
Thus, x is the superposition of two parallel SHMs of the same period : x = x₁ + x₂, where
x₁ = 8 sin (4πt) cm = A₁ sin (ωt + α) and
x₂ = 6 sin \(\left(4 \pi t+\frac{\pi}{2}\right)\) = A₂ sin (ωt + β)
∴ A₁ = 8 cm, A₂ = 6cm, ω = 4π rad/s, α = 0,
β = \(\frac{\pi}{2}\) rad

Question 30.
Show that the total energy of a particle performing linear SHM is directly proportional to
(1) the square of the amplitude
(2) the square of the frequency.
Answer:
For a particle of mass m executing SHM with angular frequency ω and amplitude A, its kinetic and potential energies are respectively,
KE = \(\frac{1}{2}\)mω²(A² – x²) … (1)
and PE = \(\frac{1}{2}\)mω²x² … (2)
Then, the total energy,
E = PE + KE
= \(\frac{1}{2}\)mω²x² + \(\frac{1}{2}\)mω²(A² – x²)
= \(\frac{1}{2}\)mω²A² …. (3)
Therefore, total energy of the particle is

1. directly proportional to the mass (E ∝ m),
2. directly proportional to the square of the amplitude (E ∝ A²)
3. proportional to the square of the frequency
   (E ∝f²), as f = ω/2π

Question 31.
State the expression for the total energy of SHM in terms of acceleration.
Answer:
The total energy of a particle of mass m performing SHM with angular frequency ω, E = \(\frac{1}{2}\)mω²A²
The maximum acceleration of the particle, a_max = ω²A²
E = \(\frac{1}{2}\) mAa_max is the required expression.

Question 32.
State the expressions for the kinetic energy and potential energy of a particle performing SHM. Find their values at
(i) an extreme position
(ii) the mean position.
Using the expressions for the kinetic energy and potential energy of a particle in simple harmonic motion at any position, show that
(i) at the mean position, total energy = kinetic energy
(ii) at an extreme position, total energy = potential energy.
Answer:
For a particle of mass m executing SHM with force constant k, amplitude A and angular frequency ω = \(\sqrt{k / m}\), its kinetic and potential energies are respectively,
KE = \(\frac{1}{2}\)k(A² – x²) and
PE = \(\frac{1}{2}\)kx²
and total energy, E = \(\frac{1}{2}\)kA²
(i) At the mean position, x = 0,
KE = \(\frac{1}{2}\)kA² = E and PE = 0

(ii) At an extreme position, x = ±A, KE = 0 and PE = \(\frac{1}{2}\)kA² = E

That is, the energy transfers back and forth between kinetic energy and potential energy, while the total mechanical energy of the oscillating particle remains constant. The total energy is entirely kinetic energy at the mean position and entirely potential energy at the extremes.

Question 33.
State the expressions for the kinetic energy (KE) and potential energy (PE) at a displacement x for a particle performing linear SHM. Find
(i) the displacement at which KE is equal to PE
(ii) the KE and PE when the particle is halfway to a extreme position.
Answer:
For a particle of mass m executing SHM with force constant k, amplitude A and angular frequency ω = \(\sqrt{k / m}\), its kinetic and potential energies are respectively,
KE = \(\frac{1}{2}\)E (A² – x²) and
PE = \(\frac{1}{2}\)kx²
and total energy, E = \(\frac{1}{2}\)kA²

∴ At x = ±\(\frac{A}{2}\), the energy is 25% potential energy and 75% kinetic energy.

Question 34.
The maximum potential energy (PE) of a particle in SHM is 2 × 10^-4 J. What will be the PE of the particle when its displacement from the mean position is half the amplitude of SHM ?
Answer:
(PE)_max = \(\frac{1}{2}\)kA², PE = \(\frac{1}{2}\)kx²
∴ PE = (PE)_max \(\left(\frac{x}{A}\right)^{2}\) = 2 × 10^-4J × \(\left(\frac{1}{2}\right)^{2}\)
= 5 × 10^-5 J is the required answer.

Question 35.
A particle performs linear SHM of amplitude 10 cm. At what displacement of the particle from its mean position will the potential energy (PE) of the particle be 1 % of the maximum PE ?
Answer:

Question 36.
Represent graphically the variations of KE, PE and TE of a particle performing linear SHM with respect to its displacement.
Answer:

Question 37.
Represent graphically the variation of potential energy, kinetic energy and total energy of a particle performing SHM with time.
Answer:
Consider a particle performing SHM, with amplitude A and period T = \(\frac{2 \pi}{\omega}\) starting from the mean position towards the positive extreme position; ω = \(\sqrt{\frac{k}{m}}\) is the appropriate constant related to the system. The total energy of the particle is E = \(\frac{1}{2}\)kA². Its displacement (x), potential energy (PE) and kinetic energy (KE) at any instant are given by
x = A sin ωt

Using the values in the table, we can plot graphs of PE, KE and total energy with times as follows:

Question 38.
Solve the following :

Question 1.
A particle of mass 10 g is performing SHM. Its kinetic energies are 4.7 J and 4.6 J when the displacements are 4 cm and 6 cm, respectively. Compute the period of oscillation.
Answer:
Data : m = 0.01 kg, KE₁ = 4.7 J, x₁ = 4 × 10^-2 m, KE₂ = 4.6 J, x₂ = 6 × 10^-2 m
Since the total energy of a particle in SHM is constant,

Question 2.
The total energy of a particle of mass 100 grams performing SHM is 0.2 J. Find its maximum velocity and period if the amplitude is 2\(\sqrt{2}\) cm.
Solution :
Data : m = 100 g = 0.1 kg, E = 0.2 J,
A = 2\(\sqrt{2}\) cm = 2\(\sqrt{2}\) × 10^-2 m
(i) The total energy,

Question 3.
An object of mass 0.5 kg performs SHM with force constant 10 N/m and amplitude 3 cm.
(i) What is the total energy of the object?
(ii) What is its maximum speed ?
(iii) What is its speed at x = 2 cm?
(iv) What are its kinetic and potential energies at x = 2 cm ?
Solution :
Data : m = 0.5 kg, A: = 10 N/m,
A = 3 cm = 3 × 10^-2m, x = 2 cm = 2 × 10^-2m

Question 4.
When the displacement in SHM is one-third of the amplitude, what fraction of the total energy is potential and what fraction is kinetic?
Solution :
Data : x = A/3

Therefore, \(\frac{1}{9}\) th of the total energy is potential and \(\frac{8}{9}\)th of the total energy is kinetic.

Question 5.
A particle executes SHM with a period of 8 s. Find the time in which half the total energy is potential.
Solution :
Data : T = 8 s, PE = \(\frac{1}{2}\)E
ω = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{8}\) = \(\frac{\pi}{4}\) rad/s
The total energy, E = \(\frac{1}{2}\)kA² and the potential
energy = \(\frac{1}{2}\)kx²
Therefore, from the data,

Assuming that the particle starts from the mean position, the equation of motion is
x = A sin ωt

Therefore, in one oscillation, the particle’s potential energy is half the total energy 1 s, 3 s, 5 s and 7 s after passing through the mean position.

Question 39.
Define practical simple pendulum.
Answer:
Practical simple pendulum is defined as a small heavy sphere, called the bob, suspended by a light and inextensible string from a rigid support.

Question 40.
Under what conditions can we consider the oscillations of a simple pendulum to be linear simple harmonic?
Answer:
The oscillations of a simple pendulum are approximately linear simple harmonic only if

1. the amplitude of oscillation is very small compared to its length
2. the oscillations are in a single vertical plane.

Question 41.
What is the effect of mass and amplitude on the period of a simple pendulum ?
Answer:
The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum. This is the law of mass.
The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small. This is the law of isochronism. If the amplitude is large, the motion is periodic but not simple harmonic.

Question 42.
From the definition of linear SHM, derive an expression for the angular frequency of a body performing linear SHM.
Answer:
When a body of mass m performs linear SHM, the restoring force on it is always directed towards the mean position and its magnitude is directly proportional to the magnitude of the displacement of the body from the mean position. Thus, if \(\vec{F}\) is the force acting on the body when its displacement from the mean position is \(\vec{x}\),
\(\vec{F}\) = m\(\vec{a}\) = – kx\(\vec{x}\)
where the constant k, the force per unit displacement, is called the force constant.
Let \(\frac{k}{m}\) = ω², a constant.
∴ Acceleration, a = –\(\frac{k}{m}\) = – ω²x
∴ The angular frequency,

Question 43.
A simple pendulum is set into oscillations in a uniformly travelling car along a horizontal road. What happens to its period if the car takes a sudden turn towards the left ?
Answer:
The equilibrium position of the string makes an angle θ = tan^-1(a_c/g) with the vertical due to the centrifugal force to the right.
The centripetal acceleration, a_c, is horizontal and towards the left. The acceleration due to gravity is vertically downward.
∴ g_eff = \(\sqrt{g^{2}+a_{\mathrm{c}}^{2}}\)
so that the period of oscillation T = \(2 \pi \sqrt{L / g_{\text {eff }}}\)
∴ As the car takes a sudden left turn, the period of oscillation decreases.

Question 44.
Define a seconds pendulum. Find an expression for its length at a given place. Show that the length of a seconds pendulum has a fixed value at a given place.
Answer:
(1) Seconds pendulum: A simple pendulum of period two seconds is called a seconds pendulum.

(2) The period of a simple pendulum is
T = \(2 \pi \sqrt{\frac{L}{g}}\)
For a seconds pendulum, T = 2s.
∴ 2 = \(2 \pi \sqrt{\frac{L}{g}}\) ∴ L = \(\frac{g}{\pi^{2}}\)
This expression gives the length of the seconds pendulum at a place where acceleration due to gravity is g.

(3) At a given place, the value of g is constant.
∴ L = g/π² = a fixed value, at a given place.

[Note : Because the effective gravitational acceleration varies from place to place, the length of a seconds pendulum should be changed in direct proportion. Since the effective gravitational acceleration increases from the equator to the poles, so should the length of a seconds pendulum be increased.]

Question 45.
Two simple pendulums have lengths in the ratio 1 : 9. What is the ratio of their periods at a given place ?
Answer:

Question 46.
If the length of a seconds pendulum is doubled, what will be the new period?
Answer:

Question 47.
Distinguish between a simple pendulum and a conical pendulum.
Answer:

Simple pendulum	Conical pendulum
1. The oscillations of the bob are in a vertical plane.	1. The bob performs UCM in a horizontal plane and the string traces out a cone of constant semivertical angle.
2. The energy of the bob transfers back and forth between kinetic energy and potential energy, while its total mech­anical energy remains con­stant.	2. The gravitational PE of the bob being constant may be taken to be zero. The total mechanical energy remains constant and is entirely kin­etic.
3. The period depends on the 3. length of the string and the acceleration due to gravity. T =2π\(\sqrt{L / g}\)	3. The period depends on the length of the string, the ac­celeration due to gravity and cosine of the semiverti­cal angle. T =2π\(\sqrt{L \cos \theta / g}\)

Question 48.
Solve the following.

Question 1.
A simple pendulum of length 1 m has a bob of mass 10 g and oscillates freely with an amplitude of 2 cm. Find its potential energy at the extreme position. [g = 9.8 m/s²]
Solution :
Data : L = 1 m, m = 10 g = 10 × 10^-3 kg = 10^-2 kg, g = 9.8 m/s², A = 2 cm = 0.02 m
Period of a simple pendulum, T = \(2 \pi \sqrt{\frac{L}{g}}\)

Question 2.
The period of oscillation of a simple pendulum increases by 20% when the length of the pendulum is increased by 44 cm. Find its
(i) initial length
(ii) initial period of oscillation at a place where g is 9.8 m/s².
Solution:
Let T and L be the initial period and length of the pendulum. Let T₁ and L₁ be the final period and length.
Data : T₁ = T + 0.2 T = 1.2 T, L₁ = L + 0.44 m

Squaring and cross-multiplying, we get,
L + 0.44 = 1.44 L
∴ 0. 44 L = 0.44
∴ L = \(\frac{0.44}{0.44}\) = 1 m
∴ T = 2π\(\sqrt{\frac{L}{g}}\) = 2 × 3.142 × \(\sqrt{\frac{1}{9.8}}\)
= 2.007 s

Question 3.
Calculate the length of a seconds pendulum at a place where g = 9.81 m/s².
Answer:
Data : T = 2 s, g = 9.81 m/s²
Period of a simple pendulum, T = \(2 \pi \sqrt{\frac{L}{g}}\)
For a seconds pendulum, 2 = \(2 \pi \sqrt{\frac{L}{g}}\)
∴ The length of the seconds pendulum,
L = \(\frac{g}{\pi^{2}}\) = \(\frac{9.81}{(3.142)^{2}}\) = 0.9937

Question 4.
A clock regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.01 m. How much will the clock gain or lose in one day ? [g = 9.8 m/s²]
Solution:
Data: L = 1.01 m, g = 9.8 m/s²

The period of a seconds pendulum is 2 seconds. Hence, the given pendulum clock will lose 0.017 s in 2.017 s during summer.
∴ Time lost in 24 hours
= \(\frac{24 \times 3600 \times 0.017}{2.017}\)s = 728.1 s
The given pendulum clock will lose 728.1 seconds per day during summer.

Question 5.
A small drop of mercury oscillates simple harmonically inside a watch glass whose radius of curvature is 2.5 m. Find the period of the motion. [g = 9.8m/s²]
Solution :
Data : R = 2.5 m, g = 9.8 m/s²
Consider a small drop of mercury on a watch glass of radius of curvature R.

Away from its equilibrium position O, its weight \(m \vec{g}\) is resolved into two perpendicular components : mg cos θ normal to the concave surface and mg sin θ tangential to the surface, mg cos θ is balanced by the normal reaction (\(\vec{N}\)) of the surface while mg sin θ constitutes the restoring force that brings the drop back to O. If θ is small and in radian,
restoring force, F = ma = – mg sin θ
= – mg θ
= -mg\(\frac{x}{R}\)
∴ The acceleration per unit displacement, |\(\frac{a}{x}\)| = \(\frac{g}{R}\)
∴ The period of the motion, T = \(\frac{2 \pi}{\sqrt{|a / x|}}\) = \(2 \pi \sqrt{\frac{R}{g}}\)
Data : R = 2.5 m, g = 9.8 m/s²
∴ The period of oscillation is
T = 2 × 3.142\(\sqrt{\frac{2.5}{9.8}}\) = 6.284 × 0.5051 = 3.174 s.

Question 49.
Explain angular or torsional oscillations.
Hence obtain the differential equation of the motion.
Answer:
Suppose a disc is suspended from its centre by a wire or a twistless thread such that the disc remains horizontal, as shown in below figure. The rest position of

the disc is marked by a reference line. When the disc is rotated in the horizontal plane by a small angular displacement 0 = 0m from its rest position (θ = θ_m), the suspension wire is twisted. When the disc is released, it oscillates about the rest position in angular or torsional oscillation with angular amplitude θ_m.

The device is called a torsional pendulum and the springiness or elasticity of the motion is associated with the twisting of the suspension wire. The twist in either direction stores potential energy in the wire and provides an alternating restoring torque, opposite in direction to the angular displacement. The motion is governed by this torque.

If the magnitude of the restoring torque (τ) is proportional to the angular displacement (θ), τ ∝ (-θ) or τ = – cθ … (1)
where the constant of proportionality c is called the torsion constant, that depends on the length, diameter and material of the suspension wire. In this case, the oscillations will be simple harmonic.

Let I be the moment of inertia (MI) of the oscillating disc.
Torque = MI × angular acceleration
τ = Iα = I\(\frac{d^{2} \theta}{d t^{2}}\)
Hence, from EQ. (1),

This is the differential equation of angular SHM.
[Note : Angular displacement being a dimensionless quantity, the SI unit of torsion constant is the same as that of torque = the newton-metre (N-m)]

Question 50.
Define angular SHM. State the differential equation of angular SHM. Hence derive an expression for the period of angular SHM in terms of
(i) the torsion constant
(ii) the angular acceleration.
Answer:
Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
I\(\frac{d^{2} \theta}{d t^{2}}\) + c θ = 0 … (1)
where I = moment of inertia of the
where I = moment of inertia of the oscillating body,
\(\frac{d^{2} \theta}{d t^{2}}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,

Question 51.
Solve the following :

Question 1.
A bar magnet of moment 10 A.m² is suspended such that it can rotate freely in a horizontal plane. The horizontal component of the Earth’s magnetic field at the place is 39 μT. Calculate the magnitude of the torque when its angular displacement with respect to the direction of the field is 10°.
Solution :
Data : μ = 10 A.m², Bh = 3.9 × 10^-5 T, θ = 10°
The magnitude of the torque is τ = – μB_h sin θ = (10)(3.9 × 10^-5) sin 10°
= (3.9 × 10^-4)(0.1736) = 6.770 × 10^-5 N.m

Question 2.
A disc, of radius 12 cm and mass 250 g, is suspended horizontally by a long wire at its centre. Its period T₁ of angular SHM is measured to be 8.43 s. An irregularly shaped object X is then hung from the same wire and its period T₂ is found to be 4.76 s. What is the rotational inertia of object X about its suspension axis ?
Solution:
Data : R = 0.12 m, M = 0.25 kg, T, = 8.43 s, T₂ = 4.76 s
The MI of the disc about the rotation axis (perpendicular through its centre) is = \(\frac{1}{2}\) MR²

Question 52.
What is meant by damped oscillations ? Draw a neat, labelled diagram of a damped spring-and-block oscillator.
Answer:
Oscillations of gradually decreasing amplitude are called damped oscillations. Oscillations of a system in the presence of dissipative frictional forces are damped.

The dissipative damping force removes energy from the system which requires external periodic force to continue.

Below Figure shows a spring-and-block oscillator attached with a light vane that moves in a fluid with viscosity. When the system is set into oscillation, the amplitude decreases for each oscillation due to the viscous drag on the vane.

Question 53.
Write the differential equation of motion for an oscillator in the presence of a damping force directly proportional to the velocity. Under what condition is the motion oscillatory? Hence, discuss the frequency, amplitude and energy of the damped oscillations.
OR
Oscillations in the presence of a force proportional to the velocity are periodic but not simple harmonic. Explain.
OR
The presence of a damping force changes the character of a simple harmonic motion. Explain this qualitatively.
Answer:
Consider the oscillations of a body in the presence of a dissipative frictional force such as viscous drag or fluid friction. Such a force is proportional to the velocity of the body and is in a direction opposite to that of the velocity. If the fluid flow past the body is streamline, then by Stokes’ law, the resistive force is
f = -βv = -β\(\frac{d x}{d t}\)… (1)
where v = \(\frac{d x}{d t}\) is the velocity and β is a positive constant of proportionality called the damping constant.
The linear restoring force on the oscillator is F = -kx … (2)
where k is the force constant. If m is the mass of the oscillator and its acceleration is \(\frac{d^{2} x}{d t^{2}}\)

where ω² = \(\frac{k}{m}\). Equation (3) is the differential equation of the oscillator in presence of a resistive force directly proportional to the velocity.
The solution of the above differential equation obtained using standard mathematical technique is

where constants A and φ can be determined in the usual way from the initial conditions. In writing this solution, it is assumed β is less than 2mω, i.e., the resistive term is relatively small.

In Eq. (4),
(1) the harmonic term, cos \(\left[\sqrt{\omega^{2}-\frac{\beta^{2}}{2 m}} t+\phi\right]\), that the motion is oscillatory with angular frequency ω’ = \(\sqrt{\omega^{2}-\frac{\beta^{2}}{4 m^{2}}}\) if β is less than 2mω. The harmonic term can also be written in terms of a sine function with the same ω’.
(2) A’ = Ae^-(β/2m)t is the amplitude of the oscillation. The exponential factor e^-(β/2m)t steadily decreases the amplitude of the motion, making it approach zero for large t. Hence, the motion is said to be damped oscillation or damped harmonic motion.
(3) the total energy, \(\frac{1}{2}\)m(ω’)², decays exponentially with time as the amplitude decreases. The energy is dissipated in the form of heat by the damping force.
(4) the period of the damped oscillations is

∴ T is greater than 2π/ω.
Thus, the motion is periodic but not simple harmonic because the amplitude steadily decreases.

Notes :

1. The energy decreases faster than the amplitude.
2. For β < 2mω, the larger the value of β, the faster the amplitude decreases. The condition is called underdamping.
3. When β = 2mω = \(2 \sqrt{k m}\) km, ω’ = 0, i.e., the system no longer oscillates. When displaced and released, it returns to its equilibrium position without oscillation. The condition is called critical damping.
4. If β > 2mω, the system is said to overdamped or dead beat. Again, the system does not oscillate but returns to equilibrium position more slowly than for critical damping.
5. All practical cases of so called free oscillations, such as that of a simple pendulum or a tuning fork, are damped. We also encounter damped oscillations in electrical circuits containing inductance, capacitance and resistance due to resistive losses. While in many cases it is desirable to minimize damping, in ammeters and voltmeters the oscillations of the pointer are designed to be dead beat.

Question 54.
Solve the following.

Question 1.
For a damped spring-and-block oscillator, the mass of the block is 0.2 kg, the spring constant is 90 N/m and the damping constant is 0.06 kg/s. Calculate
(i) the period of oscillation
(ii) the time taken for its amplitude to become half its initial value.
Solution :
Data : m = 0.2 kg, k = 90 N/m, β = 0.06 kg/s
(i) The period of the damped oscillation is

(ii) The amplitude of the damped oscillation is
A’ = Ae^-(β/2m)t
If the amplitude becomes half the initial amplitude A at time f,

Question 2.
A steel sphere of mass 0.02 kg attains a terminal speed vi = 0.5 m/s when dropped into a tall cylinder of oil. The same sphere is then attached to the free end of an ideal vertical spring of spring constant 8 N/m. The sphere is immersed in the same oil and set into vertical oscillation. Find
(i) the damping constant
(ii) the angular frequency of the damped SHM.
(iii) Hence, write the equation for displacement of the damped SHM as a function of time, assuming that the initial amplitude is 10 cm. [g = 10 m/s²]
Solution :
Data : m = 0.02 kg, v_t = 0.5 m/s, k = 8 N/m,
A = 10 cm = 0.1 m, g = 10 m/s²
When the sphere falls with terminal velocity in oil, the resultant force on it is zero. Therefore, the
The equation of motion of the damped oscillation is resistive force and its weight are equal in magnitude and opposite in direction.
∴ |F_r| = βv_t = mg
where β is the damping constant.
∴ β = \(\frac{m g}{v_{\mathrm{t}}}\) = \(\frac{0.02 \times 10}{0.5}\) = 0.4 kg/s
The angular frequency of the damped oscillation in oil,

The equation of motion of the damped oscillation is
x = Ae^(β/2m)t cos(w’t + φ)
∴ x = (0.1 m)e^-(0.4/004)t cos (17.32t + φ)
x = (0.1 m) e^-10t cos(17.32t + φ)

Question 55.
Explain
(1) free vibrations
(2) forced vibrations.
Answer:
(1) Free vibrations : A body capable of vibrations is said to perform free vibrations when it is disturbed from its equilibrium position and left to itself.

In the absence of dissipative forces such as friction due to surrounding air and internal forces, the total energy and hence the amplitude of vibrations of the body remains constant. The frequencies of the free vibrations of a body are called its natural frequencies and depend on the body itself.

In the absence of a maintaining force, in practice, the total energy and hence the amplitude decreases due to dissipative forces and the vibration is said to be damped. The frequency of damped vibrations is less than the natural frequency.

(2) Forced vibrations : The vibrations of a body in response to an external periodic force are called forced vibrations.

The external force supplies the necessary energy to make up for the dissipative losses. The frequency of the forced vibrations is equal to the frequency of the external periodic force.

The amplitude of the forced vibrations depends upon the mass of the vibrating body, the amplitude of the external force, the difference between the natural frequency and the frequency of the periodic force, and the extent of damping.

Question 56.
Distinguish between free vibrations and forced vibrations.
Answer:

Free vibrations	Forced vibrations
1. Free vibrations are pro­duced when a body is disturbed from its equilibrium position and released. Ex. Simple pendulum.	1. Forced vibrations are pro­duced by an external periodic force. Ex. Musical instrument having a sounding board.
2. The frequency of free vibra­tions depends on the body and is called its natural frequency.	2. The frequency of forced vi­brations is equal to that of the external periodic force.
3. The energy of the body remains constant only in the absence of friction, air resis­tance, etc.	3. The energy of the body is maintained constant by the external periodic force.

Question 57.
Explain resonance.
Answer:
Resonance : If a body is made to vibrate by an external periodic force, whose frequency is equal to the natural frequency (or nearly so) of the body, the body vibrates with maximum amplitude. This phenomenon is called resonance.

The corresponding frequency is called the resonant frequency.

For low damping, the amplitude of vibrations has a sharp maximum at resonance, as shown. The flatter curve without a pronounced maximum is for high damping.

Suppose several pendulums-A, B, C, D and E are coupled to a heavier pendulum Z, by suspending them from a stretched cord, and that only the length of C is the same as that of Z. When Z is set into oscillation perpendicular to the cord PQ, the others are also set into forced oscillations in parallel vertical planes. Their amplitudes vary but those of A, B, D and E never become very large because the frequency of Z is not the same as the natural frequency of any of them. On the other hand, C will be in resonant oscillation and its amplitude will be large.

Question 58.
The differential equation of SHM for a seconds pendulum is
(A) \(\frac{d^{2} x}{d t^{2}}\) + x = 0
(B) \(\frac{d^{2} x}{d t^{2}}\) + πx = 0
(C) \(\frac{d^{2} x}{d t^{2}}\) + 4πx = 0
(D) \(\frac{d^{2} x}{d t^{2}}\) + π²x = 0.
Answer:
(D) \(\frac{d^{2} x}{d t^{2}}\) + π²x = 0.

Question 59.
The phase change of a particle performing SHM between successive passages through the mean position is
(A) 2π rad
(B) π rad
(C) \(\frac{\pi}{2}\) rad
(D) \(\frac{\pi}{4}\) rad.
Answer:
(B) π rad

Question 60.
If the equation of motion of a particle performing SHM is x = 0.028 cos (2.8πt + π) (all quantities in SI units), the frequency of the motion is
(A) 0.7 Hz
(B) 1.4 Hz
(C) 2.8 Hz
(D) 14 Hz.
Answer:
(B) 1.4 Hz

Question 61.
A spring-and-block system constitutes a simple harmonic oscillator. To double the frequency of oscillation, the mass of the block must be ….. the initial mass.
(A) \(\frac{1}{4}\) times
(B) half
(C) double
(D) 4 times
Answer:
(A) \(\frac{1}{4}\) times

Question 62.
A horizontal spring-and-block system consists of a block of mass 1 kg, resting on a frictionless surface, and an ideal spring. A force of 10 N is required to compress the spring by 10 cm. The spring constant of the spring is
(A) 100 N.m^-1
(B) 10N.m^-1
(C) N.m^-1
(D) 0.1 N.m^-1.
Answer:
(C) N.m^-1

Question 63.
A vertical spring-and-block system has a block of mass 10 g and oscillates with a period 1 s. The period of SHM of a block of mass 90 g, suspended from the same spring, is
(A) \(\frac{1}{9}\)s
(B) \(\frac{1}{3}\)s
(C) 3 s
(D) 9 s.
Answer:
(C) 3 s

Question 64.
A simple harmonic oscillator has an amplitude A and period T. The time required by the oscillator to cover the distance from x = A to x = \(\frac{A}{2}\) is
(A) \(\frac{T}{2}\)
(B) \(\frac{T}{3}\)
(C) \(\frac{T}{4}\)
(D) \(\frac{T}{6}\)
Answer:
(D) \(\frac{T}{6}\)

Question 65.
The period of SHM of a particle with maximum velocity 50 cm/s and maximum acceleration 10 cm/s² is
(A) 31.42 s
(B) 6.284 s
(C) 3.142 s
(D) 0.3142 s.
Answer:
(C) 3.142 s

Question 66.
A particle executing SHM of amplitude 5 cm has an acceleration of 27 cm/s² when it is 3 cm from the mean position. Its maximum velocity is
(A) 15 cm/s
(B) 30 cm/s
(C) 45 cm/s
(D) 60 cm/s.
Answer:
(A) 15 cm/s

Question 67.
A particle performs linear SHM with a period of 6 s, starting from the positive extremity. At time t = 7 s, its displacement is 3 cm. The amplitude of the motion is
(A) 4 cm
(B) 6 cm
(C) 8 cm
(D) 12 cm.
Answer:
(B) 6 cm

Question 68.
A spring-and-block oscillator with an ideal spring of force constant 180 N/m oscillates with a frequency of 6 Hz. The mass of the block is, approximately,
(A) \(\frac{1}{8}\) kg
(B) \(\frac{1}{4}\) kg
(C) 4 kg
(D) 8 kg.
Answer:
(A) \(\frac{1}{8}\) kg

Question 69.
A particle executing linear SHM has velocities v₁ and v₂ at distances x₁ and x₂, respectively, from the mean position. The angular velocity of the particle is

Answer:
(B) \(\sqrt{\frac{v_{2}^{2}-v_{1}^{2}}{x_{1}^{2}-x_{2}^{2}}}\)

Question 70.
A particle executes linear SHM with period 12 s. To traverse a distance equal to half its amplitude from the equilibrium position, it takes
(A) 6s
(B) 4s
(C) 2s
(D) 1s.
Answer:
(D) 1s

Question 71.
The minimum time taken by a particle in SHM with period T to go from an extreme position to a point half way to the equilibrium position is
A. \(\frac{T}{12}\)
B. \(\frac{T}{8}\)
C. \(\frac{T}{6}\)
D. \(\frac{T}{4}\)
Answer:
C. \(\frac{T}{6}\)

Question 72.
In simple harmonic motion, the acceleration of a particle is zero when its
(A) velocity is zero
(B) displacement is zero
(C) both velocity and displacement are zero
(D) both velocity and displacement are maximum.
Answer:
(B) displacement is zero

Question 73.
The acceleration of a particle performing SHM is 3m/s² at a distance of 3 cm from the mean position.
The periodic time of the motion is
(A) 0.02 π s
(B) 0.04 π s
(C) 0.2 π s
(D) 2 π s.
Answer:
(C) 0.2 π s

Question 74.
A particle performing linear SHM with a frequency n is confined within limits x = ±A. Midway between an extremity and the equilibrium position, its speed is
(A) \(\sqrt{6}\)nA
(B) \(\sqrt{3}\)πnA
(C) \(\sqrt{6}\)πnA
(D) \(\sqrt{12}\)πnA
Answer:
(B) \(\sqrt{3}\)πnA

Question 75.
The total energy of a particle executing SHM is proportional to
(A) the frequency of oscillation
(B) the square of the amplitude of motion
(C) the velocity at the equilibrium position
(D) the displacement from the equilibrium position.
Answer:
(B) the square of the amplitude of motion

Question 76.
Two spring-and-block oscillators oscillate harmonically with the same amplitude and a constant phase difference of 90°. Their maximum velocities are v and v + x. The value of x is
(A) 0
(B) \(\frac{v}{3}\)
(C) 2
(D) \(\frac{v}{\sqrt{2}}\).
Answer:
(A) 0

Question 77.
If the length of a simple pendulum is increased to 4 times its initial length, its frequency of oscillation will
(A) reduce to half its initial frequency
(B) increase to twice its initial frequency
(C) reduce to \(\frac{1}{4}\) th its initial frequency
(D) increase to 4 times its initial frequency.
Answer:
(A) reduce to half its initial frequency

Question 78.
If the length of a simple pendulum is doubled keeping its amplitude constant, its energy will be
(A) unchanged
(B) doubled
(C) halved
(D) increased to four times the initial energy.
Answer:
(C) halved

Question 79.
The amplitude of oscillations of a simple pendulum of period T and length L is increased by 5%. The new period of the pendulum will be
(A) T/8
(B) T/4
(C) T/2
(D) T.
Answer:
(D) T.

Question 80.
In 20 s, two simple pendulums, P and Q, complete 9 and 7 oscillations, respectively, on the Earth. On the
Moon, where the acceleration due to gravity is \(\frac{1}{6}\)th that on the Earth, their periods are in the ratio (A) 8 : 1
(B) 9 : 7
(C) 7 : 9
(D) 3 : 14.
Answer:
(C) 7 : 9

Question 81.
If T is the time period of a simple pendulum in an elevator at rest, its time period in a freely falling elevator will be
(A) \(\frac{T}{\sqrt{2}}\)
(B) \(\sqrt{2}\)T
(C) 2T
(D) infinite.
Answer:
(D) infinite.

Question 82.
A seconds pendulum is suspended in an elevator moving with a constant speed in the downward direction. The periodic time (T) of that pendulum is
(A) less than two seconds
(B) equal to two seconds
(C) greater than two seconds
(D) very much greater than two seconds.
Answer:
(B) equal to two seconds

Question 83.
The total work done by a restoring force in simple harmonic motion of amplitude A and angular frequency ω, in one oscillation is
(A) \(\frac{1}{2}\)mA²ω²
(B) zero
(C) mA²ω²
(D) \(\frac{1}{2}\)mAω.
Answer:
(B) zero

Question 84.
Two particles perform linear simple harmonic motion along the same path of length 2A and period T as shown in the graph below. The phase difference between them is

(A) zero rad
(B) \(\frac{\pi}{4}\) rad
(C) \(\frac{\pi}{2}\) rad
(D) \(\frac{3 \pi}{4}\) rad
Answer:
(B) \(\frac{\pi}{4}\) rad

Question 85.
The average displacement over a period of SHM is
(A = amplitude of SHM)
(A) 0
(B) A
(C) 2A
(D) 4A.
Answer:
(A) 0

Question 86.
Two springs of force constants k₁ and k₂(k₁ > k₂) are stretched by the same force. If W₁ and W₂ be the work done in stretching the springs, then
(A) W₁ = W₂
(B) W₁ < W₂
(C) W₁ > W₂
(D) W₁ = W₂ = 0.
Answer:
(B) W₁ < W₂

Question 87.
Two bar magnets of identical size have magnetic moments M_A and M_B. If the magnet A oscillates at twice the frequency of magnet B, then
(A) M_A = 2M_B
(B) M_A = 8M_B
(C) M_A = 4M_B
(D) M_B = 8M_A.
Answer:
(C) M_A = 4M_B

Question 88.
A magnet is suspended to oscillate in the horizontal plane. It makes 20 oscillations per minute at a place where the dip angle is 30° and 15 oscillations per minute where the dip angle is 60°. The ratio of the Earth’s total magnetic field at the two places is
(A) 3\(\sqrt{3}\) : 16
(B) 16 : 9\(\sqrt{3}\)
(C) 4 : 9\(\sqrt{3}\)
(D) 9 : 16\(\sqrt{3}\).
Answer:
(B) 16 : 9\(\sqrt{3}\)

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 4 Thermodynamics Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 4 Thermodynamics

Question 1.
What is temperature? Explain.
Answer:
The temperature of a body is a quantitative measure of the degree of hotness or coolness of the body. According to the kinetic theory of gases, it is a measure of the average kinetic energy per molecule of the gas. Temperature difference determines the direction of flow of heat from one body to another or from one part of the body to the other. Its SI unit is the kelvin (K).

Question 2.
What is heat? Explain.
Answer:
When two bodies are in thermal contact with each other, there is a transfer of energy from the body at higher temperatures to the body at lower temperatures. The energy in the transfer is called heat. Also when two parts of a body are at different temperatures, there is a transfer of energy from the part at a higher temperature to the other part. The SI unit of heat is the joule.

[Note: Count Rumford [Benjamin Thompson] (1753-1814) Anglo-American adventurer, social reformer, inventor, and physicist, measured the relationship between work and heat. When he visited Arsenal in Munich, he found that a tremendous amount of heat was produced in a short time when a brass cannon was being bored. He found that even with a blunt borer a lot of heat can be produced from a piece of metal. At that time it was thought that heat consists of a fluid called caloric. Rumford’s experiments showed that caloric did not exist and heat is the motion of the particles of a body. He measured the relation between work done and corresponding heat produced. The result was not accurate, but important in the development of thermodynamics.]

Question 3.
What is thermodynamics ?
Answer:
Thermodynamics is the branch of physics that deals with the conversion of energy (including heat) from one form into another, the direction of energy transfer between a system and its environment with the resulting variation in temperature, in general, or changes of state, and the availability of energy to do mechanical work.

Question 4.
What is meant by thermal equilibrium ? What is meant by the expression “two systems are in thermal equilibrium” ?
Answer:
A system is in a state of thermal equilibrium if there is no transfer of heat (energy) between the various parts of the system or between the system and its surroundings.

Two systems are said to be in thermal equilibrium when they are in thermodynamic states such that, if they are separated by a diathermic (heat conducting) wall, the combined system would be in thermal equilibrium, i.e., there would be no net transfer of heat (energy) between them.
[Note :It is the energy in transfer that is called the heat.]

Question 5.
State the zeroth law of thermodynamics.
Answer:
Zeroth law of thermodynamics : If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.

[Note :The zeroth law is fundamental to the other laws of thermodynamics. That this law is assumed by the other laws of thermodynamics was realized much later. This law has no single discoverer. It was given the status of a law, following the suggestion by R. H. Fowler (in 1931), only after the first, second and third laws were named.]

Question 6.
Explain the zeroth law of thermodynamics.
Answer:
Consider three systems A, B and C. Suppose A and B are in thermal equilibrium, and A and C are also in thermal equilibrium. Then B and C are also in thermal equlibrium. Thus, A, B and C are at the same temperature and A works as a thermometer.

[Note: The arrows in the figure indicate energy exchange]

Question 7.
Define internal energy.
Answer:
Internal energy of a system is defined as the sum of the kinetic energies of the atoms and molecules belonging to the system, and the potential energies associated with the interactions between these constituents (atoms and molecules).
[Note : Internal energy does not include the potential energy and kinetic energy of the system as a whole. In the case of an ideal gas, internal energy is purely kinetic. In the case of real gases, liquids and solids, internal energy is the sum of potential and kinetic energies. For an ideal gas, internal energy depends on temperature only. In other cases, internal energy depends on temperature, as well as on pressure and volume. According to quantum theory, internal energy never becomes zero. Even at OK. particles have energy called zero-point energy.]

Question 8.
What is the internal energy of one mole of argon and oxygen ?
Answer:
Argon is a monatamic gas. In this case, with three degrees of freedom, the average kinetic energy per molecule = \(\left(\frac{3}{2}\right)\)k_BT, where k_B is the Boltzmann constant and T is the absolute (thermodynamic) temperature of the gas. Hence, the internal energy of one mole of argon = N_A\(\left(\frac{3}{2} k_{\mathrm{B}} T\right)\) = \(\frac{3}{2}\)RT, where N_A is the Avogadro number and R = N_Ak_B is the universal gas constant. Oxygen is a diatomic gas. In this case, with five degrees of freedom at moderate temperatures, the internal energy of one mole of
oxygen = \(\frac{3}{2}\)RT.

Question 9.
Find the internal energy of one mole of argon at 300 K. (R = 8.314 J/mol.K)
Answer:
The internal energy of one mole of argon at 300 K
= \(\frac{3}{2}\)RT = \(\frac{3}{2}\)(8.314)(300) J = 3741 J.

Question 10.
The internal energy of one mole of nitrogen at 300 K is 6235 J. What is the internal energy of one mole of nitrogen at 400 K ?
Answer:

This is the required quantity.
[Note : In chapter 3, the symbol E was used for internal energy.]

Question 11.
Explain the term thermodynamic system.
Answer:
A thermodynamic system is a collection of objects that can form a unit which may have ability to Surrounding

exchange energy with its surroundings. Everything outside the system is called its surroundings or environment. For example, a gas enclosed in a container is a system, the container is the boundary and the atmosphere is the environment.

Question 12.
Explain classification of thermodynamic systems.
Answer:
Depending upon the exchange of energy and matter with the environment, thermodynamic systems are classified as open, closed or isolated.

A system that can freely exchange energy and matter with its environment is called an open system. Example : water boiling in an open vessel.

A system that can freely exchange energy but not matter with its environment is called a closed system. Example : water boiling in a closed vessel.

A system that cannot exchange energy as well as matter with its environment is called an isolated system. In practice it is impossible to realize an isolated system as every object at a temperature above 0 K emits energy in the form of radiation, and no object can ever attain 0 K.

For many practical purposes, a thermos flask containing a liquid can be considered an isolated system.

These three types are illustrated in above figure.

Question 13.
What is a thermodynamic process? Give an exmple.
Answer:
A process in which the thermodynamic state of a system is changed is called a thermodynamic process.
Example : Suppose a container is partially filled with water and then closed with a lid. If the container is heated, the temperature of the water starts rising and after some time the water starts boiling. The steam produced exerts pressure on the walls of the container, Here, there is a change in the pressure, volume and temperature of the water, i.e. there is a change in the thermodynamic state of the system.

Question 14.
Explain the relation between heat and internal energy.
Answer:
Suppose a system consists of a glass filled with water at temperature T_S. Let T_E be the temperature of the environment (surroundings) such as the surrounding air in the room. There is a continuous exchange of energy between the system and the surroundings.

If T_S > T_E, the net effect of energy exchange is the net transfer of internal energy from the system to the environment till thermal equilibrium is reached, i.e., T_S and T_E became equal. This internal energy in transit is called heat (Q). The change in the temperature of the environment is usually negligible compared with the change in the temperature of the system.

For T_S < T_E, there is energy exchange between the
system and the environment, but no net transfer of
energy.

For T_S = T_E, there is energy exchange between the system and the environment, but no net transfer of energy.

Thus, the net transfer of energy takes place only when there is temperature difference.

Question 15.
Explain how the internal energy of a system can be changed.
Answer:
Consider a system (S) consisting of some quantity of gas enclosed in a cylinder fitted with a movable, massless, and frictionless piston.

Suppose the gas is heated using a burner (source of heat, environment). Let T_S = temperature of the system (gas) and T_E = temperature of the environment.

Here, T_E > T_S. Hence, there will be a net flow of energy (heat) from the environment to the system causing the increase in the internal energy of the system.

The internal energy of the gas (system) can also be increased by quickly pushing the piston inward so that the gas is compressed.

The work done on the gas raises the temperature of the gas. Thus, there is increase in the internal energy of the gas. If the gas pushes the piston outward, the work is done by the gas on the environment and the gas cools as its internal energy becomes less.

Question 16.
On the basis of the kinetic theory of gases, explain
(i) positive work done by a system
(ii) negative work done by a system.
Answer:
Consider a system consisting of some quantity of a gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.

During expansion of the gas, molecules colliding with the piston lose momentum to it. This exerts force and hence pressure on the piston, moving it outward through a finite distance. Here, the gas does a positive work on the piston. There is increase in the volume of the gas. The work done by the piston on the gas is negative.

During compression of the gas, molecules colliding with the piston gain momentum from it. The piston moving inward through a finite distance exerts force on the gas. Here, the gas does a negative work on the piston. There is decrease in the volume of the gas.

The work done by the piston on the gas is positive.

Question 17.
Obtain an expression for the work done by a gas.
OR
Show that the work done by a gas is given by
Answer:
Consider a system consisting of some quantity of a gas enclosed in a cylinder fitted with a movable, massless and frictionless piston.

During expansion of the gas, molecules colliding with the piston impart momentum to the piston. The time rate of change of momentum is the force, F exerted by the gas on the piston. If dx is the displacement of the piston, the work done by the gas, dW = F dx. If A is the area of cross section of the piston, the pressure exerted by the gas, P = \(\frac{F}{A}\).

Hence, the work done, dW = PAdx = PdV where dV = Adx is the increase in the volume of the gas. Here, dx is the infinitesimal displacement of the piston and dV is the infinitesimal increase in the volume of the gas.

If V_i is the initial volume of the gas, and V_f is the final volume, the total work done by the gas in moving the piston is given by W = \(\int_{V_{i}}^{V_{\mathrm{f}}} P d V\).

Question 18.
State the first law of thermodynamics. Express it in mathematical form.
Answer:
First law of thermodynamics : The change in the internal energy of a system (∆U) is the difference between the heat supplied to the system (Q) and the work done by the system on its surroundings (W).
Mathematically, ∆ U = Q – W, which is the same as Q = ∆ U + W.

Notes :

1. if Q is positive, it means heat is added to the system. If Q is negative, it means heat is given out by the system or removed from the system,
2. If ∆U is positive, it means there is increase in the internal energy of the system. If ∆ U is negative, it means there is decrease in the internal energy of the system,
3. If W is positive, it means it is the work done by the system on its surroundings. Negative W means work is done on the system by the surroundings,
4. The first law of thermodynamics is largely due to Joule. It is essentially the law of conservation of energy applied to the systems that are not isolated, i.e., the systems that can exchange energy with the surroundings. Thermodynamics was developed in 1850 by Rudolf Clausius (1822-88) German theoretical physicist, His ideas were developed in 1851 by William Thomson [Lord Kelvin] (1824-1907), British physicist and electrical engineer,
5. Q = ∆ U + W. Here, all quantities are expressed in the same units, e.g., cal or joule. If Q and A U are expressed in heat unit (cal, kcal) and W is expressed in mechanical unit (erg, joule) then the above equation takes the form Q = ∆ U + \(\frac{W}{J}\), where J is the mechanical equivalent of heat.]

Question 19.
What is the property of a system or a system variable ?
Answer:
The property of a system or a system variable is any measurable or observable characteristic or property of the system when the system remains in equilibrium.

Question 20.
Name the macroscopic variables of a system.
Answer:
Pressure, volume, temperature, density, mass, specific volume, amount of substance (expressed in mole) are macroscopic variables of a system.
Notes : The quantities specified above are not totally independent, e.g.,

1. density = mass/volume
2. specific volume (volume per unit mass) = 1/density.

Question 21.
What is an intensive variable ? Give examples.
Answer:
A variable that does not depend on the size of the system is called an intensive variable.
Examples : pressure, temperature, density.

Question 22.
What is an extensive variable ? Give examples.
Answer:
A variable that depends on the size of the system is called an extensive variable.
Examples : internal energy, mass.

Question 23.
What is mechanical equilibrium ?
Answer:
A system is said to be in mechanical equilibrium when there are no unbalanced forces within the system and between the system and its surroundings.
OR
A system is said to be in mechanical equilibrium when the pressure in the system is the same throughout and does not change with time.
[Note : The constituents of a system, atoms, molecules, ions, etc, are never at rest. Within a system, even in the condition of equilibrium, statistical fluctuations do occur, but the time of observation is usually very large so that these fluctuations can be ignored.]

Question 24.
What is chemical equilibrium ?
Answer:
A system is said to be in chemical equilibrium when there are no chemical reactions going on within the system.
OR
A system is said to be in chemical equilibrium when its chemical composition is the same throughout the system and does not change with time.
[Note : In this case, in the absence of concentration gradient, there is no diffusion, i.e., there is no transport of matter from one part of the system to the other.]

Question 25.
What is thermal equilibrium ?
Answer:
A system is said to be in thermal equilibrium when its temperature is uniform throughout the system and does not change with time.

Question 26.
Give two examples of thermodynamic systems not in equilibrium.
Answer:

1. When an inflated balloon is punctured, the air inside it suddenly expands and escapes into the atmosphere. During the rapid expansion, there is no uniformity of pressure, temperature and density.
2. When water is heated, there is no uniformity of pressure, temperature and density. If the vessel is open, some water molecules escape to the atmosphere.

Question 27.
What is the equation of state ? Explain.
Answer:
The mathematical relation between the state variables (pressure, volume, temperature, amount of the substance) is called the equation of state.

In the usual notation, the equation of state for an ideal gas is PV = nRT.

For a fixed mass of gas, the number of moles, n, is constant. R is the universal gas constant. Thus, out of pressure (P), volume (V) and thermodynamic temperature (T), only two (any two) are independent.

Question 28.
Draw P-V diagram for positive work at constant pressure.

Answer:
In this case, during the expansion, the work done by the gas, W = \(\int_{V_{1}}^{V_{2}} P d V\) = P(V₂ – V₁) is positive as V₂ > V₁.

Question 29.
What is a thermodynamic process ? Explain.
Answer:
A procedure by which the initial state of a system changes to its final state is called a thermodynamic process. During the process/ there may be

1. addition of heat to the system
2. removal of heat from the system
3. change in the temperature of the system
4. change in the volume of the system
5. change in the pressure of the system.

Question 30.
What is a quasistatic process ?
Answer:
A quasistatic process is an idealised process which occurs infinitely slowly such that at all times the system is infinitesimally close to a state of thermodynamic equilibrium. Although the conditions for such a process can never be rigorously satisfied in practice, any real process which does not involve large accelerations or large temperature gradients is a reasonable approximation to a quasistatic process.

Question 31.
Draw a diagram to illustrate that the work done by a system depends on the process even when the initial and final states are the same.

Answer:
In the above diagram, the initial state of a gas is characterized by (P_i, V_i) [corresponding to point A] and the final state of the gas is characterized by (P_f, V_f) [corresponding to point B]. Path 1 corresponds to constant temperature. Path 2 corresponds to the combination AC [P constant] + CB [V constant]. Path 3 corresponds to the combination AD [V constant] + DB [P constant]. The work done by the gas (W) is the area under the curve and is different in each case.

Question 32.
What is a reversible process? What is an irreversible process? Give four examples of an irreversible process. Explain in detail.
Answer:
A reversible process is one which is performed in such a way that, at the conclusion of the process, both the system and its local surroundings are restored to their initial states, without producing any change in the rest of the universe.

A process may take place reversibly if it is quasistatic and there are no dissipative effects. Such a process cannot be realized in practice.

A process which does not fulfill the rigorous requirements of reversibility is said to be an irreversible process. Thus, in this case, the system and the local surroundings cannot be restored to their initial states without affecting the rest of the universe. All natural processes are irreversible.
Examples of irreversible process :

1. When two bodies at different temperatures are brought in thermal contact, they attain the same temperature after some time due to energy exchange. Later, they never attain their initial temperatures.
2. Free expansion of a gas.
3. A gas seeping through a porous plug.
4. Collapse of a soap film after it is pricked.
5. All chemical reactions.
6. Diffusion of two dissimilar inert gases.
7. Solution of a solid in water.
8. Osmosis.

[Note : A free expansion is an adiabatic process, i.e., a process in which no heat is added to the system or removed from the system. Consider a gas confined by a valve to one half of a double chamber with adiabatic walls while the other half is evacuated.

When the gas is in thermal equilibrium, the gas is allowed to expand to fill the entire chamber by opening the valve.
No interaction takes place and hence there are no local surroundings. While rushing into a vacuum, the gas does not meet any pressure and hence no work is done by the gas. The gas only changes state isothermally from a volume V_i to a larger volume V_f.]

To restore the gas to its initial state, it would have to be compressed isothermally to the volume V_i; an amount of work W would have to done on the gas by some mechanical device and an equal amount of heat would have to flow out of the gas into a reservoir. If the mechanical device and the reservoir are to be left unchanged, the heat would have to be extracted from the reservoir and converted completely into work. Since this last step is impossible, the process of free expansion is irreversible.

It can be shown that the diffusion of two dissimilar inert gases is equivalent to two independent free expansions. It follows that diffusion is irreversible.]

Question 33.
What are the causes of irreversibility?
Answer:

1. Some processes such as a free expansion of a gas or an explosive chemical reaction or burning of a fuel take the system to non-equilibrium states.
2. Most processes involve dissipative forces such as friction and viscosity (internal friction in fluids). These forces can be minimized, but cannot be eliminated.

Question 34.
What is an isothermal process? Obtain an expression for the work done by a gas in an isothermal process.
Answer:
A process in which changes in pressure and volume of a system take place at a constant temperature is called an isothermal process.

Consider n moles of a gas enclosed in a cylinder fitted with a movable, massless and frictionless piston. Let P_i, V_i and T be the initial pressure, volume and absolute temperature respectively of the gas. Consider an isothermal expansion (or compression) of the gas in which P_f, V_f and T are respectively the final pressure, volume and absolute

temperature of the gas. Assuming the gas to behave as an ideal gas, we have, its equation of state :
PV = nRT = constant as T = constant, R is the universal gas constant. The work done by the gas,

Notes :

1. The above expression for W can be written in various forms such as W = nRT ln\(\left(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\right)\) = P_iV_i ln \(\left(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\right)\) = P_fV_f\(\left(\frac{V_{f}}{V_{\mathrm{i}}}\right)\), etc.
2. W is positive if V_f > V_i (expansion). W is negative if V_f < V_i (contraction).
3. At constant temperature, change in internal energy, ∆ U = 0.
   ∴ Q = ∆ U + W = W.
4. Isothermal process shown in P- V diagram is also called an isotherm.
5. Melting of ice is an isothermal process.

Question 35.
What is an isobaric process? Obtain the expressions for the work done, change in internal energy and heat supplied in an isobaric process in the case of a gas.
Answer:
A process in which pressure remains constant is called an isobaric process. Consider n moles of a gas enclosed in a cylinder fitted with a movable, massless and frictionless piston. We assume that the gas behaves as an ideal gas so that we can use the equation of state PV = nRT.

Consider an isobaric expansion (or compression) of the gas in which the volume of the gas changes from V_i to V_f and the temperature of the gas changes from T_i to T_f when the pressure (P) of the gas is kept constant. The work done by the gas,

Now, PV_i = nRT_i and PV_f = nRT_f
∴ PV_f – PV_i = nRT_f – nRT_i
∴ P(V_f – V_i) = nR(T_f – T_i)
∴ from Eq. (1), W = nR(T_f – T_i) … (2)
The change in the internal energy of the gas,
∆ U = nC_v(T_f – T_i) …(3)
where C_v is the molar specific heat of the gas at constant volume.
From Eqs. (2) and (3), we have, the heat supplied to the gas,
Q = ∆ U + W = nC_v(T_f – T_i) + nR(T_f – T_i)
= n(C_v + R)(T_f – T_i)
∴ Q = _nC_p(T_f – T_i) …(4)
Where C_p ( = C_v + R) is the molar specific heat of the gas at constant pressure.
[Note : P-V curve for an isobaric process is called an isobar.

Question 36.
What is an isochoric process ? Write the expressions for the work done, change in internal energy and heat supplied in this case. Also draw the corresponding P-V diagram.
Answer:
A process that takes place at constant volume is called an isochoric process (or isometric process).

As there is no change in volume in this case, the work done (W) by the system on its environment is zero. The change in the internal energy.
∆ U = _nC_v (T_f – T_i) and heat supplied,
Q = ∆ U = _nC_v(T_f – T_i)

Question 37.
What is an adiabatic process ? Obtain expressions for the work done by a system (an ideal gas) in an adiabatic process. Also draw the corresponding P-V diagram.
Answer:
A process during which there is no transfer of heat (energy) from the system to the surroundings or from the surroundings to the system is called an adiabatic process.

It can be shown that if an ideal gas is subjected to an adiabatic process, then,
PV^γ = constant = C, where γ, is \(\frac{C_{P}}{C_{V}}\). γ is called the adiabatic ratio. C_P is the molar specific heat of the gas at constant pressure and C_V is the molar specific heat at constant volume.
Let P_i = initial pressure, P_i final pressure V_f = initial volume and V_f = final volume of the gas taken through an adiabatic process.

Now, P_iV_i = nRT_i and P_fV_f = nRT_f, where n is the number of moles of the gas, T_i is the initial temperature of the gas, T_f is the final temperature of the gas and R is the universal gas constant.

[Note: We have Q = ∆ U + W = 0 in an adiabatic process.
∴ W= -∆ U = -nC_v(T_f – T_i)_nC_v(T_i – T_f)]

Question 38.
What is a cyclic process? Explain with a diagram.
Answer:
A thermodynamic process in which the system returns to its initial state is called a cyclic process. This is illustrated in below figure. Path 1 shows how the state of the system (ideal gas) is changed from (P_i, V_i) [point A] to (P_f, V_f) [point B], Path 2 shows the return of the system from point B to point A. As the system returns to its initial state, the total change in its internal energy is zero. Hence, according to the first law of thermodynamics, heat supplied,

Q = ∆ U + W = 0 + W = W. The area enclosed by the cycle in P-V plane gives the work done (W) by the system.

Question 39.
Explain the term free expansion of a gas.
Answer:
When a balloon is ruptured suddenly, or a tyre is punctured suddenly, the air inside the balloon/ tyre rushes out rapidly to the atmosphere. This process (expansion of air inside the balloon/tyre) is so quick that there is no time for transfer of heat from the system to the surroundings or from the surroundings to the system. Such an adiabatic expansion is called free expansion. It is characterized by Q = W = 0, implying ∆ U = 0. Free expansion is an uncontrolled change and the system is not in thermodynamic equilibrium. Free expansion cannot be illustrated with a P-V diagram as only the initial state and final state are known.

Question 40.
Solve the following :

Question 1.
A gas enclosed in a cylinder fitted with a movable, massless and frictionless piston is expanded so that its volume increases from 5 L to 6 L at a constant pressure of 1.013 × 10⁵ Pa. Find the work done by the gas in this process.
Solution :
Data : P = 1.013 × 10⁵ Pa, V_i = 5L = 5 × 10^-3 m³,
v_f = 6L = 6 × 10^-3 m³
The work done by the gas, in this process,
W = P(V_f – V_i)
= (1.013 × 10⁵)(6 × 10^-3 – 5 × 10^-3)J
= 1.013 × 10²J

Question 2.
The initial pressure and volume of a gas enclosed in a cylinder are respectively 1 × 10⁵ N/m² and 5 × 10^-3 m³. If the work done in compressing the gas at constant pressure is 100 J. Find the final volume of the gas.
Solution :
Data : P = 1 × 10⁵ N/m², V_i = 5 × 10^-3 m³,
W= -100 J
W = P(V_f – V_i) ∴ V_f – V_i = \(\frac{W}{P}\)
∴ V_f = V_i + \(\frac{W}{P}\) = 5 × 10^-3 + \(\frac{(-100)}{\left(1 \times 10^{5}\right)}\)
= 5 × 10^-3 -1 × 10^-3 = 4 × 10^-3 m³
This is the final volume of the gas.

Question 3.
If the work done by a system on its surroundings is 100 J and the increase in the internal energy of the system is 100 cal, what must be the heat supplied to the system? (Given : J = 4.186J/cal)
Solution :
Data : W = 100 J, ∆ U = 100 cal, J = 4.186 J / cal
The heat supplied to the system,
Q = ∆ U + W = (100 cal) (4.186 J/cal) + 100 J
= 418.6J + 100J = 518.6 J

Question 4.
Ten litres of water are boiled at 100°C, at a pressure of 1.013 × 10⁵ Pa, and converted into steam. The specific latent heat of vaporization of water is 539 cal/g. Find
(a) the heat supplied to the system
(b) the work done by the system
(c) the change in the internal energy of the system. 1 cm³ of water on conversion into steam, occupies 1671 cm³ (J = 4.186 J/cal)
Solution :
Data : P = 1.013 × 10⁵ Pa, V (water) = 10 L = 10 × 10^-3 m³, V(steam) = 1671 × 10 × 10^-3 m³, L = 539 cal/g = 539 × 10³ \(\frac{\mathrm{cal}}{\mathrm{kg}}\) = 539 × 10³ × 4.186\(\frac{\mathrm{J}}{\mathrm{kg}}\) as J = 4.186 J/cal, mass of the water (M) = volume × density = 10 × 10^-3 m³ × 10³ kg/m³ = 10 kg

(a) Q = ML = (10) (5.39 × 4.186 × 10⁵) J
= 2.256 × 10⁷J
This is the heat supplied to the system

(b) W = P∆V = (1.013 × 10⁵) (1671 – 1) × 10^-2J
= (1.013) (1670) × 10³ J = 1.692 × 10⁶ J
This is the work done by the system.

(c) ∆ U = Q – W = 22.56 × 10⁶ – 1.692 × 10⁶
= 2.0868 × 107J
This is the change (increase) in the internal energy of the system.

Question 5.
Find the heat needed to melt 100 grams of ice at 0°C, at a pressure of 1.013 × 10⁵ N/m². What is the work done in this process ? What is the change in the internal energy of the system ?
Given : Specific latent heat of fusion of ice = 79.71 cal/g, density of ice = 0.92 g/cm³, density of water = 1 g/cm³,1 cal = 4.186 J.
Solution :

1. The heat needed to melt the ice, Q = ML = (0.1) (3.337 × 10⁵) J = 3.337 × 10⁴ J
2. The work done, W = P(V_Water – V_ice)
   = (1.013 × 10⁵) (100 – 108.7) × 10^-6J = -0.8813J
3. The change in the internal energy,
   ∆ U = Q – W = 3.337 × 10⁴ J + 0.8813 J
   = 3370.8813 J

Question 6.
Find the work done by the gas when it is taken through the cycle shown in the following figure. (1 L = 10^-3 m³)
Solution :

∴ W_ABCDA = W_AB + W_BC + W_CD + W_DA
= 2000J + 0 – 1000J + 0 = 1000J

Question 7.
A gas with adiabatic constant γ = 1.4, expands adiabatically so that the final pressure becomes half the initial pressure. If the initial volume of the gas 1 × 10^-2 m³, find the final volume.
Solution:

This is the final volume of the gas.

Question 8.
In an adiabatic compression of a gas with γ = 1.4, the initial temperature of the gas is 300 K and the final temperature is 360 K. If the initial volume of the gas is 2 × 10^-3 m³, find the final volume.
Solution:
Data: γ = 1.4, T_i = 300 K, T_f = 360 K
∴ T_f/T_i = 1.2, V_i = 2 × 10^-3 m³

This is the final volume of the gas.

Question 9.
In an adiabatic compression of a gas with γ = 1.4, the final pressure is double the initial pressure. If the initial temperature of the gas is 300 K, find the final temperature.
Solution:
Data: γ = 1.4, P_f = 2P_i, T_i = 300 K

∴ log \(\frac{T_{\mathrm{f}}}{T_{\mathrm{i}}}\) = 0.2857 log 2 = 0.2857 (0.3010) = 0.086
∴ \(\frac{T_{\mathrm{f}}}{T_{\mathrm{i}}}\) = antilog 0.086 = 1.219
∴ T_f = 1219T_i
= (1.219) (300) = 365.7 K
This is the final temperature of the gas.

Question 10.
In an adiabatic compression of a gas the final volume of the gas is 80% of the initial volume. If the initial temperature of the gas is 27 °C, find the final temperature of the gas. Take γ = 5/3.
Solution :
Data: V_f = 0.8 V_i ∴ \(\frac{V_{\mathrm{i}}}{V_{\mathrm{f}}}\) = \(\frac{10}{8}\) = 1.25,
∴ T_i = 27 °C = (273 + 27) K = 300 K, γ = \(\frac{5}{3}\)

∴ x = antilog 0.0646 = 1.161
∴ T_f = (300) (1.161) = 348.3 K
= (348.3 – 273)°C = 75.3 °C
This is the final temperature of the gas.

Question 11.
In an adiabatic expansion of a gas, the final volume of the gas is double the initial volume. If the initial pressure of the gas is 105 Pa, find the final pressure of the gas. (γ = 5/3)
Solution :

Let 2^5/3 = x ∴ \(\frac{5}{3}\)log 2 = log x
∴ log x = \(\left(\frac{5}{3}\right)\) (0.3010) = 0.5017
∴ x = antilog 0.5017 = 3.175
∴ P_f = \(\frac{10^{5}}{3.175}\) = 3.15 × 10⁴ Pa
This is the final pressure of the gas.

Question 12.
In an adiabatic process, the final pressure of the gas is half the initial pressure. If the initial temperature of the gas is 300 K, find the final temperature of the gas. (Take γ = \(\frac{5}{3}\))
Solution :

This is the final temperature of the gas.

Question 13.
In an adiabatic process, the pressure of the gas drops from 1 × 10⁵ N/m² to 5 × 10⁴ N/m² and the temperature drops from 27 °C to – 46 °C. Find the adiabatic ratio for the gas.
Solution :


This is the adiabatic ratio (γ) for the gas.
[Note : This value (1.673) is slightly more than 5/3 (the value for a monatomic gas) due to error in measurement of pressure and temperature.]

Question 14.
Two moles of a gas expand isothermally at 300 K. If the initial volume of the gas is 23 L and the final volume is 46 L, find the work done by the gas on its surroundings. (R = 8.314 J/mol.K)
Solution :
Data ; n = 2, T = 300 K, V, = 23 L = 23 × 10^-3 m³, V_f = 46 L = 46 × 10^-3 m³, R = 8.314 J/mol.K
The work done by the gas on its surroundings,
W = nRT ln \(\left(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\right)\) = 2.303 nRT log₁₀ \(\left(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\right)\)
= (2.303) (2) (8.314) (300) log₁₀ \(\left(\frac{46 \times 10^{-3}}{23 \times 10^{-3}}\right)\)
= (4.606) (8.314) (300) log\(\begin{array}{r}
2 \\
10
\end{array}\)
= (4.606) (8.314) (300) (0.3010)
= 3458J

Question 15.
Four moles of a gas expand isothermally at 300 K. If the final pressure of the gas is 80% of the initial pressure, find the work done by the gas on its surroundings. (R = 8.314 J/mol.K)
Solution :
Data : n = 4, T = 300 K, P_f = 0.8 P_i
∴ \(\frac{P_{i}}{P_{f}}\) = \(\frac{10}{8}\), R = 8.314 j/mol.K
The work done by the gas on its surroundings,
W = nRT ln\(\left(\frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}\right)\)
= (4) (8.314) (300) 2.303 log₁₀ \(\left(\frac{10}{8}\right)\)
= 2.3 × 10⁴ log₁₀ (1.25) = 2.3 × 10⁴ × 0.0969
= 2.229 × 10³J

Question 16.
The molar specific heat of He at constant volume is 12.47 J/mol.K. Two moles of He are heated at constant pressure so that the rise in temperature is 10 K. Find the increase in the internal energy of the gas.
Solution :
Data : C_v = 12.47 J/mol.k, n = 2, T_f – T_i = 10 K
The increase in the internal energy of the gas,
∆ U = _nC_v (T_f – T_i)
= (2) (12.47) (10) J
= 249.4 J

Question 17.
The molar specific heat of Ar at constant volume is 12.47 J/mol.K. Two moles of Ar are heated at constant pressure so that the rise in temperature is 20 K. Find the work done by the gas on its surroundings and the heat supplied to the gas. Take R = 8.314 J/mol.K.
Solution :
Data : C_v = 12.47 j/mol.K, n = 2,T_f – T_i = 20 K,
R = 8.314 J/mol.K

1. W = nR (T_f – T_i) = (2) (8.314) (20) J = 332.6 J
   This is the work done by the gas on its surroundings.
2. Q = _nC_v(T_f – T_i) + W = (2) (12.47) (20) + 332.6
   = 498.8 + 332.6 = 831.4 J
   This is the heat supplied to the gas.

Question 18.
The molar specific heat of a gas at constant pressure is 29.11 J/mol.k. Two moles of the gas are heated at constant pressure so that the rise in temperature is 40 K. Find the heat supplied to the gas.
Solution :
Data : C_P = 29.11 J/mol.K, n = 2, T_f – T_i = 40 K.
The heat supplied to the gas,
Q = _nC_P (T_f – T_i) = (2) (29.11) (40) J
= 2329 J

Question 19.
The molar specific heat of a gas at constant volume is 20.8 J/mol.k. Two moles of the gas are heated at constant volume so that the rise in temperature is 10 K. Find the heat supplied to the gas.
Solution :
Data : C_v = 20.8 J/mol.K, n = 2, T_f – T_i = 10 K.
The heat supplied to the gas,
Q = _nC_v (T_f – T_i) = (2) (20.8) (10) J
= 416J

Question 20.
In an adiabatic expansion of 2 moles of a gas, the initial pressure was 1.013 × 10⁵ Pa, the initial volume was 22.4 L, the final pressure was 3.191 × 10⁴ Pa and the final volume was 44.8 L. Find the work done by the gas on its surroundings. Take γ = 5/3.
Solution :
Data : H = 2, P_i = 1.013 × 10⁵ Pa, P_f = 3.191 × 10⁴ Pa,
V_i = 22.4 L = 22.4 × 10^-3 m³,
V_f = 44.8 L = 44.8 × 10^-3 m³, γ = 5/3

Question 21.
In an adiabatic expansion of 2 moles of a gas, the temperature of the gas decreases from 37°C to 27°C. Find the work done by the gas on its surroundings. Take γ = 5/3 and R = 8.314 J/mol.K
Answer:
Data: n =2, T_i = (273 + 37) = 310 K,
T_f = (273 + 27)K = 300K, γ = 5/3,
R = 8.314 J/mol.K.
The work done by the gas on its surroundings,

Question 22.
A resistor of resistance 200 Ω carries a current of 2 A for 10 minutes. Assuming that almost all the heat produced in the resistor is transferred to water (mass = 5 kg, specific heat capacity = 1 kcal/kg), and the work done by the water against the external pressure during the expansion of water can be ignored, find the rise in the temperature of the water. (J = 4186 J/cal)
Solution :
Data : I = 2 A, R = 200 Ω, t = 10 min = 10 × 60 s = 600 s, M = 5 kg, S = (1 kcal/kg) (4186 J/kcal)
= 4186 J/kg
Q = ∆ U + W = MS ∆T + W \(\simeq\) MS ∆T ignoring W.
Also, Q = I²Rt ∴ I²RT = MS∆T
∴ The rise in the temperature of water = ∆T = \(\frac{I^{2} R t}{M S}\)
= \(\frac{(2)^{2}(200)(600)}{(5)(4186)}\)°C = 22.93°C

Question 23.
The initial pressure, volume and temperature of a gas are respectively 1 × 10⁵ Pa, 2 × 10^-2 m³ and 400 K. The temperature of the gas is reduced from 400 K to 300 K at constant volume. Then the gas is compressed at constant temperature so that its volume becomes 1.5 × 10^-2 m³.
Solution:
Data : P_A = 1 × 10⁵ Pa, V_A = 2 × 10^-2 m³, T_A = 400 K, V_B = v_A = 2 × 10^-2 m³, T_B = 300 K, T_C = T_B = 300 K, V_C = 1.5 x 10× 10^-2 m³, Also, P_C = P_A = 1 × 10⁵ Pa as the gas returns to its initial state.

Question 24.
If the adiabatic ratio for a gas is 5/3, find the molar specific heat of the gas at
(i) constant volume
(ii) constant pressure.
(R = 8.314 J/mol. K)
Solution :
Data : r = 5/3, R = 8.314 J/mol.K

This is the required quantity.
(ii) C_P = _γC_V = \(\left(\frac{5}{3}\right)\)(12.47) = 20.78 J/mol.

Question 41.
What is a heat engine ?
Answer:
A heat engine is a device in which a system is taken through cyclic processes that result in converting part of heat supplied by a hot reservoir into work (mechanical energy) and releasing the remaining part to a cold reservoir. At the end of every cycle involving thermodynamic changes, the system is returned to the initial state.
[Note : Automobile engine is a heat engine.]

Question 42.
What does a heat engine consist of ?
OR
What are the elements (parts) of a typical heat engine?
Answer:
The following are the parts of a typical heat engine :

(1) Working substance : It can be

1. a mixture of fuel vapour and air in a gasoline (petrol) engine or diesel engine
2. steam in a steam engine. The working substance is called a system.

(2) Hot and cold reservoirs : The hot reservoir is a source of heat that supplies heat to the working substance at constant temperature T_H. The cold reservoir, also called the sink, takes up the heat released by the working substance at constant temperature T_C < T_H.

(3) Cylinder and piston : The working substance is enclosed in a cylinder fitted with (ideally) a movable, massless, and frictionless piston. The walls of the cylinder are nonconducting, but the base is conducting. The piston is nonconducting. The piston is connected to a crankshaft so that the work done by the working substance (mechanical energy) can be transferred to the environment.

Question 43.
What are the two basic types of heat engines?
Answer:
(i) External combustion engine in which the working substance is heated externally as in a steam engine.
(ii) Internal combustion engine in which the working substance is heated internally as in a petrol engine or diesel engine.
[Note : A steam engine was invented by Thomas New-comen (1663-1729), English engineer. The first practical steam engine was constructed in 1712. The modem steam engine was invented in 1790 by James Watt (1736-1819), British instrument maker and engineer. A hot-air type engine was developed by Robert Stirling (1790-1878), Scottish engineer and clergyman.

A four-stroke internal combustion engine was devised by Nikolaus August Otto (1832-1891), German engineer. A compression-ignition internal combustion engine was devised by Rudolph (Christian Karl) Diesel (1858 – 1913), German engineer.]

Question 44.
State the basic steps involved in the working of a heat engine.
Answer:

1. The working substance absorbs heat (Q_H) from a hot reservoir at constant temperature. T_H. It is an isothermal process Q_H is positive.
2. Part of the heat absorbed by the working substance is converted into work (W), i.e. mechanical energy. In this case, there is a change in the volume of the substance.
3. The remaining heat (|Q_C| = |Q_H| – W) is transferred to a cold reservoir at constant temperature T_C < T_H. Q_C is negative.

Question 45.
Draw a neat labelled energy flow diagram of a heat engine.
Answer:

Question 46.
Define thermal efficiency of a heat engine.
Answer:
The thermal efficiency, η of a heat engine is defined W as η = \(\frac{W}{Q_{H}}\), where W is the work done (output) by Q_H the working substance and Q_H is the amount of heat absorbed (input) by it.
[Note : η has no unit and dimensions or its dimensions are [M°L°T°].]

Question 47.
Draw a neat labelled P-V diagram for a typical heat engine.
Answer:

Here, T_H is the temperature at which the work is done by the gas and T_c is the temperature at which the work is done on the gas. The area of the loop ABCDA is the work output.

Question 48.
Solve the following :

Question 1.
Find the thermal efficiency of a heat engine if in one cycle the work output is 3000 J and the heat input is 10000 J.
Solution :
Data : W = 3000 J, Q_H = 10000 J
The thermal efficiency of the engine,
η = \(\frac{W}{Q_{\mathrm{H}}}\) = \(\frac{3000 \mathrm{~J}}{10000 \mathrm{~J}}\) = 0.3 = 30%

Question 2.
The thermal efficiency of a heat engine is 25%. If in one cycle the heat absorbed from the hot reservoir is 50000 J, what is the heat rejected to the cold reservoir in one cycle ?
Solution :
Data : η = 25% = 0.25, Q_H = 50000 J W
η = \(\frac{W}{Q_{\mathrm{H}}}\)
∴ W = _ηQ_H = (0.25)(50000)J = 12500J
Now, W = Q_H – |Q_C|
∴ |Q_C| = Q_H – W
= (50000 – 12500) J
= 37500J
This is the heat rejected to the cold reservoir in one cycle.
[Notes : Q_C = – 37500 J]

Question 49.
What is a refrigerator?
Answer:
A refrigerator is a device that uses work to transfer energy in the form of heat from a cold reservoir to a hot reservoir as it continuously repeats a thermodynamic cycle. Thus, it is a heat engine that runs in the backward direction.

Question 50.
With a neat labelled energy flow diagram, explain the working of a refrigerator.
Answer:
A refrigerator performs a cycle in a direction opposite to that of a heat engine, the net result being absorption of some energy as heat from a reservoir at low temperature, a net amount of work done on the system and the rejection of a larger amount

of energy as heat to a reservoir at a higher temperature. The working substance undergoing the refrigeration cycle is called a refrigerant. The refrigerant (such as ammonia or Freon) is a saturated liquid at a high pressure and at as low a temperature as can be obtained with air or water cooling.

The refrigeration cycle comprises the following processes :

1. Throttling process : The saturated liquid refrigerant passes from the condenser through a narrow opening from a region of constant high pressure to a region of constant lower pressure almost adiabatically. It is a property of saturated liquids (not gases) that a throttling process produces cooling and partial vaporization.
2. Isothermal, isobaric vaporization-with the heat of vaporization being supplied by the materials or the region to be cooled : Heat Q_c is absorbed by the refrigerant at the low temperature T_C, thereby cooling the materials of the cold reservoir.
3. Adiabatic compression of the refrigerant by an electrical compressor, thereby raising its temperature above T_H.
4. Isobaric cooling and condensation in the condenser at T_H : In the condenser, the vapour is cooled until it condenses and is completely liquefied, i. e., heat Q_H is rejected to the surroundings which is the hot reservoir.

Question 51.
Draw a neat labelled schematic diagram of transferring heat from a cold region to a hot region.
Answer:

Question 52.
What is refrigeration?
Answer:
Refrigeration is artificial cooling of a space or substance of a system and/or maintaining its temperature below the ambient temperature.

Question 53.
Draw a neat labelled diagram to illustrate schematics of a refrigerator.
Answer:

Question 54.
What are the steps through which a refrigerant goes in one complete cycle of refrigeration ?
Answer:
In one complete cycle of refrigeration, the refrigerant, a liquid such as fluorinated hydrocarbon, goes through the following steps :

1. The refrigerant in the closed tube passes through the nozzle and expands, into a low-pressure area. This adiabatic expansion results in reduction in pressure and temperature of the fluid and the fluid turns into a gas.
2. The cold gas is in thermal contact with the inner compartment of the fridge. Here it absorbs heat at constant pressure from the contents of the fridge.
3. The gas passes to a compressor where it does work in adiabatic compression. This raises its temperature and converts it back into a liquid.
4. The hot liquid passes through the coils on the outside of the refrigerator and releases heat to the air outside in an isobaric compression process.
   The compressor, driven by an external source of energy, does work on the refrigerant during each cycle.

Question 55.
Explain the energy flow in a refrigerator and define the coefficient of performance of a refrigerator.
Answer:
In a refrigerator, Q_C is the heat absorbed by the working substance (refrigerant) at a lower temperature T_C, W is the work done on the working substance, and Q_H is the heat rejected at a higher temperature T_H. The absorption of heat is from the contents of the refrigerator and rejection of heat is to the atmosphere. Here, Q_C is positive and W and Q_H are negative. In one cycle, the total change in the internal energy of the working substance is zero.
∴ Q_H + Q_C = W ∴ Q_H = W – Q_C
∴ -Q_H = Q_C – W
Now, Q_H < 0, W < 0 and Q_C > 0
∴ |Q_H| = |Q_C| + |W|
The coefficient of performance (CoP), K, or quality factor, or Q value of a refrigerator is defined as

[Note: K does not have unit and dimensions or its dimensions are [M°L°T°.]

Question 56.
How does an air conditioner differ from a refrigerator? Define the coefficient of performance of an air conditioner and express it in terms of heat current.
Answer:
The working of an air conditioner is exactly similar to that of a refrigerator, but the volume of the chamber/room cooled by an air conditioner is far greater than that in a refrigerator. The evaporator coils of an air conditioner are inside the room, and the condenser outside the room. A fan inside the air conditioner circulates cool air in the room.

The coefficient of performance, K, of an air conditioner is defined as K = \(|\frac{Q_{\mathrm{C}}}{W}|\), where Q_C is the heat absorbed and W is the work done. The time rate of heat removed is the heat current, H = \(\frac{\left|Q_{C}\right|}{t}\), where t is the time in which heat |Q_C|, is removed.

where H = |Q_C|/t is the heat current and P( = |W|/t) is the time rate of doing work, i.e., power.

Question 57.
What is a heat pump ?
Answer:
A heat pump is a device used to heat a building by cooling the air outside it. It works like a refrigerator but cooling outside space and heating inside space. In this case, the evaporator coils are outside the building to absorb heat from the air. The condenser coils are inside the building to release the heat to warm the building.

Question 58.
Solve the following :

Question 1.
In a refrigerator, in one cycle, the external work done on the working substance is 20% of the energy extracted from the cold reservoir. Find the coefficient of performance of the refrigerator.
Solution :
Data: |W| = 0.2|Q_C|
The coefficient of performance of the refrigerator,
K = |\(\frac{Q_{C}}{W}\)| = \(\frac{\left|Q_{C}\right|}{0.2\left|Q_{C}\right|}\)
= 5

Question 2.
The coefficient of performance of a room air conditioner is 3. If the rate of doing work is 2kW, find the heat current.
Solution :
Data : K = 3, P = 2000 W
K = \(\frac{H}{P}\)
∴ Heat current, H = KP = (3) (2000) W
= 6000 W = 6kW

Question 59.
State and explain the limitations of the first law of thermodynamics.
Answer:

1. The first law of thermodynamics is essentially the principle of conservation of energy as there is a close relation between work and energy. We find that there is a net transfer of energy (heat) from a body at higher temperature to a body at lower temperature. The net transfer of energy from a body at lower temperature to a body at higher temperature is not observed though consistent with the first law of thermodynamics.
2. If two containers, one containing nitrogen and the other containing oxygen, are connected to allow diffusion, they mix readily. They never get separated into the respective containers though there is no violation of the first law of thermodynamics.
3. It is not possible to design a heat engine with 100% efficiency, though there is no restriction imposed by the first law of thermodynamics.
4. At room temperature, ice absorbs heat from the surrounding air and melts. The process in the opposite direction is not observed, though consistent with energy conservation. These examples suggest that there is some other law operative in nature that determines the direction of a process

Question 60.
State the two forms of the second law of thermodynamics.
Answer:
Second law of thermodynamics :

1. It is impossible to extract an amount of heat Q_H from a hot reservoir and use it all to do work W. Some amount of heat Q_C must be exhausted (given out) to a cold reservoir. This prohibits the possibility of a perfect heat engine.
   This statement is called the first form or the engine law or the engine statement or the Kelvin-Planck statement of the second law of thermodynamics.
2. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this. This prohibits the possibility of a perfect refrigerator.
   This statement is called the second form or the Clausius statement of the second law of thermodynamics.

Notes :

1. Max Planck (Karl Ernst Ludwig) (1858-1947) German physicist, put forward quantum therory of radiation.
2. Rudolf Clausius (1822-88) German theoretical physicist, made significant contribution to thermodynamics.

Question 61.
Draw neat labelled diagrams to illustrate
(i) energy flow diagram for engine statement.
(ii) energy flow diagram for a perfect refrigerator.
Answer:

Question 62.
State the difference between a reversible process and an irreversible process.
OR
Distinguish between a reversible process and an irreversible process.
Answer:
A reversible process is a bidirectional process, i.e., its path in P-V plane is the same in either direction. In contrast, an irreversible process is a undirectional process, i.e., it can take place only in one direction.

A reversible process consists of a very large number of infinitesimally small steps so that the system is all the time in thermodynamic equilibrium with its environment. In contrast an irreversible process may occur so rapidly that it is never in thermodynamic equilibrium with its environment.

Question 63.
Draw a neat labelled diagram of a Carnot cycle and describe the processes occurring in a Carnot engine. Write the expression for the efficiency of a Carnot engine.
Answer:
Basically, two processes occur in a Carnot engine :

(1) Exchange of heat with the reservoirs : In isothermal expansion AB, the working substance takes in heat Q_H from a lot reservoir (source) at constant temperature T_H. In isothermal compression CD, the working substance gives out heat Q_C to a cold reservoir (sink) at constant temperature T_C.

(2) Doing work : In adiabatic expansion BC, the working substance does work on the environment and in adiabatic compression DA, work is done on the working substance by the environment.
All processes are reversible. It can be shown that \(\frac{\left|Q_{C}\right|}{Q_{\mathrm{H}}}\) = \(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\). Hence, the efficiency of a Carnot engine,

Question 64.
What is a Carnot refrigerator? State the expressions for the coefficient of performance of a Carnot refrigerator.
Answer:
A Carnot refrigerator is a Carnot engine operated in the reverse direction. Here, heat Q_C is absorbed from a cold reservoir at temperature T_C, work W is provided externally, and heat Q_H is given out to a hot reservoir at temperature T_H.
The coefficient of performance of a Carnot refrigerator is

[Note: K is large if T_H-T_C is small. It means a large quantity of heat can be removed from the body at lower temperature to the body at higher temperature by doing small amount of work. K is small if T_H – T_C is large.
It means a small quantity of heat can be removed from the body at lower temperature to the body at higher temperature even with large amount of work.]

Question 65.
Solve the following :

Question 1.
A Carnot engine receives 6 × 10⁴ J from a reservoir at 600 K, does some work, and rejects some heat to a reservoir at 500 K. Find the
(i) the heat rejected by the engine
(ii) the work done by the engine
(iii) the efficiency of the engine.
Solution :
Data : Q_H = 6 × 10⁴J, T_H = 600K, T_C = 500K
(i) The heat rejected by the engine

Question 2.
A Carnot engine operates between 27 °C and 87 °C. Find its efficiency.
Solution :
Data : T_C = 27 °C = (273 + 27) K = 300 K,
T_H = 87 °C = (273 + 87) = 360 K
The efficiency of the engine,

Question 3.
The coefficient of performance of a Carnot refrigerator is 4. If the temperature of the hot reservoir is 47 °C, find the temperature of the cold reservoir.
Solution :
Data : K = 4, T_H = 47°C = (273 + 47) K = 320
K = \(\frac{T_{\mathrm{c}}}{T_{\mathrm{H}}-T_{\mathrm{c}}}\) ∴ KT_H – KT_c = T_c
∴ KT_H = (1 + K)T_c
∴ T_c = \(\frac{K T_{\mathrm{H}}}{1+K}\) = \(\frac{(4)(320)}{1+4}\)K = (0.8)(320)K
= 256K = (256 – 273)°C = – 17°C
This is the temperature of the cold reservoir.

[Note : A hot-air type engine consisting of two cylinders, was developed by Robert Stirling (1790 -1878), a Scottish engineer and clergyman. He developed the concept in 1816 and obtained a patent for his design in 1827. Some engines were made in 1844. He also used helium and hydrogen in some engines developed thereafter. Stirling engines are used in submarines and spacecrafts.]

Question 66.
Draw a neat labelled diagram of a Sterling cycle and describe the various processes taking place in a Sterling engine.

Answer:
The working substance can be air or helium or hydrogen or nitrogen. All processes are reversible.

1. AB is isothermal expansion, at temperature T_H, in which heat Q_H is absorbed from the source and useful work is done by the working substance.
2. BC is isochoric process in which some heat is released by the gas (working substance) to the refrigerator and the gas cools to temperature T_c < T_H.
3. CD is isothermal compression, at temperature T_c, in which heat Q_c is rejected to the coolant (sink).
4. DA is isochoric process in which heat is taken in by the gas and its temperature rises to T_H.

[Note : Stirling engine operated in reverse direction is used in the field of cryogenics to obtain extremely low – temperatures to liquefy air or other gases.]

Question 67.
Refer above figure and answer the following questions.
(i) What is the work done in process AB ?
(ii) What is the change in internal energy and heat released in process BC ?
Answer:
(i) In this case, the change in the internal energy is zero, as the temperature of the gas remains constant. Hence, the work done, W = heat absorbed, Q_H.

(ii) In this case, the change in the internal energy, ∆ U = _nC_V (T_C – T_H), where n = number of moles of the gas used in the Stirling engine and C_V = molar specific heat of the gas. As W = 0 at constant volume, heat released’= ∆ U.

Question 68.
Choose the correct option :

Question 1.
According to the first law of thermodynamics, in the usual notation,
(A) Q = ∆U + W
(B) Q = ∆U – W
(C) Q = W – ∆U
(D) Q= -(∆ U + W).
Answer:
(A) Q = ∆U + W

Question 2.
In an isothermal process, in the usual notation,
(A) PV = constant
(B) V/T = constant
(C) P/T = constant
(D) Q = 0.
Answer:
(A) PV = constant

Question 3.
In an isothermal process, in the usual notation,
(A) W = nRT (V_f/V_i)
(B) W = ∆ U
(C) W = Q
(D) W = 0.
Answer:
(C) W = Q

Question 4.
In an adiabatic process, in the usual notation,
(A) TV^γ = constant
(B) PT^γ = constant
(C) W = 0
(D) PV^γ = constant.
Answer:
(D) PV^γ = constant.

Question 5.
In an isobaric process, in the usual notation,
(A) W = P (V_f – V_i)
(B) W = Q
(C) W = – ∆U
(D) ∆T = 0.
Answer:
(A) W = P (V_f – V_i)

Question 6.
In an adiabatic process, in the usual notation,
(A) ∆ P = 0
(B) ∆ V = 0
(C) Q = 0
(D) ∆U = 0.
Answer:
(C) Q = 0

Question 7.
In an isothermal process, in the usual notation,
(A) W = P(V_f – V_i)
(B) W = 0
(C) W = V(P_f – P_i)
(D) W = nRT In(V_f/V_i).
Answer:
(D) W = nRT In(V_f/V_i).

Question 8.
In an isobaric process, in the usual notation,
(A) W = _nC_V (T_f – T_i)
(B) Q = _nC_P (T_f – T_i)
(C) ∆U = nR(T_f – T_i)
(D) W = 0.
Answer:
(B) Q = _nC_P (T_f – T_i)

Question 9.
In the usual notation, the isothermal work, W =
(A) P(V_f – V_i)
(B) nRT(P_i/ P_f)
(C) nRT ln(P_i/P_f)
(D) nRT(P_f/P_i).
Answer:
(C) nRT ln(P_i/P_f)

Question 10.
If Q and ∆u are expressed in cal and W is expressed in joule, then,
(A) \(\frac{Q}{J}\) = \(\frac{\Delta U}{J}\) + W
(B) Q = ∆U – (W/J)
(C) \(\frac{Q}{J}\) = \(\frac{\Delta U}{J}\) + W
(D) Q = ∆U + (W/J)
Answer:
(D) Q = ∆U + (W/J)

Question 11.
In an adiabatic process, in the usual notation, W =

Answer:
(A) \(\frac{P_{\mathrm{i}} V_{\mathrm{i}}-P_{\mathrm{f}} V_{\mathrm{f}}}{\gamma-1}\)

Question 12.
In a cyclic process,
(A) ∆U = Q
(B) Q = 0
(C) W = 0
(D) W = Q
Answer:
(D) W = Q

Question 13.
The efficiency of a heat engine is given by η =
(A) Q_H/W
(B) W/Q_c
(C) W/Q_H
(D) Q_c/W.
Answer:
(C) W/Q_H

Question 14.
In a cyclic process, the area enclosed by the loop in the P – V plane corresponds to
(A) ∆U
(B) W
(C) Q – W
(D) W – Q.
Answer:
(B) W

Question 15.
The efficiency of a Carnot engine is given by K =
(A) T_c/(T_H – T_c)
(B) (T_H – T_c)/T_c
(C) T_H/(T_H – T_c)
(D) (T_H – T_c)/T_H
Answer:
(A) T_c/(T_H – T_c)

Question 16.
The coefficient of performance of a Carnot refrigerator is given by K =
(A) T_c(T_H-T_c)
(B) (T_H-T_c)/T_c
(C) T_H/(T_H-T_c)
(D) (T_H-T_c)/T_H.
Answer:
(A) T_c(T_H-T_c)

Question 17.
The efficiency of a Carnot engine working between T_H = 400 K and T_c = 300 K is
(A) 75%
(B) 25%
(C) 1/3
(D) 4/7.
Answer:
(B) 25%

Question 18.
If a Carnot engine receives 5000 J from a hot reservoir and rejects 4000 J to a cold reservoir, the efficiency of the engine is
(A) 25%
(B) 10%
(C) 1/9
(D) 20%.
Answer:
(D) 20%.

Question 19.
If a Carnot refrigerator works between 250 K and 300 K, its coefficient of performance =
(A) 6
(B) 1.2
(C) 5
(D) 10.
Answer:
(C) 5

Question 20.
The coefficient of performance of a Carnot refrigerator is 4. If T_c = 250 K, then T_H =
(A) 625 K
(B) 310 K
(C) 312.5 K
(D) 320 K.
Answer:
(C) 312.5 K

Question 21.
The coefficient of performance of a Carnot refrigerator working between T_H and T_c is K and the efficiency of a Carnot engine working between the same T_H and T_c is η. Then
(A) ηk = \(\frac{Q_{\mathrm{H}}}{Q_{\mathrm{c}}}\)
(B) η/k = Q_c/Q_H
(C) η/k = Q_H/Q_c
(D) ηk = \(\frac{Q_{c}}{Q_{H}}\)
Answer:
(D) ηk = \(\frac{Q_{c}}{Q_{H}}\)

Question 22.
The internal energy of one mole of organ is
(A) \(\frac{5}{2}\)RT
(B) RT
(C) \(\frac{3}{2}\)RT
(D) 3RT.
Answer:
(C) \(\frac{3}{2}\)RT

Question 23.
The internal energy of one mole of oxygen is
(A) \(\frac{5}{2}\)RT
(B) 5RT
(C) \(\frac{3}{2}\)RT
(D) 3RT.
Answer:
(C) \(\frac{3}{2}\)RT

Question 24.
The internal energy of one mole of nitrogen at 300 K is about 6225 J. Its internal energy at 400 K will be about
(A) 8300J
(B) 4670J
(C) 8500J
(D) 8000J.
Answer:
(A) 8300J

Question 25.
The adiabatic constant γ for argan is
(A) 4/3
(B) 7/5
(C) 6/5
(D) 5/3.
Answer:
(D) 5/3.

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 3 Kinetic Theory of Gases and Radiation Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 3 Kinetic Theory of Gases and Radiation

Question 1.
What is an ideal or perfect gas? Explain.
Answer:
An ideal or perfect gas is a gas that obeys the gas laws (Boyle’s law, Charles’ law, and Gay-Lussac’s law) at all pressures and temperatures. An ideal gas cannot be liquefied by the application of pressure or lowering the temperature.

A molecule of an ideal gas is an ideal particle having only mass and velocity. Its structure and size are ignored. Also, intermolecular forces are zero except during collisions.

Question 2.
Derive the ideal gas equation, PV = nRT.
OR
Derive the equation of state for an ideal gas
Answer:
Let P₁, V_1, and T₁ be the pressure, volume, and absolute temperature (thermodynamic temperature) of n moles of gas (assumed to be ideal).
Suppose the gas is heated at constant pressure (P₁) so that its temperature becomes T and volume becomes V₂. Then by Charles’ law,\(\frac{V_{2}}{V_{1}}\) = \(\frac{T}{T_{1}}\).
∴ T = \(\frac{V_{2}}{V_{1}}\)T₁ … (1)
Now, suppose that the gas is heated at constant volume (V₂) so that its temperature becomes T₂ and pressure becomes P₂. Then by Gay—Lussac’s law,
\(\frac{P_{2}}{P_{1}}\) = \(\frac{T_{2}}{T}\)
∴ T = \(\frac{P_{1}}{P_{2}}\)T₂ … (2)
From equations (1) and (2), we get,
\(\frac{V_{2}}{V_{1}}\)T₁ = \(\frac{P_{1}}{P_{2}}\)T₂
∴ \(\frac{P_{2} V_{2}}{T_{2}}\) = \(\frac{P_{1} V_{1}}{T_{1}}\)
When pressure and temperature are constant, V ∝ n. Hence, in general, we can write \(\frac{P V}{T}\) = nR, where R is a proportionality constant, called the universal gas constant.
Thus, PV = nRT for an ideal gas. This is the equation of state for an ideal gas.

[ Note : By Boyle’s law, for a fixed mass of gas at constant temperature, PV = constant. Writing PV ∝ nT, we get PV = nRT.]

Question 3.
Express the ideal gas equation in terms of
(i) the Avogadro number
(ii) The Boltzmann constant.
Answer:
In the usual notation, PV = nRT;
(i) Now, the number of moles, n = \(\frac{N}{N_{\mathrm{A}}}\), where N is the number of molecules corresponding to n moles of the gas and N_A is the Avogadro number.
∴ PV = \(\frac{N}{N_{\mathrm{A}}}\)RT

(ii) PV = \(\frac{N}{N_{\mathrm{A}}}\)RT = N(\(\frac{R}{N_{\mathrm{A}}}\))T
= Nk_BT, where k_B = \(\frac{R}{N_{\mathrm{A}}}\) is the Boltzmann constant.

Question 4.
Express the ideal gas equation in terms of the density of the gas.
Answer:
In the usual notation, PV = nRT. Now, the number of moles n = \(\frac{M}{M_{0}}\), where M is the mass of the gas M₀ corresponding to n moles and M₀ is the molar mass of the gas.

is the density of the gas.

Notes :

(1) Charles’ law : At a constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (thermodynamic temperature).
If V is the volume of n moles of an ideal gas at a pressure P and T denotes its absolute temperature, P and n remaining constant,
V ∝ T or \(\frac{V}{T}\) = constant

The law was deduced experimentally in 1787 by Jacques Alexandre Cesar Charles (1746 -1823), French physicist.

From Charles’ law, V = KT, where K is a constant which depends on the mass and pressure. If M and ρ are the mass and mass per unit volume of a gas, V = M/ρ.

Charles’ law was properly established only after the publication in 1802 of more accurate experimental results by Joseph Gay-Lussac (1778-1850), French chemist. Hence, the law is also known as Gay-Lussac’s law.

(2) The Boltzmann constant is the ratio of the molar gas constant (R) to the Avogadro constant (Avogadro number) (NA). Its symbol is k_B or k.
k_B = 1.380 648 8(13) × 10^-23 JK^-1

It is a fundamental physical constant relating the average energy of a molecule (at the microscopic level) with temperature (a state variable, at the macroscopic level). It can be called the gas constant per molecule. It is named after Ludwig Eduard Boltzmann (1844-1906), Austrian physicist.

Question 5.
State the basic assumptions of the kinetic theory of gases.
Answer:
The basic assumptions of the kinetic theory of an ideal gas :

1. A gas of a pure material consists of an extremely large number of identical molecules.
2. A gas molecule behaves as an ideal particle, i.e., it has mass but its structure and size can be ignored as compared with the intermolecular separation in a dilute gas and the dimensions of the container.
3. The molecules are in constant random motion with various velocities and obey Newton’s laws of motion.
4. Intermolecular forces can be ignored on the average so that the only forces between the molecules and the walls of the container are contact forces during collisions. It follows that between successive collisions, a gas molecule travels in a straight line with constant speed.
5. The collisions are perfectly elastic conserving total momentum and kinetic energy, and the duration of a collision is very small compared to the time interval between successive collisions.

Notes :

(1) The walls of the container holding the gas are assumed to be rigid and infinitely massive so that they do not move. (2) Assumption (2) allows us to ignore intermolecular collisions because if they are truly point like (i.e., of negligible extent) they cannot make contact with each other. Therefore, the only collisions we consider are those with the walls of the container. If these collisions are perfectly elastic [by assumption (5)], they only change the direction of the velocity of a gas molecule and a gas molecule possesses only kinetic energy by assumption
(4). The kinetic theory of gases was developed by Daniel Bernoulli (1700 – 82), Swiss mathematician, Rudolf Clausius (1822 – 88), German theoretical physicist, James Clerk Maxwell (1831-79), British physicist, Ludwig Eduard Boltzmann, Josiah Willard Gibbs (1839-1903), US physical chemist.]

Question 6.
Distinguish between an ideal gas and a real gas.
Answer:

Ideal Gas	Real Gas
1. Molecules of an ideal gas behave as ideal particles, i.e., they are like geometrical points without size and structure.	1. Molecules of a real gas have finite size and structure.
2. Molecules of an ideal gas have only translational motion.	2. Polyatomic molecules have in general, translational, vibrational and rotational motion.
3. There are no intermolecular forces in this case.	3. Intermolecular forces are non-zero in this case.
4. An ideal gas cannot be liquefied.	4. A real gas can be liquefied and also solidified in many cases, by application of pressure and reducing temperature.
5. There are no intermolecular collisions in this case.	5. There are intermolecular collisions in this case, and under normal conditions the collision frequency is very high.
6. In the absence of intermolecular forces, the internal energy of the gas is only kinetic.	6. Due to the intermolecular forces, the internal energy of the gas is (potential energy + kinetic energy.)

Question 7.
The pressure exerted by a certain mass of enclosed gas at 300 K is 5 × 10⁴ Pa. What will be the pressure exerted by the gas at 600 K if the volume of the gas is kept constant ?
Answer:
Here, P ∝ T ∴\(\frac{P_{2}}{P_{1}}\) = \(\frac{T_{2}}{T_{1}}\) = \(\frac{600}{300}\) = 2
∴ P₂ = 2P₁ = 2 × 5 × 10⁴ Pa = 10⁵ Pa is the required pressure.

Question 8.
Solve the following :

Question 1.
16 g of oxygen occupy 0.025 m3 at 27 °C. Find the pressure exerted by it.
[Molar mass of oxygen = 32 g/mol, universal gas constant R = 8.3 J/mol.K]
Solution :
Data : M = 16 g, M₀ = 32 g/mol, V = 0.025 m³,
T = 27 °C = (273 + 27) K = 300 K,R = 8.3 J/mol.K
PV = nRT, where number of moles, n = \(\frac{M}{M_{0}}\)
∴ The pressure exerted by the gas,

[Note : 1 Pa = 1 N/m²]

Question 2.
Two tanks of equal volume contain equal masses of oxygen and nitrogen at 127 °C. Find the ratio of
(i) the number of molecules of the gases
(ii) the pressure exerted by the gases in the two tanks. [Molar mass of oxygen = 32 g/mol, molar mass of nitrogen = 28 g/mol]
Solution :
Data : M₀ = M_N (equal masses of the two gases), equal volumes and the same temperature,
M₀₁ (Oxygen) = 32 g/ mol,
M₀₂(nitrogen) = 28 g/mol
Let N₁ and N₂ be the number of molecules of oxygen and nitrogen, n₁ and n₂ the respective number of moles, and N_A the Avogadro number.

This is the ratio of the number of oxygen-molecules to that of nitrogen.

(ii) Using the ideal gas equation, PV = nRT,
P₁ = n₁\(\frac{R T}{V}\) and P₂ = n₂\(\frac{R T}{V}\) … (3)
since the gases occupy the same volume V and are at the same temperature T.

This is the corresponding pressure ratio.

Question 3.
A room is to be prepared for a birthday party filled with helium balloons. Some balloons are filled to occupy 0.240 m³ when the pressure inside them is 0.038 atm and the constant temperature of the room is 70 °F. With what pressure should the larger balloons be filled with helium so that they occupy 0.400 m³ ?
Solution :
Data : V₁ = 0.240 m³, P₁ = 0.038 atm, temperature = 70 °F, V₂ = 0.400 m³
At constant temperature, by Boyle’s law,
P₂V₂ = P₁V₁
∴ Hence, the pressure of helium inside the larger balloons,
∴ P₂ = \(\frac{P_{1} V_{1}}{V_{2}}\) = \(\frac{(0.038)(0.240)}{0.400}\)
= 0.038 × 0.6 = 0.0228 atm

Question 4.
Determine the pressure exerted by 4 grams of hydrogen occupying a volume of 16 litres at 10 °C. (R = 8.314 J/mol.K, molar mass of hydrogen = 2 g/mol)
Solution :
Data : Mass of hydrogen = 4 grams,
V = 16 litres = 16 × 10^-3 m³,
T = 273 + 10 = 283 K, M₀ = 2 g/mol,
R = 8.314 J/mol.K
Number of moles (n) = \(\frac{\text { mass }}{M_{0}}\) = \(\frac{4}{2}\) = 2
PV = nRT ∴ P = \(\frac{n R T}{V}\) = \(\frac{2 \times 8.314 \times 283}{16 \times 10^{-3}}\)
P = 2.941 × 10⁵ N/m²
This is the pressure exerted by the gas.

Question 5.
Find the number of molecules per m3 of a gas at STP.
[Given : 1 atmosphere = 1.013 × 10⁵ N/m² N_A = 6.022 × 10²³ per mole, R = 8.314 J/mol.K]
Solution :
Data : P = 1.013 × 10⁵ N/m²,
N_A = 6.022 × 10²³ per mole,
R = 8.314 J/mol.K, T = 273 K
Number of molecules per m³,

= 2.688 × 10²⁵ molecules/m³
[ Note : STP means Standard temperature (0°C) and pressure (1 atmosphere).]

Question 6.
Explain the concept of free path of a gas molecule. Define free path and mean free path. State the expression for mean free path.
Answer:
Molecules of a real gas’ are not point like particles. For simplicity, molecules are assumed to be spherical. Because of their finite size, two molecules collide with each other when they come within a molecular diameter of each other, strictly speaking, within the sphere of influence of each molecule.
The molecular collisions, though perfectly elastic, result in significant changes in the speeds and

directions of motion of the molecules. Hence, every molecule follows a zigzag path, with abrupt changes in its speed and direction of motion at short and random time intervals. This is called Brownian motion.

Free path : The straight line path of a molecule, i.e., the distance covered by it between successive collisions, is called a free path.

Mean free path : The average distance travelled by a gas molecule between successive collisions, the average being taken over a large number of free paths (or collisions) is called the mean free path.

Maxwell, on the basis of the law of distribution of molecular speeds, obtained the formula for the mean free path (λ) as λ = \(\frac{1}{\sqrt{2} \pi d^{2}(N / V)}\), where N is the number of molecules in volume V of the gas and d is the diameter of a molecule.

[Note : Brownian motion is named after Robert Brown (1773-1858), British botanist, who, in 1827, observed under the microscope the random movement of pollen grains suspended in water. ]

Question 7.
Express the mean free path of a gas molecule in terms of pressure and temperature.
Answer:
In the usual notation, the mean free path of a gas molecule,

Question 8.
Express the mean free path of a gas molecule in terms of the density of the gas
Answer:
In the usual notation, the mean free path of a gas molecule,
λ = \(\frac{1}{\sqrt{2} \pi d^{2}(N / V)}\)
Density, ρ, of the gas = \(\frac{\text { mass }}{\text { volume }}\)
= \(\frac{m N}{V}\), where m is the molecular mass.
∴ \(\frac{N}{V}\) = \(\frac{\rho}{m}\)
∴ λ = \(\frac{m}{\sqrt{2} \pi d^{2} \rho}\)

Question 9.
State the factors on which the mean free path of a gas molecule depends.
Answer:
The mean free path of a gas molecule depends upon the number of molecules per unit volume of the gas and molecular diameter.

Question 10.
What happens to the mean free path of a gas molecule if there is

1. increase in temperature at constant pressure
2. increase in pressure at constant temperature?

Answer:

1. The mean free path increases,
2. The mean free path decreases.

Question 9.
Solve the following :

Question 1.
Calculate the mean free path of a gas molecule if the molecular diameter is 5A and the number of molecules per unit volume of the gas is 2 × 10²⁴ per m³. Compare it with molecular diameter.
Solution :
Data : d = 5Å = 5 × 10^-10m, \(\frac{N}{V}\) = 2 × 10²⁴ per m³

[Note : 1Å = 10^-10m, 1 nm = 10^-9 m = 10 Å]

Question 2.
If the mean free path of a gas molecule under certain conditions is 5000Å and the molecular speed is 500 mIs, find
(i) the time interval between successive collisions
(ii) the collision frequency (number of collisions per unit time) of a gas molecule.
Solution :
Data : λ = 5000 Å = 5000 × 10^-10 m
= 5 × 10^-7 m, v = 500 m/s
(i) speed (v) = \(\frac{\text { distance }(\lambda)}{\text { time }(T)}\)
∴ The time interval between successive collisions of a gas molecule, T = \(\frac{\lambda}{v}\)
= \(\frac{5 \times 10^{-7} \mathrm{~m}}{500 \mathrm{~m} / \mathrm{s}}\) = 10^-9 s

(ii) The collision frequency (number of collisions per unit time) of a gas molecule,
f = \(\frac{1}{T}\) = \(\frac{1}{10^{-9} \mathrm{~s}}\) = 10⁹ collisions per second

Question 3.
Calculate the mean free path of a gas molecule with diameter 4 Å if the pressure of the gas is 1.013 × 10⁵ N/m² and the temperature is 27° C. (The Boltzmann constant, k_B = 1.38 × 10^-23 J/K)
Solution :
Data : d = 4 Å = 4 × 10^-10m, P = 1.013 × 10⁵ N/m², T = (273 + 27) K = 300 K_B = 1.38 × 10^-23 J/K
The mean free path of a gas molecule,

Question 4.
Find the mean free path of a gas molecule if the diameter of a molecule is 4 A, the mass of a molecule is 5.316 × 10^-26 kg and density of the gas is 1.429 kg/m³.
Solution :
Data : d = 4 Å = 4 × 10^-10 m, m = 5.316 × 10^-26 kg, ρ = 1.429 kg/m³
The mean free path of a gas molecule,

Question 5.
The mean free path of a gas molecule is 60 nm when the density of the gas is 1.2 kg/m³. What will be the mean free path if the density is reduced to 0.8 kg/m³ ?
Solution :
Data : λ₁ = 60 nm, ρ₁ = 1.2 kg/m3, ρ₂ = 0.8 kg/m³
λ = \(\frac{m}{\sqrt{2} \pi d^{2} \rho}\)
∴ \(\frac{\lambda_{2}}{\lambda_{1}}\) = \(\frac{\rho_{1}}{\rho_{2}}\) as m and d are not changed.
∴ λ₂ = \(\lambda_{1} \frac{\rho_{1}}{\rho_{2}}\) = 60\(\left(\frac{1.2}{0.8}\right)\) nm = 90 nm
This is the required quantity.

Question 10.
What is the origin of pressure exerted by a gas on the walls of a container?
Answer:
The pressure exerted by a gas on the walls of a container results from the momentum transfer during the collisions of the gas molecules with the walls. The normal force on a wall by all the molecules, per unit wall area, is the gas pressure on the wall.

Question 11.
Define
(1) the mean (or average) speed
(2) the mean square speed
(3) the root-mean-square speed of gas molecules. State the expressions for the same.
Answer:
(1) Mean (or average) speed of molecules of a gas : The mean speed of gas molecules is defined as the arithmetic mean of the speeds of all molecules of the gas at a given temperature.

(2) Mean square speed of molecules of a gas : The mean square speed of gas molecules is defined as the arithmetic mean of the squares of the speeds of all molecules of the gas at a given temperature.

(3) Root-mean-square speed of molecules of a gas : The root-mean-square (rms) speed of gas molecules is defined as the square root of the arithmetic mean of the squares of the speeds of all molecules of the gas at a given temperature.

If there are N molecules in an enclosed pure gas and v₁, v₂, v₃, …, v_N are the speeds of different molecules,

1. the mean speed, \(\bar{v}\) = \(\frac{v_{1}+v_{2}+\ldots+v_{N}}{N}\)
2. the mean square speed,
   \(\overline{v^{2}}\) = \(\frac{v_{1}^{2}+v_{2}^{2}+\ldots+v_{N}^{2}}{N}\)
3. the rms speed, v_rms = \(\sqrt{\overline{v^{2}}}\)

[Note : The mean square velocity is numerically equal to the mean square speed. Similarly, the rms velocity is numerically equal to the rms speed. But in random motion, the mean velocity would be statistically zero, but the mean speed cannot be zero. ]

Question 12.
On the basis of the kinetic theory of gases, derive an expression for the pressure exerted by the gas.
Answer:
The pressure exerted by a gas on the walls of a container results from the momentum transfer during the collisions of the gas molecules with the walls. The normal force on the wall by all the molecules, per unit wall area, is the gas pressure on the wall.

Consider a very dilute gas at a steady-state temperature enclosed in a cubical container of side L. The coordinate axes are aligned with the edges of this cube, as shown in above figure. Suppose that there are N molecules in the container, each of mass m. Consider a molecule moving with velocity \(\).
\(\overrightarrow{v_{1}}\) = \(v_{1 x} \hat{\mathbf{i}}\) + \(v_{1 y} \hat{\mathrm{j}}\) + \(v_{1 z} \hat{\mathbf{k}}\) …. (1)

where v, v_1x and v_1z are the velocity components along the x-, y- and z-axes, respectively, so that
\(v_{1}^{2}\) = \(v_{1 x}^{2}\) + \(v_{1 y}^{2}\) + \(v_{1 z}^{2}\)
The change in momentum of the molecule on collision with the right wall
= final momentum – initial momentum
= (-m|v_1x|) – m|v_1x| = -2m|v_1x|
Assuming the collision to be elastic, the change in momentum of the right wall due to this collision is 2m|v_1x|.

Since the distance between the right and left walls is L, the time between successive collisions with the right wall is,
∆t = \(\frac{2 L}{\left|v_{1 x}\right|}\)
Therefore, the frequency with which the molecule collides with the right wall is \(\frac{\left|v_{1 x}\right|}{2 L}\).

∴ Rate of change of momentum of the right wall = (change in momentum in one collision)
(frequency of collision)
= (2m|v_1x|).\(\frac{\left|v_{1 x}\right|}{2 L}\) = \(\frac{m v_{1 x}^{2}}{L}\) …. (2)
By Newton’s second law of motion, this is the force f_1x exerted by the molecule on the right wall.
Thus, the net force on the right wall by all the N molecules is
F_x = f_1x + f_2x + … + f_Nx

The pressure P_x exerted by all the molecules on a wall perpendicular to the x-axis is

where V = L³ is the volume of the container.

As pressure is the same in all directions,

By definition, the mean square speed of the gas molecules is

where v_rms is the root-mean-square speed of molecules of the gas at a given temperature. Equation (12) gives the pressure exerted by an ideal gas in terms of its density and the root-mean-square speed of its molecules.

Question 13.
Assuming the expression for the pressure P exerted by an ideal gas, prove that the kinetic energy per unit volume of the gas is \(\frac{3}{2}\)P.
Answer:
According to the kinetic theory of gases, the pressure P exerted by the gas is
P = \(\frac{1}{3} \rho v_{r m s}^{2}\) = \(\frac{1}{3} \frac{M}{V} v_{\mathrm{rms}}^{2}\)
where v_rms is the rms speed (root-mean-square speed) of the gas molecules; M, V and ρ are the mass, volume and density of the gas, respectively.

Thus, the kinetic energy per unit volume of an ideal gas is \(\frac{3}{2}\)P.

Question 14.
What is the kinetic energy per unit volume of a gas if the gas pressure is 10⁵ N/m²?
Answer:
Kinetic energy per unit volume of a gas
= \(\frac{3}{2}\)P = \(\frac{3}{2}\) × 10⁵ = 1.5 × 10⁵ J/m³.

Question 15.
Assuming the expression for the pressure exerted by an ideal gas, prove that (1) the kinetic energy per mole of the gas = \(\frac{3}{2}\)RT (2) the rms speed of a gas molecule, u_rms = \(\sqrt{3 R T / M_{0}}\).
OR
Assuming the expression for the pressure exerted by an ideal gas, show that the rms speed of a gas molecule is directly proportional to the square root of its absolute temperature.
OR
Show that the rms velocity of gas molecules is directly proportional to the square root of its absolute temperature.
Answer:
According to the kinetic theory of gases, the pressure P exerted by a gas is
ρ = \(\frac{1}{3} \rho v_{\mathrm{rms}}^{2}\) = \(\frac{1}{3} \frac{M}{V} v_{\mathrm{rms}}^{2}\)
∴ PV = \(\frac{1}{3} m v_{\mathrm{rms}}^{2}\)
where v_rms is the rms speed (root-mean-square speed) of the gas molecules; M, V and ρ are the mass,
volume and density of the gas, respectively. If there are n moles of the gas and M₀ is the molar mass,
M = nM₀, so that PV = \(\frac{1}{3} n M_{0} v_{\mathrm{rms}}^{2}\) … (1)
The equation of state of an ideal gas is
PV = nRT … (2)
where T is the absolute temperature of the gas and R is the molar gas constant.
From Eqs. (1) and (2), we get,

where the term on the left-hand side is the kinetic energy of one mole of the gas.
∴ Kinetic energy per mole of the gas = \(\frac{3}{2}\)RT… (5)
From Eq. (3),

In Eq. (6), R and M0 are constant so that v_rms ∝ \(\sqrt{T}\). Thus, the rms speed of a gas molecule is directly proportional to the square root of the absolute temperature of the gas.

Question 16.
The rms speed of oxygen molecules at a certain temperature is 400 m/s. What is the rms speed of nitrogen molecules at the same temperature ?
[M₀₁ (oxygen) = 32 × 10^-3 kg/mol,
M₀₂ (nitrogen) = 28 × 10^-3 kg/mol]
Answer:

Question 17.
The kinetic energy per unit mass of a certain gas at 300 K is 1.3 × 10⁵ J/kg. What will be the kinetic energy per unit mass of the gas at 600 K ?
Answer:
KE per unit mass of a gas (KE/M) ∝ T

This is the required quantity.

Question 18.
Prove that the kinetic energy per molecule of an ideal gas is \(\frac{3}{2}\)K_BT.
Give the interpretation of temperature according to the kinetic theory of gases.
Answer:
Consider n moles of an ideal gas in a container of volume V. If m is the mass of a gas molecule and v_rms is the root-mean-square speed of the gas molecules then, by the kinetic theory, the pressure exerted by the gas is

Therefore, the kinetic energy per molecule of an ideal gas is directly proportional to its absolute temperature.

This equation is the relation between the kinetic per molecule of a gas and the absolute temperature which is the macroscopic parameter of the gas. The absolute temperature of a gas is a measure of the kinetic energy per molecule of the gas. This result is called the kinetic intepretation of temperature, i.e., the interpretation of temperature according to the kinetic theory of gases.

Question 19.
The rms speed of molecules of a certain gas at 300 K is 400 m/s. What will be the rms speed if the gas is heated to 600 K?
Answer:

gives the required rms speed.

Question 20.
State Boyle’s law. Deduce it on the basis of the kinetic theory of an ideal gas.
OR
Deduce Boyle’s law using the expression for pressure exerted by an ideal gas.
Answer:
Boyle’s law : At a constant temperature, the pressure exerted by a fixed mass of gas is inversely proportional to its volume. If P and V denote the pressure and volume of a fixed mass of gas, then, PV = constant at a constant temperature, for a fixed mass of gas.
According to the kinetic theory of gases, the pressure exerted by the gas is
P = \(\frac{1}{3} \frac{N m v_{\mathrm{rms}}^{2}}{V}\)
where N is the number of molecules of the gas, m is the mass of a single molecule, v_rms is the rms speed of the molecules and V is the volume occupied by the gas.
∴ PV = (\(\frac{1}{2}\)mv²rms) × \(\frac{2}{3}\) N
= (KE of a gas molecule) \(\frac{2}{3}\) N … (1)

For a fixed mass of gas, N is constant. Further, intermolecular forces are ignored so that the corresponding potential energy of the gas molecules may be assumed to be zero. Therefore, \(\frac{1}{2} m v_{\mathrm{rms}}^{2}\) is the total energy of a gas molecule and N\(\left(\frac{1}{2} m v_{\mathrm{rms}}^{2}\right)\) is the total energy of the gas molecules, which is proportional to the absolute temperature of the gas. Then, the right-hand side of EQ. (1) will be constant if its temperature is constant. Hence, it follows that PV = constant for a fixed mass of gas at constant temperature, which is Boyle’s law.

Question 21.
Solve the following :

Question 1.
The speeds of five molecules are 200 m/s, 300 m/s, 400 m/s, 500 m/s and 600 m/s respectively. Find
(i) the mean speed
(ii) the mean square speed
(iii) the root mean square speed of the molecules.
Solution :
Data : v₁ = 200 m/s, v₂ = 300 m/s, v₃ = 400 m/s,. v₅ = 500 m/s, v₆ = 600 m/s
(i) Mean speed,

Question 2.
Find the number of molecules in 1 cm³ of oxygen at a pressure of 105 N/m² if the rms speed of oxygen molecules is 426 m/s. What is the temperature of the gas? (Mass of an oxygen molecule = 5.28 × 10^-26 kg, k_B = 1.38 × 10^-23 J/K)
Solution :
Data : m = 5.28 × 10^-26 kg, v_rms = 426 m/s,
V = 1 cm³ = 10^-6 m³, P = 10⁵ Pa, k_B = 1.38 × 10^-23 J/K

This is the number of molecules in 1 cm3 of oxygen under given conditions.

(ii) PV = nRT

This is the required temperature.

Question 3.
A cylinder filled with hydrogen at 400 K exerts a pressure of 3 atmospheres. If hydrogen is replaced by an equal mass of helium at the same temperature, find the
(i) relative number of molecules of hydrogen and helium occupying the cylinder
(ii) pressure exerted by helium. (Molar mass of hydrogen = 2 g/mol, molar mass of helium = 4 g/mol)
Solution:
Let the subscript 1 refer to hydrogen and the subscript 2 to helium.
From the data in the example,
T = 400 K, P₁ = 3 atmospheres, molar mass of hydrogen, M₁ = 2 g/mol
Molar mass of helium, M₂ = 4 g/mol
∴ \(\frac{M_{2}}{M_{1}}\) = \(\frac{4}{2}\) = 2

(i) Since both the gases have the same mass, N₁m₁ = N₂m₂, where is the mass of a molecule of hydrogen and m2 is the mass of a molecule of helium.

∴ P₂ = \(\frac{2}{3}\) × \(\frac{N_{2}}{V}\) × \(\left(\frac{1}{2} m_{2} v_{2 \mathrm{rms}}^{2}\right)\)
As the temperature is the same in both the cases, the average kinetic energy of gas molecules will be the same in both the cases.

Pressure exerted by helium,
P₂ = 1.5 atmospheres

Question 4.
Calculate the rms speed of oxygen molecules at 127 °C. (Density of oxygen at STP = 1.429 kg/m³ and one atmosphere = 1.013 × 10⁵ N/m²)
Solution:
By the data, ρ = 1.429 kg/m³,
P = 1.013 × 10⁵N/m²

The root mean square speed of oxygen molecules at 127° C = 558.1 m/s.

Question 5.
At what temperature will helium molecules have the same rms speed as that of hydrogen molecules at STP ? (Molar mass of hydrogen = 2 g/mol, molar mass of helium = 4 g/mol)
Solution :
Data : M₀₁ (hydrogen) = 2 g/mol
M₀₂ (helium) = 4 g/mol, T₁ (hydrogen) = 273 K,
v_rms (hydrogen) = v_2rms (helium)

The temperature of helium = 546 K = 273 °C

Question 6.
Compute the temperature at which the rms speed of nitrogen molecules is 831 m/s. [Universal gas constant, R = 8310 J/kmol K, molar mass of nitrogen = 28 kg/kmol]
Solution :
Data : v_rms = 831 m/s, R = 8310 J/kmol.K,
M₀ = 28 kg/kmol

This is the required temperature.

Question 7.
If the rms speed of oxygen molecules at STP is 460 m/s, determine the rms speed of hydrogen molecules at STP. [Molar mass of oxygen = 32 g/mol, molar mass of hydrogen = 2 g/mol]
Solution :
Data : v_rms (oxygen) = 460 m/s, same temperature, M₀₁ (oxygen) = 32 × 10^-3 kg/mol,
M₀₂ (hydrogen) = 2 × 10^-3 kg/mol

This is the rms speed of hydrogen molecules at STP.

Question 8.
Calculate the rms speed of helium atoms at 27 °C. [Density of helium at STP = 0.1785 kg/m3, one atmosphere = 1.013 × 10⁵ Pa]
Solution :
Data : ρ = 0.1785 kg/m³, P = 1.013 × 10⁵ Pa,
T₀ = 273 K, T = (273 + 27) = 300 K
The rms speed of helium atoms at STP,

Question 9.
Determine the pressure of oxygen at 0°C if the density of oxygen at STP = 1.44 kg/m³ and the rms speed of the molecules at STP = 456.4 m/s.
Solution :
Data : ρ = 1.44 kg/m³, v_rms = 456.4 m/s
The pressure of oxygen,
P = \(\frac{1}{3} \rho v_{\mathrm{rms}}^{2}\)
= \(\frac{1}{2}\)(1.44)(456.4)² = 9.98 × 10⁴ Pa

Question 10.
Calculate the rms speed of hydrogen molecules at 373.2K. [Molar mass of hydrogen, M₀ = 2 × 10^-3 kg/mol, R = 8.314 J/mol.K]
Solution :
Data : T = 373.2 K, M₀ = 2 × 10^-3 kg/mol,
R = 8.314 J/mol.K
The rms speed of hydrogen molecules,

= 2157 m/s = 2.157 km/s

Question 11.
Find the rms speed of hydrogen molecules if its pressure is 10⁵ Pa and density is 0.09 kg/m³.
Solution :
Data : ρ = 0.09 kg/m³, P = 10⁵ Pa
The rms speed of hydrogen molecules,

Question 12.
The temperature of matter in interstellar space has an average value of about 3 K. Find the rms speed of a proton in the space. [m_p = 1.673 × 10^-27 kg, k_B = 1.38 × 10^-23 J/K]
Solution :
Data : T = 3 K, mp = 1.673 × 10^-27 kg, k_B = 1.38 × 10^-23 J/K
The rms speed of a proton in interstellar space,

Question 13.
Find the temperature at which
(i) the rms speed
(ii) the average kinetic energy of the molecules of an ideal gas are double their respective values at STP.
Solution:
Data : T₀ = 273 K, v = 2v₀ (rms speeds),
KE = 2KE₀ (average kinetic energies)
(i) v_rms ∝ \(\sqrt{\mathrm{T}}\) for a given gas.
∴\(\frac{v}{v_{\mathrm{o}}}\) = \(\sqrt{\frac{T}{T_{0}}}\)
∴\(\sqrt{\frac{T}{T_{0}}}\) = 2
∴ Temperature, T = 4T₀ = 4 × 273 = 1092 K

(ii) Average Kinetic Energy ∝ T
∴ \(\frac{\mathrm{KE}}{\mathrm{KE}_{0}}\) = \(\frac{T}{T_{0}}\)
∴ Temperature, T = 2T₀ = 2 × 273 = 546 K

Question 14.
At what temperature is the rms speed of an argon atom equal to that of a helium atom at – 20 °C ? [Atomic mass of Ar = 39.9 amu, atomic mass of He = 4.0 amu]
Solution :
Data : T₁ = -20 °C = 273 – 20 = 253 K,
M₀₁ (He) = 4 amu, M₀₂ (Ar) = 39.9 amu

∴ T₂ = \(\frac{M_{02}}{M_{01}}\) T₁ = \(\frac{39.9}{4} \times 253\)
= 2523 K
This is the required temperature.

Question 15.
Calculate the kinetic energy
(i) per mole
(ii) per unit mass
(iii) per molecule of nitrogen at STP. [Molar mass of nitrogen = 28 × 10^-3 kg/mol, R = 8.314 J/mol.K, k_B = 1.38 × 10^-23 J/K]
Solution :
Data : T = 273 K, M₀ = 28 × 10^-3 kg/mol,
R = 8.314 J/mol.K, k_B = 1.38 × 10^-23 J/K
(i) The KE per mole = \(\frac{3}{2}\)RT
= \(\frac{3}{2}\)(8.314)(273) = 3.404 × 10³ J/mol

(ii) The KE per unit mass = \(\frac{3}{2} \frac{R T}{M_{0}}\)
= \(\frac{3}{2} \cdot \frac{(8.314)(273)}{28 \times 10^{-3}}\) = 1.216 × 10⁵ J/kg

(iii) The KE per molecule

Question 16.
Calculate the average molecular kinetic energy
(i) per kilomole
(ii) per kilogram of oxygen at 27 °C. [R = 8310 J/kmol.K, Avogadro’s number = 6.02 × 10²⁶ molecules/kmol]
Solution :
Data : T = 273 + 27 = 300 K, M₀ = 32 kg / kmol,
R = 8310 J/kmol.K
(i) The average molecular kinetic energy per kilomole of oxygen
= \(\frac{3}{2}\) RT = \(\frac{3}{2}\) (8310) (300)
= 4155 × 900 = 3739 × 1000
= 3.739 × 10⁶ J/kmol

(ii) The average molecular kinetic energy per unit mass of oxygen

Question 17.
Calculate the kinetic energy of 10 grams of argon molecules at 127 °C. [Universal gas constant R = 8310 J/kmol.K, atomic weight of argon = 40]
Solution:
Data : M = 10 g = 10 × 10^-3 kg, T = 273 + 127 = 400 K,R = 8310 J/kmol.K, M₀ = 40 kg/kmol Kinetic energy per unit mass

= 15 × 8310 = 1.247 × 10⁵ J/kg
∴ The kinetic energy of 1 × 10^-3 kg of argon at 127°C = 1.247 × 10⁵ × 10 × 10^-3
= 1247 J

Question 18.
Calculate the kinetic energy of 2 kg of nitrogen at 300 K. [Molar mass of nitrogen = 28 × 10^-3 kg/mol, R = 8.314 J/mol.K]
Solution :
Data : Mass of nitrogen, M = 2 kg, T = 300 K,
M₀ = 28 × 10^-3 kg/mol, R = 8.314 J/mol.K
Kinetic energy per kg = \(\frac{3}{2} \frac{R T}{M_{0}}\)
∴ The kinetic energy of 2 kg of nitrogen

Question 19.
The kinetic energy of 1 kg of oxygen at 300 K is 1.169 × 10⁵ J. Find the kinetic energy of 4 kg of oxygen at 400 K.
Solution :
Data : Masses of oxygen, M₁ = 1 kg and M₂ = 4 kg, T₁ = 300 K, T₂ = 400 K, kinetic energy,
K₁ = 1.169 × 10⁵ J
Kinetic energy of a given mass (M) of a gas,

This is the kinetic energy of 4 kg of oxygen at 400 K.

Question 20.
The kinetic energy per unit mass of nitrogen at 300 K is 2.5 × 10⁶J/kg. Find the kinetic energy of 4 kg of oxygen at 600 K. [Molar mass of nitrogen = 28 kg/kmol, molar mass of oxygen = 32 kg/kmol]
Solution:
Data : T₁ = 300 K, K₁ = 2.5 × 10⁶ J/kg,
M₁ (nitrogen) = 1 kg, M₂(oxygen) = 4 kg,
M₀₁ (nitrogen) = 28 kg/kmol, T₂ = 600 K,
M₀₂(oxygen) = 32 kg/kmol
Kinetic energy of a given mass M of a gas,

This is the required quantity.

Question 21.
The pressure of a gas in a 0.1 litre container is 200 kPa and the KE per molecule is 6 × 10^-21J. Find the number of gas molecules in the container. How many moles are in the container? [Avogadro number = 6.022 × 10²³ mol^-1]
Solution:
Data : V = 0.1 litre = 10^-4 m³, P = 200 kPa
= 2 × 10⁵ Pa, KE per molecule = 6 × 10^-21 J,
N_A = 6.022 × 10²³ mol^-1
The KE per molecule of a gas = \(\frac{3}{2}\)k_BT

Question 22.
At what temperature will the average kinetic energy of a gas be exactly half of its value at NTP ?
Solution :
Data : KE₂/KE₁ = 1/2, T₁ = 273K
Average molecular kinetic energy per mole of the gas = \(\frac{3}{2}\)RT

This is the required temperature.

Question 23.
Find the kinetic energy per unit volume of nitrogen at a pressure of 76 cm of mercury. Hence, find the kinetic energy of 10 cm³ of the gas under the same condition. Take ρ (mercury) = 13.6 g/cm³ and g = 9.8 m/s².
Solution :
Data :h = 76 cm = 0.76 m, ρ = 13.6 g/cm³ = 13.6 × 10³ kg/m³, g = 9.8 m/s²,
V = 10 cm³ = 10 × 10^-6 cm³

(i) The kinetic energy per unit volume of nitrogen

(ii) The kinetic energy of 10 cm³ of the gas = \(\frac{3}{2}\) PV= (1.519 × 10⁵) (10 × 10^-6)
= 1.519 J

Question 22.
What is meant by degrees of freedom ? Explain the degrees of freedom for
(i) an atom
(ii) a diatomic molecule.
Answer:
The concept of degrees of freedom as used in the kinetic theory specifies the number of independent ways in which an atom or molecule can take up energy. It depends only on the possibilities of motion of the atom or molecule.

Gas molecules of all types have x-, y- and z-components of velocity that are entirely independent of one another. Thus, they have three ways to move in translation, i.e., three degrees of translational freedom.

An atom (or a monatomic molecule, i.e., a molecule containing a single atom, e.g., He) treated as a point mass, has no rotational motion. Hence, it has only three degrees of translational freedom.

A diatomic molecule, in addition to translation, can rotate about axes perpendicular to the line connecting the atoms, as shown in below figure, but not about that line itself. Therefore, it has only two degrees of rotational freedom.

Further, the two atoms may oscillate alternately toward and away from one another along the line joining them, as if connected by a spring. As a, i harmonic oscillator can have potential energy as well as kinetic energy, a diatomic molecule is regarded to have two degrees of vibrational freedom. Thus, at high enough temperatures, a diatomic molecule has seven degrees of freedom : three of translation, and two each of rotation and vibration.

Notes :

(1) That a monatomic gas molecule does not have rotational energy, and that a diatomic molecule does not have a third rotational degree of freedom corresponding to rotation about the line joining the atoms, are explained by quantum theory.

(2) Also according to quantum theory, rotational and oscillatory motions begin at certain higher temperatures. For a molecule of a diatomic gas (like hydrogen), only translation is possible at very low temperatures (below about 100 K). As the temperature increases, rotational motion can begin; so that, at room temperature, a diatomic molecule has only five degrees of freedom-behaving like a pair of atoms rigidly connected like a dumbbell. Oscillatory motion can begin only at quite high temperatures substantially above room temperature (usually of the order of thousands of kelvin).

Question 23.
Derive Mayer’s relation between the molar specific heat of a gas at constant pressure and that at constant volume.
OR
Using the first law of thermodynamics, show that for an ideal gas, the difference between the molar specific heat capacities at constant pressure and at constant volume is equal to the molar gas constant R.
Answer:
Consider a cylinder of volume V containing n moles of an ideal gas at pressure P, fitted with a piston of area A. Suppose, the gas is heated at constant pressure which raises its temperature by dT. The gas exerts a total force F = PA on the piston which moves outward a small distance dx.

dW = Fdx = PAdx = PdV … (1)
where Adx = dV is the increase in volume of the gas during the expansion. dW is the work done by the gas on the surroundings as a result of the expansion. If the heat supplied to the gas is dQ_P and the increase in its internal energy is dE then, by the first law of thermodynamics,
dQ_P = dE + dW=dE + PdV
If C_P is the molar specific heat capacity of the gas at constant pressure, dQ_P = nC_P dT.
∴ nC_PdT = dE + PdV …(2)

On the other hand, if the gas was heated at constant volume (instead of at constant pressure) from the initial state such that its temperature increases by the same amount dT, then dW=0. Since the internal energy of an ideal gas depends only on the temperature, the increase in internal would again be dE. If dQv was the heat supplied to the gas in this case, by the first law of thermodynamics and the definition of molar specific heat capacity at constant volume (C_V),

This is Mayer’s relation between C_P and C_V.

Here, heat and work are expressed in the same units. If heat is expressed in calorie or kilo calorie and work is expressed in erg or joule, the above relation becomes
C_P – C_V = \(\frac{R}{J}\) …. (7)
Where J is the mechanical equivalent of heat.

Question 24.
Express Mayer’s relation in terms of the principal specific heats, S_P and S_V.
Answer:
Mayer’s relation : C_P – C_V = R. Let M₀ = molar mass of the gas, S_P = specific heat of the gas at constant pressure and S_V = specific heat of the gas at constant volume.

Now, C_P = M₀S_P and C_V = M₀S_V
∴ M₀S_p – M₀S_V = R
∴ S_P – S_V = \(\frac{R}{M_{0}}\) when heat and work are expressed
in the same units. If heat is expressed in calorie or kilo calorie and work is expressed in erg or joule, we
get,
S_p – S_v = \(\frac{R}{M_{0} J}\), where J is the mechanical equivalent of heat.

Question 25.
Explain : Each translational and rotational degree of freedom contributes only one quadratic term to the energy but one vibrational mode contributes two quadratic terms.
Answer:
With three translational degrees of freedom, the average translational energy per molecule of a gas is

where m is the mass of the molecule and v_x, v_y and v_z are the x-, y- and z-components of the molecular velocity.

A diatomic molecule has two rotational degrees of freedom. If ω₁ and ω₂ are the angular speeds about the two axes and I₁ and I₂ are the corresponding moments of inertia, the rotational energy of a diatomic molecule,

A diatomic molecule is regarded to have two degrees of vibrational freedom for the vibrational mode in which the two atoms vibrate relative to, and without affecting, the centre of mass of the molecule. Comparing this system with a vibrating body of mass m connected to a spring of force constant k, the vibrational energy has two terms corresponding to the kinetic and potential energies :

where x is the displacement from the mean position.

From Eqs. (1), (2) and (3), each translational and rotational degree of freedom contributes only one quadratic term to the average energy of a gas molecule while one vibrational mode contributes two quadratic terms.

[ Note : For a gas at an absolute temperature T, the

Question 26.
Explain the degrees of freedom a polyatomic molecule can have. Hence write the expressions for
(i) the energy per molecule
(ii) the energy per mole
(iii) C_V
(iv) C_P
(v) the adiabatic constant, for a polyatomic gas.
Answer:
A polyatomic molecule has three degrees of translational freedom like any particle. Each molecule can also rotate about its centre of mass with an angular velocity components along each of the three axes. Therefore, each molecule has three degrees of rotational freedom. Additionally, if the molecule is soft at high enough temperatures, it can vibrate easily with many different frequencies, say, f, because there are many interatomic bonds. Hence, it has 2f degrees of vibrational freedom. Then, according to the law of equipartition of energy, for each degree of freedom of translation and rotation,
the molecule has the average energy \(\frac{1}{2}\)k_BT, but for each frequency of vibration the average energy is k_BT, since each vibration involves potential energy and kinetic energy. k_B is the Boltzmann constant and T is the thermodynamic temperature.
(i) The energy per molecule

[Note : As /increases, y decreases.]

Question 27.
The top of a cloud of smoke holds together for hours. Why?
Answer:
According to the law of equipartition of energy, the smoke particles have the same average kinetic energy of random motion as the air molecules. But smoke particles have a much larger mass than air molecules and therefore move slowly, i.e., the average speed of diffusion of smoke particles is small. Hence, in the absence of significant turbulence in the atmosphere, a smoke cloud is relatively stable.

Question 28.
Solve the following :

Question 1.
Find the kinetic energy per molecule of a monatomic gas at 300 K.
Solution :
Data : T = 300 K, k_B = 1.38 × 10^-23 J/K
The kinetic energy per molecule of the gas
= \(\frac{3}{2}\)k_BT
= \(\frac{3}{2}\) (1.38 × 10^-23) (300) J/molecule
= (4.5) (1.38 × 10^-21) J/molecule
= 6.21 × 10^-21 J/molecule
[Note : Here, the number of degrees of freedom is 3.]

Question 2.
Find the kinetic energy of two moles of a mon-atomic gas at 400 K.
Solution :
Data : n = 2, T = 400 K, R = 8.314 J/mol.K
The kinetic energy of two moles of the gas
= \(\frac{3}{2}\)nRT = \(\frac{3}{2}\)(2) (8.314) (400) J
= (12) (8.314 × 10²) J
= 9.977 × 10³ J.

Question 3.
Find the kinetic energy of 10 kg of a monatomic gas at 500 K if the molar mass of the gas is 4 × 10^-3 kg/mol.
Solution :
Data : M = 10 kg, M₀ = 4 × 10^-3 kg/mol, T = 500 K, R = 8.314 J/mol.K.
The kinetic energy of 10 kg of the gas

Question 4.
Find the kinetic energy per molecule of a diatomic gas at 300 K.
Solution :
Data : T = 300 K, k_B = 1.38 × 10^-23 J/K
The kinetic energy per molecule of the gas
= \(\frac{5}{2}\) k_BT = \(\frac{5}{2}\)(1.38 × 10^-23) (300) J/molecule
= (7.5) (1.38 × 10^-23) J/molecule
= 1.035 × 10^-22 J/molecule.
[Note : Here, the number of degrees of freedom is 5.]

Question 5.
Find the kinetic energy of four moles of a diatomic gas at 400 K.
Solution :
Data : n = 4, T = 400 K, R = 8.314 J/mol.K
The kinetic energy of four moles of the gas
= \(\frac{5}{2}\)nRT = \(\left(\frac{5}{2}\right)\) (8.314) (400) J
= (40)(8.314 × 10²) J
= 3.326 × 10⁴ J

Question 6.
Find the kinetic energy of 56 kg of a diatomic gas at 500 K if the molar mass of the gas is 28 × 10^-3 kg/mol.
Solution :
Data : M = 56 kg, M₀ = 28 × 10^-3 kg/mol, T = 500 K, R = 8.314 J/mol.K
The kinetic energy of 56 kg of the gas

= 25(8.314 × 10⁵)
= 2.079 × 10⁷ J

Question 7.
When two kilocalories of heat are supplied to a system, the internal energy of the system increases by 5030 J and the work done by the gas against the external pressure is 3350 J. Calculate J, the mechanical equivalent of heat.
Solution :
Data : From the data in the example,
dQ = 2 kcal, dE = 5030 J, dW = 3350 J.
dQ = \(\frac{d E+d W}{J}\), where J = mechanical equivalent of heat

Question 8.
Find the increase in the internal energy of a gas of mass 10 grams when it is heated from 300 K to 305 K.
Given : S_V = 0.16 kcal/kg.K, J = 4186 J/kcal
Solution :
Data : M = 10 grams = 10 × 10^-3 kg,
S_V = 0.16 kcal/kg.K, J = 4186 J/kcal
Rise in the temperature of the gas,
∆T = 305 – 300 = 5K
∆E = J S_V M∆T = (4186)(0.16)(10 × 10^-3) × (5) J
= 33.49 J
The increase in the internal energy of the gas, ∆E = 33.49 J

Question 9.
The molar specific heat of helium at constant volume is 12.5 J/mol.K. Find its molar specific heat at constant pressure. Take R = 8.31 J/mol.K.
Solution :
Data : C_V = 12.5 J/mol.K, R = 8.31 J/mol.K
The molar specific heat of helium at constant press-ure,
C_P = C_V + R = (12.5 + 8.31) J/mol.K
= 20.81 J/mol.K.

Question 10.
Find the principal specific heats of helium and hence the universal gas constant. Given : C_P = 20.81 J/mol.K, C_V = 12.5 J/mol.K. M₀ (He) = 4 × 10^-3 kg/mol.
Solution :
Data : C_P = 20.81 J/mol.K, C_V = 12.5 J/mol.K,
M₀ = 4 × 10^-3 kg/mol.
(i) The specific heat of helium at constant pressure,

[Note : R = C_P – C_V = (20.81 – 12.5) J/mol.K = 8.31 J/mol.K. The difference between the two values of R is due to the approximation involved in calculation.]

Question 11.
The molar specific heat of nitrogen at constant volume is 4.952 cal/mol.K and that at constant pressure is 6.933 cal/mol.K. Find the mechanical equivalent of heat. Take R = 8.31 J/mol.K.
Solution :
Data : C_V = 4.952 cal/mol.K, C_P = 6.933 cal/mol.K, R = 8.31 J/mol.K

Question 29.
Name and define the modes of heat transfer.
Answer:
The three modes of heat transfer are conduction, convection and radiation.

1. Conduction is the mode of heat transfer within a body or between two bodies in contact, from a region of high temperature to a region of lower temperature without the migration of the particles of the medium.
2. Convection is the mode of heat transfer from one part of a fluid to another by the migration of the particles of the fluid.
3. Radiation is the mode of heat transfer by electromagnetic waves / quanta.

Question 30.
What is thermal radiation or heat radiation? State its characteristic properties.
Answer:
Radiation is the mode of heat transfer or in general, energy transfer by electromagnetic waves / quanta. Thermal radiation or heat radiation is the radiation produced by thermal agitation of the particles of a body, and its spectrum, i.e., frequency distribution or wavelength distribution, is continuous from the far infrared to the extreme ultraviolet region depending on the temperature of the body.

Properties :

1. Thermal radiations are electromagnetic waves/ quanta extending from the far infrared to the extreme ultraviolet region. In this spectrum, the infrared waves (wavelengths ranging from about 700 nm to about 1 mm) are sensed as heat.
2. They have the same speed in free space as that of light, nearly 3 × 10⁸ m/s, which makes radiation the most rapid mode of heat transfer.
3. They exhibit all the optical phenomena of light, viz., reflection, absorption, refraction, interference, diffraction and polarization.
4. Radiation incident on a body is, in general, partly reflected, partly, absorbed and partly transmitted.
5. Thermal radiation obeys the inverse-square law of intensity, i.e., the intensity at a point is inversely proportional to the square of its distance from a point source of radiation.

Question 31.
Define the coefficients of absorption, reflection and transmission. Obtain the relation between them.
Answer:

1. The coefficient of absorption or absorptive power or absorptivity of a body is the ratio of the quantity of radiant energy absorbed by the body to the quantity of radiant energy incident on the body in the same time.
2. The coefficient of reflection or reflectance or reflective power of the surface of a body is the ratio of the quantity of radiant energy reflected by the surface to the quantity of radiant energy incident on the surface in the same time.
3. The coefficient of transmission or transmittance or transmissive power of a body is the ratio of the quantity of radiant energy transmitted by the body to the quantity of radiant energy incident on the body in the same time.

Let Q be the quantity of radiant energy incident on a body and Q_a, Q_r and Q_t be the quantities of radiant energy absorbed, reflected and transmitted by the body respectively, in the same time. Since the total energy is conserved, we have,

Hence, a + r + t = 1
This is the required relation.

[Note : The coefficients of absorption, reflection and transmission are, respectively, the measures of the ability of the body or material to absorb, reflect or transmit radiation.

They are dimensionless quantities and have no units. They depend on the material and physical conditions of the body as well as on the frequency of the incident radiation.]

Question 32.
If for a certain body, under certain conditions, the coefficient of absorption is 0.2 and the coefficient of reflection is 0.5, what will be the coefficient of transmission ?
Answer:
In the usual notation, a + r +1 = 1
Hence, the coefficient of transmission of the body = 1 – (a + r) = 1 – (0.2 + 0.5) = 1 – 0.7 = 0.3.

Question 33.
Give four exmaples of

1. athermanous substance
2. diathermanous substances.

Answer:

1. Water, wood, iron, copper, moist air, benzene are athermanous substances.
2. Quartz, sodium chloride, hydrogen, oxygen, dry air, carbon tetrachloride are diathermanous substances.

Question 34.
A substance may be athermanous or diathermanous for certain wavelength ranges while good absorber for other wavelength ranges. Explain.
Answer:
For interaction of light with matter, it is necessary to consider the atomic nature of matter. Interaction of an atom with light depends on the frequency or equivalently on the photon energy.

An atom absorbs light if the photon’s energy equals one of the excitation energies of the atom. In the dense atomic neighbourhood of ordinary gases at pressures above 100 Pa, solids and liquids, the discrete atomic energy states widen into bands. Thus, bulk matter, depending on its nature, possesses absorption bands in specific regions within the electromagnetic frequency spectrum.

Radiant energy of other frequencies is elastically scattered so that the material is transparent at these frequencies.

[Note : Colourless, transparent materials have their absorption bands outside the visible region of the spectrum, which is why they are, in fact, colourless and transparent. Glasses have absorption bands in the ultraviolet region (~ 100 nm-200 nm), where they become opaque. At even longer wavelengths of radiowaves glass is again transparent. In comparison, a stained glass has absorption band in the visible region where it absorbs out a particular range of frequencies, transmitting the complementary colour. Semiconductors, such as ZnSe, CdTe, GaAs and Ge, which are opaque in the visible region of the spectrum are highly transparent in the infrared region (2 µm to 30 µm).]

Question 35.
In Fery’s blackbody, the hole is the blackbody but not the inner coated or outer sphere alone. Explain.
Answer:
Any space which is almost wholly closed – e.g., an empty closed tin can with a tiny hole punched-approximates to a blackbody. The hole is a good absorber and looks black because any light which enters through it is almost completely absorbed after multiple reflections inside. This absorption is quickened in Fery’s blackbody by the inner coat of lampblack. Thus, the hole acts as a perfect absorber.

Also, the relative intensities of radiation at different wavelengths is determined only by the temperature of the blackbody, not by the nature of its surfaces. This is so because the radiation coming out from a small part of the inner surface is made up of

1. the radiation emitted by that area
2. the radiation from other parts reflected at that area. Since the hole in Fery’s blackbody is very small, the radiations from the inner surface are well mixed up by reflection before they can escape. Hence, when Fery’s blackbody is placed in a high-temperature bath of fused salts, the hole serves as the source of blackbody radiation.

The outer surface of Fery’s blackbody is made highly reflective and lampblack too has a small coefficient of reflection. Hence, neither of them alone is a blackbody.

Question 36.
What is
(i) a cavity radiator
(ii) cavity radiation?
Answer:
(i) A cavity radiator is a block of material with internal cavity. The inner and outer surfaces of the block are connected by a small hole. Most of the radiant energy entering the block through the hole cannot escape from the hole. The block, therefore, acts almost like a blackbody.

(ii) When the cavity radiator is heated to high temperature, radiation coming out from the hole resembles blackbody radiation. It is called cavity radiation. It depends only on the temperature of the radiator and not on the shape and size of the cavity as well as the material of the walls of the cavity.

Question 37.
‘If r = 1, then it is a white body’. Is this true ? Explain.
Answer:
No.
Since a + r + t = l,a = 0 and t = 0 for r = 1. Therefore, by Kirchhoff’s law, e = a = 0 which is impossible as every body at temperature above 0 K does emit radiant energy, and T = 0 K is impossible.

A blackbody being a full radiator, when heated to high enough temperature it would emit thermal radiation at all the wavelengths and thus appear white. A perfect reflector (r = 1), on the other hand, is a poor emitter and thus would not necessarily appear white when heated.

Question 38.
State and explain Prevost’s theory of exchange of heat.
Answer:
In 1792, Pierre Prevost put forward a theory of exchange of heat. According to this theory, all bodies at all temperatures above the absolute zero temperature (0 K) radiate thermal energy to the surroundings and at the same time receive radiant energy from the surroundings. Thus, there is continuous exchange of radiant energy between a body and its surroundings.

The quantity of radiant energy (thermal energy) emitted by a body per unit time depends upon the nature of the emitting surface, the area of the surface and the temperature of the surface. The quantity of radiant energy absorbed by a body per unit time depends upon the nature of the absorbing surface, the area of the surface and the time rate at which the radiant energy is incident on the body.

If the time rate of emission of thermal energy is greater than the time rate of absorption of thermal energy, the temperature of the body falls. If the emission rate is less than the absorption rate, the temperature of the body increases. If the emission rate equals the absorption rate, the temperature of the body remains constant.

[Note : A body appears red if its temperature is around 800 °C, and white hot if its temperature is around 3000 °C]

Question 39.
Define
(1) emissive power or radiant power
(2) coefficient of emission of a body
Answer:
(1) Emissive power or radiant power of a body (symbol, R) : The emissive power or radiant power of a body at a given temperature is defined as the quantity of radiant energy emitted by the body per unit time per unit surface area of the body at that temperature.

(2) Coefficient of emission (or emissivity) of a body (symbol, e) : The coefficient of emission (or emissivity) of a body is defined as the ratio of the emissive power of the body (R) to the emissive power of a perfect blackbody (R_b) at the same temperature as that of the body.
e = \(\frac{R}{R_{\mathrm{b}}}\)

[Note : The SI unit and dimensions of emissive power are the watt per square metre (W/m² Or Js^-1m^-2) and [M¹L°T^-3]. The coefficient of emission is a dimensionless and unitless quantity. For a perfect blackbody, e = 1 and for a perfect reflector, e = 0.]

Question 40.
If the emissive power of a certain body at a certain temperature is 2000 W/m² and the emissive power of a perfect blackbody at the same temperature is 10000 W/m², what is the coefficient of emission of the body?
Answer:
The coefficient of emission of the body,
e = \(\frac{R}{R_{\mathrm{b}}}\) = \(\frac{2000}{1000}\) = 0.2

Question 41.
State the characteristics of blackbody radiation spectrum.
Answer:
Characteristics of blackbody radiation spectrum :

1. The emissive power R_λ for every wavelength λ increases with increasing temperature.
2. Each curve has a characteristic form with a maximum for R_λ at a certain wavelength λ_m.
3. λ_m depends on the absolute temperature of the body and, with increasing temperature, shifts towards shorter wavelengths, i.e., towards the ultraviolet end of the spectrum.
4. λ_mT = a constant.
5. The area under each curve gives the total radiant power per unit area of a blackbody at that temperature and is proportional to T⁴, (Stefan-Boltzmann law).

Notes :

1. Experimental work on the distribution of energy in blackbody radiation, was carried out by German physicists Otto Lummer (1860-1925), Wilhelm Wien (1864-1928) and Ernst Pringsheim (1859-1917).
2. Explanation of the radiation spectrum, given by Wien on the basis of thermodynamics could account only for the short wavelength region. The formula obtained by Rayleigh and Jeans, on the basis of the equipartition of energy could account only for long wavelength region. Planck’s empirical formula, put forward in 1900, could account for the entire spectrum.

Question 42.
State the significance of Wien’s displacement law.
Answer:
Significance :

1. It can be used to estimate the surface temperature of stars.
2. It explains the common observation of the change of colour of a solid on heating-from dull red (longer wavelengths) to yellow (smaller wavelengths) to white (all wavelengths in the visible region).

Question 43.
Explain the Stefan-Boltzmann law.
Answer:
The power per unit area radiated from the surface of a blackbody at an absolute temperature T is its emissive power or radiant power R_b at that temperature. According to the Stefan-Boltzmann law,
R_b ∝ T⁴ ∴ R_b = σT⁴
where the constant a is called Stefan’s constant.

If A is the surface area of the body, its radiant power, i.e., energy radiated per unit time, is AσT⁴.

[Note : This law was deduced by Josef Stefan (1835-93), Austrian physicist, from the experimental results obtained by John Tyndall (1820-93), British physicist. The theoretical derivation of this law is due to Boltzmann in 1884. Hence, the law is known as the Stefan- Boltzmann law.

Question 44.
What is the emissive power of a perfect black-body at 1000 K? (σ = 5.67 × 10^-8 W/m².K⁴)
Answer:
R_b = σT⁴ = 5.67 × 10^-8 × (10³)⁴
= 5.67 × 10⁴\(\frac{\mathrm{W}}{\mathrm{m}^{2}}\) is the required emissive power.

Question 45.
In the above case (Question 43), if the body is not a perfect blackbody and e = 0.1, what will be the emissive power?
Answer:
R = eRh = 0.1 × 5.67 × 104 = 5.67 × 10³\(\frac{\mathrm{W}}{\mathrm{m}^{2}}\)

Question 46.
Derive an expression for the net rate of loss of heat per unit area by a perfect blackbody in a cooler surroundings.
Answer:
Consider a perfect blackbody at absolute temperature T. We assume its surroundings also to act as a perfect blackbody at absolute temperature T₀, where T₀ < T.

The power per unit area radiated from the surface of a blackbody at temperature T is its emissive power R_b at that temperature. According to the Stefan-Boltzmann law,
R_b = σT⁴
where σ is the Stefan constant.
At the same time, the body absorbs radiant energy from the surroundings. The radiant energy absorbed per unit time per unit area by the black-body is \(\sigma T_{0}^{4}\).

Therefore, the net rate of loss of radiant energy per unit area by the blackbody is σ(T⁴ – \(T_{0}^{4}\)).

[Note : If the body at temperature T has emissivity e < 1, (i.e., it is not a perfect blackbody) the net rate of loss of radiant energy per unit area is eσ(T⁴ – \(T_{0}^{4}\)).]

Question 47.
Compare the rates of radiation of energy by a metal sphere at 600 K and 300 K.
Answer:

Question 48.
Solve the following :

Question 1.
Radiant energy is incident on a body at the rate of 2000 joules per minute. If the reflection coefficient of the body is 0.1 and its transmission coefficient is 0.2, find the radiant energy
(i) absorbed
(ii) reflected
(iii) transmitted by the body in 2 minutes.
Solution :
Data : r = 0.1, t = 0.2
a + r + t = 1
∴ a = 1 – r – f = 1 – 0.1 – 0.2 = 0.7
Radiant energy incident per minute on the body is 2000 J. Hence, in 2 minutes, the radiant energy (Q) incident on the body is 4000 J. Let Q_a, Q_r and Q_t be the quantities of radiant energy absorbed, reflected and transmitted in 2 minutes by the body, respectively.

Question 2.
Heat is incident at the rate of 10 watts on a completely opaque body having emissivity 0.8. Find the quantity of radiant heat reflected by it in 1 minute.
Answer:
Data : e = 0.8, time =1 min =60 s, \(\frac{d Q}{d t}\)= 10 W
For a completely opaque body, t =0.
Also, a = e = 0.8
a + r + t = 1
∴ 0.8 + r + 0 = 1
∴ r = 1 – 0.8 = 0.2
Total radiant heat incident on the body in 1 minute
Q = \(\left(\frac{d Q}{d t}\right)\) × time = 10 × 60 = 600J
∴ The quantity of radiant heat reflected by the body in 1 minute is
Q_r = rQ = 0.2 × 600 = 120J

Question 3.
A metal cube of side 2 cm emits 672 J of heat in 100 s at a certain temperature. Calculate its emissive power (radiant power) at that temperature.
Solution :
Data : L = 2 cm = 2 × 10^-2 m, Q = 672 J, t = 100 s
The surface area of the cube is
A = 6L² = 6(2 × 10^-2)² = 24 × 10^-4 m²
The energy radiated by the cube is Q = RAt
where R ≡ emissive power (radiant power).

Question 4.
The energy of 6000 J is radiated in 5 minutes by a body of surface area 100 cm². Find the emissive power (radiant power) of the body.
Solution :
Data : Q = 6000 J, t = 5 minutes = 5 × 60 s = 300 s,
A = 10 cm² = 10 × 10^-4 m² = 10^-3 m²

Emissive power (radiant power), R = \(\frac{Q}{A t}\)
= \(\frac{6000}{10^{-3} \times 300}\)
= 2000 J/m².s 2000 W/m²

Question 5.
An ordinary body A and a perfect blackbody B are maintained at the same temperature. If the radiant power of A is 2 × 10⁴ W/m² and that of B is 5 × 10⁴ W/m², what is the coefficient of emission (emissivity) of A ?
Solution :
Data : R = 2 × 10⁴ W/m², R_b = 5 × 10⁴ W/m²
The coefficient of emission (emissivity) of A,
e = \(\frac{R}{R_{\mathrm{b}}}\) = \(\frac{2 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2}}{5 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2}}\) = 0.4

Question 6.
A body of surface area 100 cm² radiates energy 3000 J in 10 minutes at a certain constant temperature. The radiant power of a perfect blackbody maintained at the same temperature is 2500 W/m². Find the the radiant power and emissivity of the body.
Solution:
Data : A = 100 cm² = 100 × 10^-4m² = 10^-2m², t = 10 minutes = 10 × 60 s = 600 s, Q = 3000 J.
R_b = 2500 W/m².

(i) The radiant power of the body,
R = \(\frac{Q}{A t}\) = \(\frac{3000 \mathrm{~J}}{\left(10^{-2} \mathrm{~m}^{2}\right)(600 \mathrm{~s})}\)
= 5 × 10² W/m²

(ii) The emissivity of the body,
e = \(\frac{R}{R_{\mathrm{b}}}\) = \(\frac{500 \mathrm{~W} / \mathrm{m}^{2}}{2500 \mathrm{~W} / \mathrm{m}^{2}}\) = 0.2

Question 7.
For a certain body, the coefficient of absorption (absorptive power, absorptivity) is 0.4. The body is maintained at a constant temperature. The radiant power of a perfect blackbody maintained at the same temperature is 5 × 10⁴ W/m². Find the radiant power of the body at that temperature.
Solution:
Data: a = 0.4, R_b = 5 × 10⁴ W/m²
As the emissivity, e = a,
we have, e = 0.4
∴ R = eR_b = (0.4) (5 × 10⁴ W/m²)
= 2 × 10⁴ W/m²
This is the required quantity.

Question 8.
Calculate the wavelength in angstrom at which the emissive power is maximum for a blackbody heated to 3727 °C.
[Wien’s constant, b = 2.898 × 10^-3 m.K]
Solution :
Data : T = 3727 °C = 3727 + 273 = 4000 K,
b = 2.898 × 10^-3 m.K

Question 9.
The maximum radiant power of the Sun is at wavelength 500 nm. The Wien displacement law constant is 2.898 × 10^-3 m.K. Estimate the temperature of the surface of the Sun. Assume the Sun to be a blackbody radiator.
Solution :
Data : λ_m = 500 nm = 5 × 10⁷ m, b = 2.898 × 10^-3 m.K
By Wien’s displacement law,
λ_mT = b
∴ The estimated surface temperature of the Sun,

Question 10.
What is the peak wavelength of the radiation emitted by a blackbody at 35 °C ? [Wien’s displace-ment law constant, b = 2.898 × 10^-3 m.K]
Solution :
Data : t = 35 °C, b = 2.898 × 10^-3 m.K
The absolute temperature of the blackbody,
T = t + 273 = 35 + 273 = 308 K
By Wien’s displacement law, λ_mT = b
∴ The peak wavelength,

Question 11.
Calculate the emissive power of a perfect black-body at 127°C.
Solution:
Data: T = 127°C = (127 + 273) K = 400 K,
= 5.67 × 10^-8 W/m².K⁴
Emissive power, R_b = σT⁴
= 5.67 × 10^-8 × (400)⁴
= 1.452 × 10³ W/m²

Question 12.
What is the temperature at which a blackbody radiates heat at the rate of 1 kilowatt per square metre?
Solution :
Data : R_b = 1 kW/m² = 1000 W/m²,
σ = 5.67 × 10^-8 W/m²K⁴
R_b = σT⁴
∴ Temperature, T = \(\left(\frac{R_{\mathrm{b}}}{\sigma}\right)^{1 / 4}\)

Question 13.
A perfect blackbody, maintained at 27 °C, radiates energy at the rate of 551.1 W. Find the surface area of the body.
Solution :
Data : dQ/dt = 551.1 W,
T = 27 °C = (27 + 273) K = 300 K,
σ = 5.67 × 10^-8 W/m².K⁴
Let A be the surface area of the body. Energy radiated per unit time,

Question 14.
Calculate the energy radiated in two minutes by a perfect black sphere of radius 2 cm, maintained at 427 °C.
Solution :
Data : T = 427 °C = (427 + 273) K = 700 K, r = 2 cm = 0.02 m, σ = 5.67 × 10^-8 W/m².K⁴, t = 2 minutes = 120 seconds
∴ A = 4πr² = 4 × 3.142 × (0.02)² = 5.027 × 10^-3 m²
Energy radiated in time t = AσT⁴t
∴ Energy radiated in 120 seconds
= 5.027 × 10^-3 × 5.67 × 10^-8 × (700)⁴ × 120 J
= 8.212 × 10³ J

Question 15.
The temperature of the filament of a 100 watt electric lamp is 2727 °C. Calculate its emissivity if the length of the filament is 8 cm and its radius is 0.5 mm.
Solution :

Question 16.
A 60 watt electric bulb loses its energy entirely by radiation from the surface of its filament. If the surface area of the filament is 4 cm² and its coefficient of emission is 0.4, calculate the temperature of the filament.
Solution :

Question 17.
Energy is emitted from a hole in an electric furnace at the rate of 20 watts when the furnace is at 227 °C. Find the area of the hole.
Solution :
Data : T = 227 °C = (227 + 273) K = 500 K,
P = 20 W, σ = 5.67 × 10^-8 W/m².K⁴
Let A be the area at the hole. Then,
P = AσT⁴

Question 18.
A body of surface area 10 cm2 and temperature 727 °C emits 300 J of energy per minute. Find its emissivity.
Solution :
Data : A = 10 cm² = 10 × 10^-4m² = 10^-3m²,
T = 273 + 727 = 1000 K, Q = 300 J,

Question 19.
A pinhole is made in a hollow sphere of radius 5 cm whose inner wall is at 727 °C. Find the power radiated per unit area. [Stefan’s constant, a = 5.7 × 10^-8J/m².s.K⁴, emissivity (e) = 0.2]
Solution :
Data : r = 5 × 10^-2m, T = 727 + 273 = 1000 K, e = 0.2,
σ = 5.7 × 10^-2 J/m².s.K⁴
The power radiated per unit area, i.e., emissive power,
R = eσT⁴
= (0.2) (5.7 × 10^-8) (10³)⁴
= 1.14 × 10^-8 × 10¹² = 1.14 × 10⁴ W/m²

Question 20.
A body of surface area 400 cm² and absorption coefficient 0.5 radiates energy 1.5 kcal in 2 minutes when the temperature of the body is kept constant. Find the temperature of the body. (Given : J = 4186 J/kcal, σ = 5.67 × 10^-8 J/s.m.K⁴)
Solution :
Data : A = 400 cm² = 400 × 10^-4 4 m²
= 4 × 10^-2 m², absorption coefficient, a = 0.8
But a = e ∴ e = 0.8, J = 4186 J/kcal
Q = 1.5 kcal = 1.5 × 4186 J = 6279 J,
t = 2 minutes = 120 s, σ = 5.67 × 10^-8 J/s.m. K⁴
Energy radiated, Q = σ AeT⁴t
∴ 6279 = (5.67 × 10^-8) × (4 × 10^-2) × 0.8 × T⁴ × 120
∴ T⁴ = \(\frac{6279 \times 10^{8}}{21.77}\) = 288.4 × 10⁸
∴ T = 4.121 × 10² K
This is the temperature of the body.

Question 21.
The filament of an evacuated light bulb has length 10 cm, diameter 0.2 mm and emissivity 0.2. Calculate the power it radiates at 2000 K. (a = 5.67 × 10^-8 W/m².K⁴)
Solution :
Data : 1 = 10 cm = 0.1 m, d = 0.2 mm
∴ r = 0.1 mm = 0.1 × 10^-3 m, e = 0.2,
T = 2000 K, σ = 5.67 × 10^-3 W/m².K⁴
Surface area of the filament, A = 2πrl

This is the required quantity.

Question 22.
A body of surface area 10 cm² and temperature 727 °C emits 300 J of energy per minute. Find its emissivity.
Solution :
Data : A = 10 cm² = 10 × 10^-4 m² = 10^-3 m²,
T = 273 + 727 = 1000 K, Q = 300 J,
t = 1 minute = 60 s, σ = 5.67 × 10^-8 J/m².s.K⁴
Q = σAeT⁴t

Question 23.
A metal cube of each side of length 1 m loses all its energy at the rate of 3000 watts. If the emissivity of the material of the cube is 0.4, estimate its temperature.
Solution :
Data : L = 1 m, e = 0.4, \(\frac{d Q}{d t}\) = 3000 W, dt
a = 5.67 × 10^-8 W/m².K⁴
Surface area of the cube, A = 6L² = 6 m²

Question 24.
Assuming the Stefan-Boltzmann law, compare the rate of radiation from a metal ball at 727 °C with its rate of radiation at 527 °C. Also compare its rate of loss of heat at the two temperatures, if the temperature of the surroundings is 27 °C. Solution :
Data : T₀ = 273 + 27 = 300 K,
T₁ = 273 + 727 = 1000 K, T₂ = 273 + 527 = 800 K
(i) Rate of radiation (radiant power),
P = σAeT⁴
If P₁ and P₂ are the radiant powers at the temperatures T₁ and T₂, respectively,

Question 25.
A black body at 1000°c radiates 14.89 watt per square centimetre of its surface. The surfaceof a certain star radiates 10 kW per square centimetre. Assuming that the star’s surface behaves as a perfect blackbody, estimate its temperature.
Solution:
Data: T₁ = 273 + 1000 = 1273K,
R₁ = 14.89 watt/cm² = 14.89 × 10⁴ watt/m²,
R₂ = 10 kW/cm² = 10⁷ watt/cm²
Emissive power of a blackbody, R = σT⁴

∴ This is the temperature of the star’s surface.

Question 26.
A blackbody with initial temperature of 300 °C is allowed to cool inside an evacuated enclosure surrounded by melting ice at the rate of 0.35 °C/second. If the mass, specific heat and surface area of the body are 32 grams, 0.10 cal/g °C and 8 cm² respectively, calculate Stefan’s constant. (Take J = 4200 j/kcal.)
Solution :
Data : T = 273 + 300 = 573 K, T₀ = 273 K,
\(\frac{d Q}{d t}\) = 0.35 °C/s = 0.35 K/s, at
M = 32 g = 32 × 10^-3 kg, A = 8 cm² = 8 × 10^-4 m² C = 0.10 cal/g.°C = 0.10 kcal/kg.K = 420 j/kg.K
since J = 4200 J/kcal

Question 27.
A blackbody at 327 °C, when suspended in a black enclosure at 27 °C, loses heat at a certain rate. Find the temperature of the body at which its rate of loss of heat by radiation will be half of the above rate. Assume that the other conditions remain unchanged.
Solution :
Data : T₁ = 273 + 327 = 600 K, T₀ = 273 + 27 = 300 K,

Question 49.
Choose the correct option.

Question 1.
The rms speed of a gas molecule is directly proportional to
(A) its absolute temperature
(B) the square root of its absolute temperature
(C) the square of its absolute temperature
(D) its molar mass.
Answer:
(B) the square root of its absolute temperature

Question 2.
Temperature remaining constant, if you double the number of molecules in a box, the pressure will
(A) remain unchanged
(B) double
(C) become one-half
(D) quadruple.
Answer:
(B) double

Question 3.
The pressure P of an ideal gas having volume V is \(\frac{2 E}{3 V}\). Then E is
(A) translational kinetic energy
(B) rotational kinetic energy
(C) vibrational kinetic energy
(D) potential energy.
Answer:
(A) translational kinetic energy

Question 4.
According to the kinetic theory of gases, at a given temperature, molecules of all gases have the same
(A) rms speed
(B) momentum
(C) energy
(D) most probable speed.
Answer:
(C) energy

Question 5.
The kinetic energy per molecule of a gas at temperature T is
(A) \(\frac{3}{2} R T\)
(B) \(\frac{3}{2} k_{\mathrm{B}} T\)
(C) \(\frac{2}{3} R T\)
(D) \(\frac{3}{2} \frac{R T}{M_{0}}\)
Answer:
(B) \(\frac{3}{2} k_{\mathrm{B}} T\)

Question 6.
When the temperature of an enclosed gas is increased by 2 °C, its pressure increases by 0.5%. The initial temperature of the gas was
(A) 250 K
(B) 275 K
(C) 300 K
(D) 400 K.
Answer:
(D) 400 K.

Question 6.
For a given gas at 800 K, the rms speed of the ‘ molecules is
(A) four times the rms speed at 200 K
(B) half the rms speed at 200 K
(C) twice the rms speed at 200 K
(D) twice the rms speed at 400 K.
Answer:
(C) twice the rms speed at 200 K

Question 7.
If the absolute temperature of a gas becomes three times the initial absolute temperature, the rms speed of the gas molecules
(A) becomes \(\frac{1}{3}\) times the initial rms speed
(B) becomes \(\frac{1}{\sqrt{3}}\) times the initial rms speed
(C) becomes \(\sqrt{3}\) times the initial rms speed
(D) becomes 3 times the initial rms speed.
Answer:
(C) becomes \(\sqrt{3}\) times the initial rms speed

Question 8.
An ideal gas occupies 2 m³ at a pressure of 2 atm. Taking 1 atm = 10⁵ Pa, the energy density of the gas is
(A) 3 × 10⁵ J/m³
(B) 1.5 × 10⁵ J/m³
(C) 300 J/m³
(D) 150 J/m³.
Answer:
(A) 3 × 10⁵ J/m³

Question 9.
An ideal gas is confined to a cylinder with a movable piston. As it is heated to twice its initial absolute temperature, the gas is allowed to expand freely against the atmospheric pressure. The average thermal energy of the molecules
(A) quadruples
(B) doubles
(C) increases by a factor of \(\sqrt{2}\)
(D) remains unchanged.
Answer:
(B) doubles

Question 10.
Equal volumes of hydrogen and oxygen (relative molar masses 2 and 32, respectively) in separate containers are equimolar and exert equal pressure. The rms speeds of hydrogen and oxygen molecules are in the ratio
(A) 1 : 32
(B) 1 : 16
(C) 1 : 4
(D) 4 : 1.
Answer:
(D) 4 : 1.

Question 11.
At what temperature will the rms velocity of a gas molecule be double its value at NTP ?
A) 273 °C
(B) 546 °C
(C) 819 °C
(D) 1092 °C.
Answer:
(C) 819 °C

Question 12.
The rms speed of the molecules of a gas is 200 m/s at 27 °C and 1 atmosphere pressure. The rms speed at 127 °C and double the pressure is

Answer:
(C) \(\frac{400}{\sqrt{3}}\) m/s

Question 13.
The temperature at which helium molecules have the same rms speed as hydrogen molecules at STP is
[M_oH = M_oHe = 4g/mol]
A) 1092 K
(B) 546 K
(C) 300 K
(D) 273 K.
Answer:
(B) 546 K

Question 14.
The number of degrees of freedom for a rigid diatomic molecule is
(A) 3
(B) 5
(C) 6
(D) 7
Answer:
(B) 5

Question 15.
For polyatomic molecules having / vibrational modes, the ratio of two specific heats is

Answer:
(C) \(\frac{4+f}{3+f}\)

Question 16.
A nonlinear triatomic molecule has …… degree(s) of freedom of rotational motion.
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(D) 3

Question 17.
The wavelength range of thermal radiation is
(A) from 4000 Å to 7000 Å
(B) from 7700 Å to 4 × 10⁶ Å
(C) from 10⁶ Å to 10⁸ Å
(D) from 4 × 10^-12 Å to 4 × 10⁸ Å.
Answer:
(B) from 7700 Å to 4 × 10⁶ Å

Question 18.
The coefficient of reflection of a perfectly opaque body is 0.16. Its coefficient of emission is
(A) 0.94
(B) 0.84
(C) 0.74
(D) 0.64.
Answer:
(B) 0.84

Question 19.
Which of the following materials is diathermanous ?
(A) Wax
(B) Glass
(C) Quartz
(D) Porcelain
Answer:
(C) Quartz

Question 20.
Which of the following substances is opaque to radiant energy?
(A) Carbon tetrachloride
(B) Sodium chloride
(C) Benzene
(D) Potassium bromide
Answer:
(C) Benzene

Question 21.
The substance which allows heat radiations to pass through it is
(A) iron
(B) water vapour
(C) wood
(D) dry air.
Answer:
(D) dry air.

Question 22.
A perfect blackbody is the one that
(A) absorbs all incident radiation
(B) reflects all incident radiation
(C) transmits all incident radiation
(D) both reflects and transmits incident radiation.
Answer:
(A) absorbs all incident radiation

Question 23.
The conical projection in Fery’s blackbody is
(A) used to support the spheres
(B) used to transmit incident radiation to outer sphere
(C) used to prevent reflected radiation to escape outside
(D) used for all of the above purposes.
Answer:
(C) used to prevent reflected radiation to escape outside

Question 24.
The emissive power of a body is
(A) the energy emitted by the body in a given time
(B) the radiant energy emitted by the body per unit area of the body
(C) the radiant energy emitted by the body per unit time
(D) the radiant energy emitted by the body per unit time per unit area of the body.
Answer:
(D) the radiant energy emitted by the body per unit time per unit area of the body.

Question 25.
The dimensions of emissive power are
(A) [M¹L^-2T^-3]
(B) [M¹L²T^-3]
(C) [M¹L⁰T^-3]
(D) [M¹L^0-2]
Answer:
(C) [M¹L⁰T^-3]

Question 26.
The emissive power per wavelength interval R_λ of a blackbody at an absolute temperature T₁ is maximum at λ₁ = 1.1 µm. At an absolute temperature T₂, its R_λ is maximum at λ₂ = 0.55 µm. Then, \(\frac{T_{1}}{T_{2}}\) is equal to
(A) \(\frac{1}{2}\)
(B) 1
(C) 2
(D) 4.
Answer:
(A) \(\frac{1}{2}\)

Question 27.
The temperature of the photosphere of the Sun is about 6000 K. Wien’s displacement law constant is 2.898 × 10^-3 m.K. The photosphere has maximum emissive power at wavelength
(A) 483 nm
(B) 496.7 nm
(C) 4830 nm
(D) 4967 nm.
Answer:
(A) 483 nm

Question 28.
A sphere and a cube made of the same metal have equal volumes, identical surface characteristics and are at the same temperature. If they are allowed to cool in identical surroundings, the ratio of their rates of loss of heat will be
(A) \(\frac{4 \pi}{3}\) : 1
(B) 1 : 1
(C) \(\left(\frac{\pi}{6}\right)^{\frac{2}{3}}\) : 1
(D) \(\left(\frac{\pi}{6}\right)^{\frac{1}{3}}\) : 1
Answer:
(D) \(\left(\frac{\pi}{6}\right)^{\frac{1}{3}}\) : 1

Question 29.
If the absolute temperature of a blackbody is increased by a factor 3, the energy radiated by it per unit time per unit area will increase by a factor of
(A) 9
(B) 27
(C) 81
(D) 243.
Answer:
(C) 81

Question 30.
Two spheres P and Q, having radii 8 cm and 2 cm, and of the same surface characteristics are maintained at 127 °C and 527 °C, respectively. The ratio of the radiant powers of P to Q is
(A) 0.0039
(B) 0.0156
(C) 1
(D) 2.
Answer:
(C) 1

Question 31.
Two copper spheres of radii 6 cm and 12 cm, respectively, are suspended in an evacuated enclosure. Each of them is at a temperature of 15 °C above the surroundings. The ratio of their rate of loss of heat is
(A) 2 : 1
(B) 1 : 4
(C) 1 : 8
(D) 8 : 1.
Answer:
(B) 1 : 4

Question 32.
The peak of the radiation spectrum of a blackbody occurs at 2 µn. Taking Wien’s displacement law constant as 2.9 × 10^-3 m.K, the approximate temperature of the blackbody is
(A) 15 K
(B) 150 K
(C) 750 K
(D) 1500 K.
Answer:
(D) 1500 K.

Question 33.
The light from the Sun is found to have a maximum intensity near the wavelength of 470 nm. Assuming the surface of the Sun as a blackbody, the temperature of the Sun is [Wien’s constant b = 2.898 × 10^-3 m.K]
(A) 5800 K
(B) 6050 K
(C) 6166 K
(D) 6500 K.
Answer:
(C) 6166 K

Question 34.
The amount of energy radiated per unit time by a body does not depend upon the
(A) nature of its surface
(B) area of its surface
(C) mass of the body
(D) temperature difference of the surface and surroundings.
Answer:
(C) mass of the body

Question 35.
Two gases exert pressure in the ratio 3 : 2 and their densities are in the ratio 2 : 3. Then the ratio of their rms speeds is
(A) 2 : 3
(B) 3 : 2
(C) 2 : 1
(D) 1 : 2.
Answer:
(B) 3 : 2

Question 36.
Find the wavelength at which a blackbody radiates maximum energy, if its temperature is 427 °C.
[Wien’s constant b = 2.898 × 10^-3 m.K]
(A) 0.0414 × 10^-6 m
(B) 4.14 × 10^-6 m
(C) 41.4 × 10^-6 m
(D) 414 × 10^-6 m
Answer:
(B) 4.14 × 10^-6 m

Question 37.
If the total kinetic energy per unit volume of gas enclosed in a container is E, the pressure exerted by the gas is
(A) E
(B) \(\frac{3}{2}\)E
(C) \(\sqrt{3}\)E
(D) \(\frac{2}{3}\)E.
Answer:
(D) \(\frac{2}{3}\)E.

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 2 Mechanical Properties of Fluids Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 2 Mechanical Properties of Fluids

Question 1.
What is fluid? Give two examples.
Answer:
A fluid is a substance that can flow. A fluid has shear modulus O and yields to shear. Under shear stress and a pressure gradient, fluid begins to flow. Liquids, gases, and plasmas are collectively called fluids.

Examples: All gases, all liquids, molten glass and lava, honey, etc.

Question 2.
What is an ideal fluid?
OR
State the characteristics of an ideal fluid.
Answer:
An ideal fluid is one that has the following properties:

1. It is incompressible, i.e., its density has a constant value throughout the fluid.
2. Its flow is irrotational, i.e., the flow is steady or laminar. In an irrotational flow, the fluid doesn’t rotate like in a whirlpool arid the velocity of the moving fluid at a specific point doesn’t change over time. (Many fluids change from laminar to turbulent flow as the speed of the fluid increases above some specific value. This can dramatically change the properties of the fluid.)
3. Its flow is nonviscous or inviscid, i.e., internal friction or viscosity is zero so that no energy lost due the motion of the fluid.

Question 3.
How does a fluid differ from a solid ?
Answer:
In response to a shear as well as normal force, a solid deforms and develops a restoring force. Within the elastic limit, both types of deformation is reversible. A solid changes its shape under a shear. A normal force causes a change in its length or volume. If the elastic limit is exceeded, the solid gets an irreversible deformation called a permanent set.

A fluid, on the other hand, can only be subjected to normal compressive stress, called pressure. A fluid does not have a definite shape, so that under a shear it begins to flow, Real fluids, with non-zero viscosity, display a weak resistance to shear.

Question 4.
State the properties of a fluid.
Answer:
Properties of a fluid :

1. They do not resist deformation and get permanently deformed.
2. They are capable of flowing.
3. They take the shape of the container.

Question 5.
Define pressure. State its SI and CGS units and dimensions.
Answer:
Definition : The pressure at a point in a fluid in hydrostatic equilibrium is defined as the normal force per unit area exerted by the fluid on a surface of infinitesimal area containing the point.
Thus, the pressure, p = \(\lim _{\Delta A \rightarrow 0} \frac{F}{\Delta A}\)
where F is the magnitude of the normal force on a surface of area ∆A. The pressure is defined to be a scalar quantity.

SI unit: the pascal (Pa), 1 Pa = 1 N∙m^-2
CGS unit: the dyne per square centimetre (dyn/cm²)
Dimensions : [p] = [F][A^-1] = [MLT^-2, L^-2]
= [ML^-1 T^-2]

Question 6.
State two non-SI units of pressure.
Answer:
Two non-SI units, which are either of historical interest, or are still used in specific fields are the bar and the torr.
1 bar = 0.1 MPa = 100 kPa = 1000 hPa = 10⁵Pa
1 torr = (101325/760) Pa = 133.32 Pa
[Note : Their use in modern scientific and technical work is strongly discouraged.]

Question 7.
If a force of 200 N is applied perpendicular to a surface of area 10 cm², what is the corresponding pressure ?
Answer:
Pressure, p = \(\frac{F}{A}=\frac{200 \mathrm{~N}}{10 \times 10^{-6} \mathrm{~m}^{2}}\) = 2 × 10⁷ N/m²

Question 8.
Explain why the forces acting on any surface within a fluid in hydrostatic equilibrium must be normal to the surface.
Answer:
In a fluid, the molecules are in a state of random motion and the intermolecular cohesive forces are weak. If a fluid is subjected to a tangential force (shear) anywhere within it, the layers of the fluid slide over one another, i.e., the fluid begins to flow. Thus, a fluid cannot sustain a tangential force. So, in

turn, a fluid at rest cannot exert a tangential force on any surface with which it is in contact. It can exert only a force normal to the surface. Hence, if a fluid is in hydrostatic equilibrium (i.e., at rest), the force acting on any surface within the fluid must be normal to the surface.

Question 9.
Would you rather have someone wearing studs step on your foot or have someone wearing tennis shoes step on your foot ?
Answer:
A person would exert the same downward force regardless of whether he or she was wearing studs or tennis shoes. However, if the person were wearing studs, the force would be applied over a much smaller area, so the pressure would be greater (and so would be more painful).

Question 10.
Would you rather have an elephant stand on your foot directly or have an elephant balance on a thumbtack on top of your foot?
Answer:
The downward force of the elephant’s weight would be applied over a much smaller area if it were balancing on a thumbtack, so the pressure would be greater.

Question 11.
Derive an expression for pressure exerted by a liquid column.
Answer:
At a point at depth h below the surface of a liquid of uniform density ρ, the pressure due to the liquid is due to the weight per unit area of a liquid column of height h above that point.

In above figure to find the pressure due to the liquid at point P, consider the cylindrical liquid column, of cross section A and height h, above that point.

The weight of this liquid column = volume × density × acceleration due to gravity
= (Ah)(ρ)(g)
∴ Pressure due to the liquid at depth h
= \(\frac{\text { weight of the liquid column }}{\text { cross sectional area }}\)
= \(\frac{A h \rho g}{A}\) = hpg
If the free surface of the liquid is open to the atmosphere, the pressure on the surface is the atmosphere pressure p₀. Then, the absolute pressure within the liquid at a depth h is p = p₀ + hρg

Question 12.
State the characteristics of pressure due to a liquid at rest at a point within it.
Answer:
Characteristics of pressure due to a liquid at rest at a point within it:

1. Within a liquid of constant density, the pressure is directly proportional to the depth.
2. At the same depth within liquids of different densities, the pressure is directly proportional to the density of the liquid.
3. Within a liquid of constant density, the pressure at a given depth is directly proportional to the acceleration due to gravity.
4. The pressure at a point within a given liquid is the same in all directions.
5. The pressure at all points at the same horizontal level within a given liquid is the same.

Question 13.
How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force ?
Answer:
Pressure p = F/A. Therefore, the force on one side is F = ρ ∙ A = (1.013 × 10⁵ Pa) (8.50 × 11.0 × 10^-4 m²) = 947.2 N.

The pressure at a point within a fluid being the same in all directions, the same force acts on the other side of the paper. Thus, the net force on the paper is zero.

Question 14.
What is the pressure exerted by a water column of height 1 m?[ρ = 10³ kg/m³, g = 9.8 m/s²]
Answer:
Pressure exerted by the water column = hρg
= 1 m(10³ kg/m³) × (9.8 m/s²)
= 9.8 × 10³Pa

Question 15.
Would you rather breathe through a 2 m long tube to the surface in 1.5 m of water in the ocean or breathe at the beach near the ocean?
Answer:
The pressure on one’s lungs would be much greater under water than standing on the beach because the force exerted by the water on the lungs ‘ is greater than the force exerted by the air. Because the pressure of the water on the lungs is so much greater than the outward pressure of the air inside, it would be difficult to take a breath under 1.5 m of water than on the beach.

Question 16.
What is atmospheric pressure ? Define standard atmospheric pressure.
Answer:
The Earth’s surface is covered with a layer of atmosphere, with more than 99% of the atmosphere lying within 31 km of the surface. The weight of the atmosphere exerts a downward thrust on any surface lying within it. This gives rise to atmospheric pressure. The atmospheric pressure at any height above the Earth’s surface is the weight of a column of air of unit cross section from that altitude to the top of the atmosphere.

Definition : Standard atmospheric pressure, or one atmosphere of pressure, is defined as the pressure equivalent of a column of mercury that is exactly 0.7600 m in height at 0 °C.

We can calculate this equivalent pressure in SI unit by using the density of mercury
ρ = 13.6 × 10³ kg/m³ and g = 9.80 m/s².
1 atm = (0.76 m) . (13.6 × 10³ kg/m³) . (9.80 m/s²)
= 1.013 × 10⁵ Pa = 101.3 kPa
[Note : 1000 mbar = 100 kPa. Therefore, 1 atm = 1013 mbar.]

Question 17.
Explain gauge pressure and absolute pressure within a liquid open to the atmosphere.
OR
Explain the effect of gravity on fluid pressure.
Answer:
Consider a cylindrical fluid column of uniform density ρ, area of cross section A and height h,

The mass of the fluid within the column is
m = density × volume
= ρAh

If p₁ and p₂ are the pressures at the top and bottom faces of the column, the forces on the top and bottom faces are respectively.
F₁ = p₁A + mg (downward)
and F₂ = p₂A (upward)
If the column is in equilibrium,
F₂ = F₁
∴ p₂A = p₁A + mg = p₁A + ρAhg
∴ P₂ ~ P₁ = ρhg
If p₁ = p₀ = atmospheric pressure, the gauge pressure
P₂ – P₀ = ρhg

In the absence of gravity, p₂ = p₀ But since atmospheric pressure is equal to the weight per unit area of the entire air column above, even p0 will be zero in the absence of gravity.

Question 18.
Define gauge pressure.
When is gauge pressure (i) positive (ii) negative ?
Give two examples where gauge pressure is more relevant.
Answer:
Definition : Gauge pressure is the pressure exerted by a fluid relative to the local atmospheric pressure.

Gauge pressure, p_g = p – p₀

where p is the absolute pressure and p₀ is the local atmospheric pressure.

When the pressure inside a closed container or tank is greater than atmospheric pressure, the pressure reading on a pressure gauge is positive. The pressure inside a ‘vacuum chamber’-a rigid chamber from which some of the air is pumped out-is less than the atmospheric pressure, so a pressure gauge on the chamber designed to measure negative pressure reads a negative value.

At a depth within a liquid of density ρ, the gauge pressure is p_g = p – p₀ = hpg

Examples : Tyre pressure gauge, blood pressure gauge, pressure gauge on an oxygen or scuba tank.

Question 19.
Define absolute pressure.
Answer:
Definition : The absolute pressure, or total press-ure, is measured relative to absolute zero on the pressure scale-which is a perfect vacuum-and is the sum of gauge pressure and atmospheric press-ure. It is the same as the thermodynamic pressure.

Absolute pressure accounts for the atmospheric pressure, which in effect adds to the pressure in any fluid which is not enclosed in a rigid container i.e., the fluid is open to the atmosphere.
p = p₀ + P_g
where p₀ and p_g are respectively atmospheric pressure and the gauge pressure.

Absolute pressure can be never negative.

Question 20.
If your tyre gauge reads 2.31 atm (234.4 kPa), what is the absolute pressure ?
Answer:
The absolute pressure, p = p₀ + p_g = 1 atm + 2.31 atm = 3.31 atm (≅ 335 kPa).

Question 21.
State and explain the hydrostatic paradox. OR Explain hydrostatic paradox.
Answer:
Hydrostatic paradox : The normal force exerted by a liquid at rest on the bottom of the containing vessel is independent of the amount of liquid or the shape of the container, but depends only on the area of the base and its depth from the liquid surface.

Consider several vessels of the same base area as shown in figure (a). A liquid is poured into them to the same level, so that the pressure is the same at the bottom of each vessel. Then it must follow that the normal force on the base of each vessel is also the same. However, when placed on a scale balance they are found to have different weights. Herein lies the paradox.

Explanation : Since a liquid always exerts a normal force on a wall of the container, in turn, the wall exerts an equal and opposite reaction on the liquid. In the case of tube A, this reaction is everywhere horizontal; so that the normal force at the base of A is only due to the weight of the liquid column above.

The reaction of the slanted wall of vessel C has a – vertical component, as shown in figure (b), which supports the weight of the liquid above the slanted side. Hence, the normal force at the base of C is only due to the weight of the vertical liquid column above the base, shown by dashed lines. Since the vessels A and C are filled to the same height and have the same base area, the pressures at the bases of the two vessels are also same. However, the volume of the liquid being clearly different, they have different weights.

In the case of vessel B, the downward vertical component of the reaction of the wall provides an extra normal force at the base, as shown in figure (c).

Question 22.
Can pressure in a fluid be increased by pushing directly on the fluid ? Give an example.
Answer:
Yes, but it is much easier if the fluid is enclosed.

Examples : (1) The heart increases the blood pressure by pushing on the blood in an enclosed ventricle.
(2) Hydraulic brakes, lifts and cranes operate by pushing on oil in an enclosed system.

Question 23.
State Pascal’s law.
Answer:
Pascal’s law : A change in the pressure applied to an enclosed fluid at rest is transmitted un-diminished to every point of the fluid and to the walls of the container, provided the effect of gravity can be ignored.

[Note : The law does not say that ‘the pressure is the same at all points of a fluid’ – rightly so, since the pressure in a fluid near Earth varies with height. Rather, the law applies to the change in pressure. According to Pascal’s law, if the pressure on an enclosed static fluid is changed by a certain amount, the pressure at all points within the fluid changes by the same amount.

The above law is due to Blaise Pascal (1623 – 62), French mathematician and physicist.]

Question 24.
Describe an experimental proof of Pascal’s law.
Answer:
Consider a spherical vessel having four cylindrical tubes A, B, C and D each fitted with air-tight

frictionless pistons of areas of cross section A, A/2, 2A and 3A, respectively, as shown in above figure. The vessel is filled with an incompressible liquid such that there is no air between the liquid and the pistons.

If the piston A is pushed with a force F, the pressure on the piston and the liquid in the vessel is p_A = F/A. It is seen that the other three pistons are pushed outwards. To keep these pistons at their respective original positions, forces of F/2, IF and 3F, respectively are required to be applied on pistons B, C and D respectively to hold them. Then, the pressures on the respective pistons are
p_B = \(\frac{F / 2}{A / 2}\) = F/A, p_C = 2F/2A = F/A, and
p_D = 3F/3A = F/A
∴ p_A = p_B = p_C = p_D = F/A

This indicates that the pressure applied is trans-mitted equally to all parts of liquid. This proves Pascal law.

Question 25.
Explain the principle of multiplication of thrust.
Answer:
Principle of multiplication of thrust by transmission of fluid pressure : The normal force exerted by a fluid on any surface in contact with it is called the thrust. Consider two hydraulically connected cylinders, one of cross section a and the other A, as in figure. If a force F_a is exerted on the smaller piston, pressure p = \(\frac{F_{\mathrm{a}}}{a}\) is produced and transmitted undiminished throughout the liquid. Then, the thrust F_A on the larger piston is
F_A = pA = \(\frac{A}{a}\) F_a
If A = na, F_A = nF_a, i.e., the thrust on the larger piston is multiplied n times. This is known as the principle of multiplication of thrust by transmission of fluid pressure.

Question 26.
State any two applications of Pascal’s law.
Briefly explain their working.
Ans.
Applications of Pascal’s law :

1. Hydraulic car lift and hydraulic press
2. Hydraulic brakes.

All the above applications work on the principle of multiplication of thrust by transmission of fluid pressure.

(1) Working of a hydraulic lift : Two hydraulically connected cylinders, one of cross section a and the other A, are such that A is many times larger than a : A = na. If a force F_a is exerted on the smaller piston, a pressure p = \(\frac{F_{\mathrm{a}}}{a}\) is produced and transmitted undiminished throughout the liquid. Then, the thrust F_A on the larger piston
F_A = pA = \(\frac{A}{a}\) F_a = nF_a
is n times greater than that on the smaller piston. A platform attached to the larger piston can lift a car (as in a hydraulic car lift), or press bales of cotton or paper against a fixed rigid frame (as in Brahma’s.)

(2) Working of hydraulic brakes in a car: Brakes which are operated by means of hydraulic pressure are called hydraulic brakes. An automobile hydraulic brake system, shown schematically in below figure, has fluid-filled master and slave cylinders connected by pipes. When the brake pedal is pushed, it depresses the piston of the pedal or master cylinder through a lever. The change in pressure in the master cylinder is transmitted to the four wheel or slave cylinders. Since the brake fluid is incompressible, the pistons of the slave cylinders are pushed out, pressing braking pads onto the braking discs on the wheels. Note that we can add as many wheel cylinders as we wish.

The master cylinder has a much smaller area of cross section A_m compared to the combined area A_s

of the slave cylinders. Hence, with a small force F_m on the master cylinder, a force F_s = \(\frac{A_{\mathrm{s}}}{A_{\mathrm{m}}}\) F_m which is greater than F_m is applied on each slave cylinder. Consequently, the master piston has to travel sev-eral inches to move the slave pistons the fraction of an inch it takes to apply the brakes. But the arrangement allows great force to be exerted at the brake pads.

[Note : (1) Pascal’s law laid the foundation for hydraulics, the use of a liquid under pressure to transfer force or motion, or to increase an applied force. It is one of the most important branches in modern engineering. (2) A hydraulic system, as an example of a simple machine, can increase force but cannot do more work than is done on it. Work being force times the distance moved, the piston in a wheel cylinder moves through a smaller distance than that in the pedal cylinder. Power brakes in modern automobiles have a motorized pump that does most of the work in the system.]

Question 27.
Why are liquids used in hydraulic systems but not gases?
Answer:
Liquids are used in a hydraulic system because liquids are incompressible and transmit a change in pressure undiminished to all parts of the system. On the other hand, on increasing the pressure, a gas will be compressed into a smaller volume due to which there will be no transmission of force or motion.

Question 28.
State one advantage of hydraulic brakes in an automobile.
Answer:
Advantages of

1. By Pascal’s law, equal braking effort is applied to all the wheels.
2. It is easily possible to increase or decrease the applied force-during the design stage-by changing the size of piston and cylinder relative to other.

Question 29.
What is a barometer? Explain the use of a simple mercury barometer to measure atmospheric pressure.
Answer:
A barometer is an instrument to measure atmospheric pressure. The mercury barometer was in-vented by Evangelista Torricelli (1609-47). Italian physicist and mathematician.

A strong glass tube, about one metre long and closed at one end, is filled with mercury. With a finger over the open end, the tube is inverted and the open end is immersed into a bowl of mercury. When the finger is removed, the mercury level in the tube drops. The mercury column in the tube stands at a height h for which the pressure at point A inside the tube due to the weight of the mercury column is equal to the atmospheric pressure p0 outside (at point B).

The space at the closed end of the tube, after the mercury level drops, is nearly a vacuum, known as the Torricellian vacuum, so the pressure there can be taken as zero. It, therefore, follows that p₀ = pgh Where p is the density of mercury and h is the height of the mercury column.

Question 30.
What is an open tube manometer? Briefly describe its function with a neat diagram.
Answer:
An open tube manometer is a device to measure the pressure of a gas in a vessel. It consists of a U-shaped tube containing a liquid (say, mercury) of density p, as shown in below figure.

One end of the tube is connected to the vessel while the other end is open to the atmosphere. The pressure p at point A is the (unknown) pressure of the gas in the vessel. The pressure on the mercury column in the open tube is the atmospheric pressure p₀.

A point B, at the same horizontal level as A, is at a depth h from the surface of mercury in the open tube. Therefore, the pressure at B is p₀ + ρgh.

The pressures at points A and B at the same liquid level being the same, equating the unknown pressure p (at A) to the pressure at B.
p = p₀ + ρgh
The pressure p is called the absolute pressure, and the difference in pressure p – p₀ is called the gauge pressure.

Question 31.
An open tube manometer is connected to (i) a vacuum-packed candy jar, with the atmospheric pressure in the open tube supporting a column of fluid of height h (ii) a gas tank, with the absolute pressure in the tank supporting a column of fluid of height h. Is the absolute pressure in the jar and the gas tank greater than or less than the atmospheric pressure ? By how much ?
Answer:
In the first case, p_abs is less than the atmospheric pressure, whereas in the second case, p_abs is greater than the atmospheric pressure. In both cases, p_abs differs from the atmospheric pressure by the gauge pressure hρg, where ρ is the density of the fluid in the manometer.

32. Solve the following

Question 1.
For diver’s safety, a 10 m platform diving pool should be 5 m deep. However, with an excellent dive, a diver usually reaches a maximum depth of 2.5 m.
(i) Calculate the pressure due to the weight of the water at the depth of 2.5 m.
(ii) Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.0 atm. [Density of water = 10³ kg/m³, 1 atm = 101.3 kPa]
Solution:
Data : h = 250 m, ρ = 1000 kg/m³, g = 9.8 m/s², 1 atm = 101.3 kPa
(i) ρ = hpg = (250)(1000)(9.8) = 2.45 mPa
= \(\frac{2.45 \times 10^{6}}{1.013 \times 10^{5}}\) = 24.18 atm
This gives the pressure at a depth of 250 m.

(ii) h = \(\frac{p}{\rho g}=\frac{1.013 \times 10^{5}}{10^{3} \times 9.8}\) = 10.34 m
This gives the required depth.

Question 2.
Suppose a dam is 250 m wide and the water is 40 m deep at the dam. What is
(i) the average pressure on the dam
(ii) the force exerted against the dam due to the water?
Solution :
Data : Width, L = 250 m, depth H = 40 m, ρ = 1000 kg/m³, g = 9.8 m/s²
Since pressure increases linearly with depth, the average pressure p_av due to the weight of the water is the pressure at the average depth h of 20 m. The force exerted on the dam by the water is the average pressure times the area of contact, F = p_av A = p_av LH.
(i) p _av = hρg = (20)(1000)(9.8) = 1.96 × 10⁵Pa
(ii) F = p_avA = p_avLH = (1.96 × 10⁵)(250)(40)
= 1.96 × 10⁵ N

Question 3.
A car lift at a service station has a piston of diameter 30 cm. The lift and piston weigh 800 kg wt. What pressure (in excess of the atmospheric pressure) must be exerted on the piston to raise a car weighing 1700 kg wt at a constant speed? [g = 9.8 m/s²]
Solution:
Data : Piston diameter, D = 30 cm = 0.3 m. mass of lift and piston, m = 800 kg, mass of car, M = 1700 kg
Cross-sectional area of the piston.
A = \(\frac{\pi D^{2}}{4}=\frac{3.142(0.3 \mathrm{~m})^{2}}{4}\) = 7.07 × 10^-2 m ²
Total weight of the car and lift,
W = (m + M)g
= (800 kg + 1700 kg) (9.8 m/s²)
= 2.45 × 10⁴ N
Therefore, the pressure on the piston
p = \(\frac{F}{A}=\frac{W}{A}=\frac{2.45 \times 10^{4} \mathrm{~N}}{7.07 \times 10^{-2} \mathrm{~m}^{2}}\)
= 3.465 × 10⁵ Pa
A pressure of 3.465 × 10⁵ Pa must be exerted on the piston.

Question 4.
The diameters of two pistons in a hydraulic press are 5 cm and 25 cm respectively. A force of 20 N is applied to the smaller piston. Find the force exerted on the larger piston.
Solution:
Data : D₁ = 5 cm, D₂ = 25 cm, F₁ = 20 N
By Pascal’s law,

Question 5.
In a hydraulic lift, the input piston has surface area 20 cm². The output piston has surface area 1000 cm². If a force of 50 N is applied to the input piston, it raises the output piston by 2 m. Calculate the weight of the support on the output piston and the work done by it.
Solution:
Data : A₁ = 20 cm² = 2 × 10^-3 m²,
A₂ = 1000 cm² = 10^-1 m², F₁ = 50 N, s₂ = 2m
(i) By Pascal’s law,

This gives the weight of the support on the output piston.

(ii) The work done by the force transmitted to the output piston is
F₂S₂ = (2500 N) (2 m)
= 5000 J

Question 6.
A driver pushes the brake pedal of a car exerting a force of 100 N that is increased by the simple lever to a force of 500 N on the pedal (master) cylinder. The hydraulic system transmits this force to the four wheel (slave) cylinders. If the pedal cylinder has a diameter of 0.5 cm and each wheel cylinder has a diameter of 2.5 cm, calculate the magnitude of the force F_s on each of the wheel cylinder.
Solution:
Data : F_m = 500 N, D_m = 1 cm, D_s = 2.5 cm
\(\frac{F_{\mathrm{s}}}{A_{\mathrm{s}}}=\frac{F_{\mathrm{m}}}{A_{\mathrm{m}}}\)
∴ The magnitude of the force on each of the wheel cylinders,
F_s = \(\frac{A_{\mathrm{s}}}{A_{\mathrm{m}}}\) F_m = (\(\frac{D_{\mathrm{s}}}{D_{\mathrm{m}}}\))² F_m = (\(\frac{2.5}{0.5}\))² (500)
= 25 × 500 = 12.5 kN

Question 7.
Mercury manometers are often used to measure arterial blood pressure. The typical blood pressure of a young adult raises the mercury to a height of 120 mm at systolic and 80 mm at diastolic. Express these values in pascal and bar. [Density of mercury = 13600 kg/m³, 1 mbar = 100 Pa]
Solution:
Data : p_max = p_syst = 120 mm of Hg, p_min = p_dias = 80 mm of Hg, ρ = 13600 kg/m³, g = 9.8 m/s², 1 mbar = 100 Pa
p = hρg
∴ p_syst = (0.120)(1.36 × 10⁴)(9.8)
= 1.6 × 10⁴ Pa = 16 kPa
= 1600 mbar = 1.5 bar

and P_dias = (0.08)(1.36 × 10⁴)(9.8)
= 1.066 × 10⁴ Pa = 10.66 kPa
= 1066 mbar = 1.066 bar

Question 33.
Describe the phenomenon of surface tension, giving four examples.
Answer:
Surface tension is a unique property of liquids that arises because the net intermolecular force of attraction on the liquid molecules at or near a liquid surface differs from that on molecules deep in the interior of the liquid. This results in the tendency of the free surface of a liquid to minimize its surface area and behave somewhat like a stressed elastic membrane.

Surface tension is important in understanding the peculiar behaviour of the free surface of a liquid in many cases as illustrated below :

1. Small quantities of liquids assume the form of spherical droplets, as in mist, or a mercury droplet on a flat surface. This is because the stressed surface ‘skin’ tends to contract and mould the liquid into a shape that has minimum surface area for its volume, i.e., into a sphere.
2. Surface tension is responsible for the spherical shape of freely-falling raindrops and the behaviour of bubbles and soap films.
3. The bristles of a paint brush cling together when it is drawn out of water or paint.
4. A steel needle or a razor blade can, with care, be supported on a still surface of water which is much less dense than the metal from which these objects are made of.
5. Many insects like ants, mosquitoes, water striders, etc., can walk on the surface of water.

Question 34.
Define (1) cohesive force (2) adhesive force.
Give one example in each case.
Answer:
(1) Cohesive force : The intermolecular force of attraction between two molecules of the same material is called the cohesive force.
Example : The force of attraction between two water molecules.

(2) Adhesive force : The intermolecular force of attraction between two molecules of different materials is called the adhesive force.
Example : The force of attraction between a water molecule and a molecule of the solid surface which is in contact with water.

Question 35.
Define (1) range of molecular attraction or molecular range (2) sphere of influence.
Answer:
1) Range of molecular attraction or molecular range : Range of molecular aftraction or molecular range is defined as the maximum distance between two molecules up to which the intermolecular force of attraction is appreciable.

[Note : The intermolecular force is a short range force, LeV, it is effective over a very short range-about 10^-9 m. Beyond this distance, the force is negligible. The inter molecular force does not obey inverse square law.]

2) Sphere of influence : The sphere of influence of a molecule is defined as an imaginary sphere with the molecule as the centre and radius equal to the range of molecular attraction.

[Note : All molecules lying within the sphere of influence of a molecule are attracted by (as well as attract) the molecule at the centre. For molecules which lie outside this sphere, the intermolecular force due to the molecule at the centre is negligible.]

Question 36.
What is meant by a surface film?
Answer:
The layer of the liquid surface of thickness equal to the range of molecular attraction is called a surface film.

Question 37.
What is meant by free surface of a liquid ?
Answer:
The surface of a liquid open to the atmosphere is called the free surface of the liquid.

Question 38.
Explain the phenomenon of surface tension on the basis of molecular theory.
Answer:
The phenomenon of surface tension arises due to the cohesive forces between the molecules of a liquid. The net cohesive force on the liquid molecules within the surface film differs from that on molecules deep in the interior of the liquid.

Consider three molecules of a liquid : A molecule A well inside the liquid, and molecules B and C lying within the surface film, shown in figure. The figure also shows their spheres of influence of radius R.

(1) The sphere of influence of molecule A is entirely inside the liquid and the molecule is surrounded by its nearest neighbours on all sides. Hence, molecule A is equally attracted from all sides, so that the resultant cohesive force acting on it is zero. Hence, it is free to move anywhere within the liquid.

(2) For molecule B, a part of its sphere of influence is outside the liquid surface. This part contains air molecules whose number is negligible compared to the number of molecules in an equal volume of the liquid. Therefore, molecule B experiences a net cohesive force downward.

(3) For molecule C, the upper half of its sphere of influence is outside the liquid surface. Therefore, the resultant cohesive force on molecule C in the
downward direction is maximum.

(4) Thus, all molecules lying within a surface film of thickness equal to R experience a net cohesive force directed into the liquid.

(5) The surface area is proportional to the number of molecules on the surface. To increase the surface area, molecules must be brought to the surface from within the liquid. For this, work must be done against the cohesive forces. This work is stored in the liquid surface in the form of potential energy. With a tendency to have minimum potential energy, the liquid tries to reduce the number of molecules on the surface so as to have minimum surface area. This is why the surface of a liquid behaves like a stressed elastic membrane.

Question 39.
Define surface tension.
State its formula and CGS and SI units.
Answer:
The surface tension of a liquid is defined as the tangential force per unit length, acting at right angles on either side of an imaginary line on the free surface of the liquid.

If F is the force on one side of a line of length Z, drawn on the free surface of a liquid, the surface tension (T) of the liquid is

The CGS unit of surface tension : The dyne per centimetre (dyn/cm) or, equivalently, the erg per square centimetre (erg/cm²).

The SI unit of surface tension : The newton per metre (N/m) or, equivalently, the joule per square metre (J/m²).

Question 40.
Obtain the dimensions of surface tension.
Answer:
Surface tension is a force per unit length.

Question 41.
Define and explain surface energy of a liquid.
OR
Define surface energy.
OR
State its dimensions and SI unit.
OR
Why do molecules of a liquid in the surface film possess extra energy?
Answer:
Surface energy : The surface energy is defined as the extra (or increased) potential energy possessed by the molecules in a liquid surface with an isothermal increase in the surface area of the liquid.

A liquid exerts a resultant cohesive force on every molecule of its surface, trying to pull it into the liquid. To increase the surface area, it is necessary to bring more molecules from inside the liquid to the liquid surface. For this, external work must be done against the net cohesive forces on the molecules. This work is stored in the liquid surface in the form of potential energy.

This extra potential energy that the molecules in the liquid surface have is called the surface energy. Thus, the molecules of a liquid in the surface film possess extra energy.
Dimensions : [surface energy] = [ML²T^-2]
SI unit: the joule (J).

Question 42.
Why is the surface tension of paints and lubricating oils kept low?
Answer:
For better wettability (surface coverage), the surface tension and angle of contact of paints and lubricating oils must be low.

Question 43.
Derive the relation between the surface tension and surface energy of a liquid.
OR
Derive the relation between surface tension and surface energy per unit area.
OR
Show that the surface tension of a liquid is numerically equal to the surface energy per unit area.
Answer:
Suppose a soap film is isothermally stretched over the area enclosed by a U-shaped frame ABCD and a w cross-piece PQ that can slide smoothly along the frame, as shown in the figure. Let T be the surface tension of the soap solution and l, the length of wire PQ in contact with the soap film.

The film has two surfaces, both of which are in contact with the wire. The film tends to contract by exerting a force on wire PQ. As each surface exerts a force Tl, the net force on the wire is 2Tl.

Suppose that wire PQ is pulled outward very slowly through a distance dx to the position P’Q’ by an external force of magnitude 2T l. The work done by the external force against the force due to the film is
W = applied force × displacement
∴ W = Fdx = ITldx (∵ F = 2Tl)
This work is stored in the unit surface area in the form of potential energy. This potential energy is called the surface energy.

Due to the displacement dx, the surface area of the film increases. As the film has two surfaces, the increase in its surface area is
A = 2ldx
Thus, the work done per unit surface area is
\(\frac{W}{A}=\frac{2 T l d x}{2 l d x}\) = T
Thus, the surface energy per unit area of a liquid is equal to its surface tension.

Question 44.
Two soap bubbles of the same soap solution have diameters in the ratio 1 : 2. What is the ratio of work done to blow these bubbles ?
Answer:
Work done oc surface area.
∴ W₁/W₂ = (r₁/r₂)² = (\(\frac{1}{2}\))² = \(\frac{1}{4}\)
∴ W₁ : W₁ = 1 : 4.

Question 45.
If the surface tension of a liquid is 70 dyn/cm, what is the total energy of the free surface of the liquid drop of radius 0.1 cm ?
Answer:
E = 4πr²T = 4 × \(\frac{22}{7}\) × (0.1)² × 70
= 88 × 10^-2 × 10 = 8.8 ergs

Question 46.
The total energy of the free surface of a liquid drop of radius 1 mm is 10 ergs. What is the total energy of a liquid drop (of the same liquid) of radius 2 mm ?
Answer:
E = 4πr²T ∴ \(\frac{E_{2}}{E_{1}}=\left(\frac{r_{2}}{r_{1}}\right)^{2}=\left(\frac{2}{1}\right)^{2}\) = 4
∴ E₂ = 4E₁ = 4 × 10 = 40 ergs is the required

47. Solve the following

Question 1.
Calculate the work done in blowing a soap bubble of radius 4 cm. The surface tension of the soap solution is 25 × 10^-3 N/m.
Solution:
Data : r = 4 cm = 4 × 10^-2 m, T = 25 × 10^-3 N/m
Initial surface area of soap bubble = 0
Final surface area of soap bubble = 2 × 4πr²
Increase in surface area = 2 × 4πr² The work done
= surface tension x increase in surface area
= T × 2 × 4πr²
= 25 × 10^-3 × 2 × 4 × 3.142 × (4 × 10^-2)2
= 1.005 × 10^-3 J

Question 2.
Two soap bubbles have radii in the ratio 4 : 3. What is the ratio of work done to blow these bubbles?
Solution:
Data : \(\frac{r_{1}}{r_{2}}=\frac{4}{3}\)
Work done, W = 2TdA
∴ W₁ = 2T(4πr₁²), W₂ = 2T(4πr₂²)
∴\(\frac{W_{1}}{W_{2}}=\frac{2 T\left(4 \pi r_{1}^{2}\right)}{2 T\left(4 \pi r_{2}^{2}\right)}=\left(\frac{r_{1}}{r_{2}}\right)^{2}\)
= (\(\frac{4}{3}\))² = \(\frac{16}{9}\)

Question 3.
Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of the soap solution is 30 dyn/cm.
Solution:
Data : r₁ = 1 cm, r₂ = 2 cm, T = 30 dyn/cm
Initial surface area = 2 × 4πr₁²
Final surface area = 2 × 4πr₂²
∴ Increase in surface area
= 2 × 4πr₂² – 2 × 4πr₁² = 8π(r₂² – r₁²)
∴ The work done
= surface tension × increase in surface area
= T × 8π(r₂² – r₁²)
= 30 × 8 × 3.142 × [(2)² – (1)²]
= 2262 ergs

Question 4.
A mercury drop of radius 0.5 cm falls from a height on a glass plate and breaks into one million droplets, all of the same size. Find the height from which the drop fell. [Density of mercury = 13600 kg/m3, surface tension of mercury = 0.465 N/m]
Solution:
Data : R = 0.5 cm = 0.5 × 10^-2 m, n = 10⁶, ρ = 13600 kg m³, T = 0.465 N/m, g = 9.8 m/s²
\(\frac{4}{3}\) πR³ = n × \(\frac{4}{3}\) πr³
as the volume of the mercury remains the same.
∴ r = \(\frac{R}{\sqrt[3]{n}}=\frac{0.5 \times 10^{-2}}{\sqrt[3]{10^{6}}}\) = 0.5 × 10^-4 m
This gives the radius of a droplet.
By energy conservation, if h is the height from which the drop of mass m falls,

This gives the required height.

Question 5.
Eight droplets of mercury, each of radius 1 mm, coalesce to form a single drop. Find the change in the surface energy. [Surface tension of mercury = 0.472 J/m²]
Solution:
Data : r = 1 mm = 1 × 10^-3 m, T = 0.472 J/m²
Let R be the radius of the single drop formed due to the coalescence of 8 droplets of mercury.
Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant.
∴ 8 × \(\frac{4}{3}\) πr³ = \(\frac{4}{3}\) πR³
∴ 8r³ = R³
∴ 2r = R
Surface area of 8 droplets = 8 × 4πr²
Surface area of single drop = 4πR²
∴ Decrease in surface area = 8 × 4πr² – 4πR²
= 4π(8r² – R²)
= 4π[8r² – (2r)²]
= 4π × 4r²
∴ The energy released
= surface tension × decrease in surface area
= T × 4π × 4r²
= 0.472 × 4 × 3.142 × 4 × (1 × 10^-3)²
= 2.373 × 10^-5 J

Question 6.
The total energy of the free surface of a liquid drop is 2 × 10^-4 π times the surface tension of the liquid. What is the diameter of the drop ? (Assume all terms in SI unit.)
Solution:
Data : 4πr²T = 2 × 10^-4 πT (numerically)
∴ 2r² = 10^-4
∴ r = \(\frac{10^{-2}}{\sqrt{2}}=\frac{10^{-2}}{1 \cdot 414}\)
= 0.7072 × 10^-2 m
∴ d = 2r = 2 × 0.7072 × 10^-2
= 1.4144 × 10^-2 m
This gives the diameter of the liquid drop.

Question 48.
Define angle of contact.
Answer:
The angle of contact for a liquid-solid pair (a liquid in contact with a solid) is defined as the angle between the surface of the solid and the tangent drawn to the free surface of the liquid at the extreme edge of the liquid, as measured through the liquid.

Question 49.
Draw neat diagrams to show the angle of contact in the case of a liquid which
(i) completely wets
(ii) partially wets
(iii) does not wet the solid. State the characteristics of the angle of contact in each case, giving one example of each.
Answer:
Characteristics :
(1) For a liquid, which completely wets the solid, the angle of contact is zero.
For example, pure water completely wets clean glass. Therefore, the angle of contact at the water glass interface is zero [from figure (a)].

(2) For a liquid which partially wets the solid, the angle of contact is an acute angle. For example, kerosine partially wets glass, so that the angle of contact is an acute angle at the kerosine glass interface [from figure (b)].
(3) For a liquid which does not wet the solid, the angle of contact is an obtuse angle. For example, mercury does not wet glass at all, so that the angle of contact is an obtuse angle at the mercury-glass interface [from figure (c)].
(4) The angle of contact for a given liquid solid pair is constant at a given temperature, provided the liquid is pure and the surface of the solid is clean.

Question 50.
State any two characteristics of angle of contact.
Answer:
Characteristics of angle of contact:

1. It depends upon the nature of the liquid and solid in contact, and is constant for a given liquid-solid pair, other factors remaining unchanged.
2. It depends upon the medium (gas) above the free surface of the liquid.
3. It is independent of the inclination of the solid to the liquid surface.
4. It changes with surface tension and, hence, with the temperature and purity of the liquid.

Question 51.
Explain why the free surface of some liquids in contact with a solid is not horizontal.
OR
Explain the formation of concave and covex surface of a liquid on the basis of molecular theory.
Answer:
For a molecule in the liquid surface which is in contact with a solid, the forces on it are largely the solid-liquid adhesive force \(\vec{F}_{\mathrm{A}}=\overrightarrow{P A}\) and the liquid- liquid cohesive force \(\vec{F}_{\mathrm{C}}=\overrightarrow{P C} \vec{F}_{A}\) is normal to the solid surface and \(\vec{F}_{\mathrm{C}}\) is at 45° with the horizontal, from figure (a). The free surface of a liquid at rest is always perpendicular to the resultant \(\vec{F}_{\mathrm{R}}=\overrightarrow{P R}\) of these forces.

If F_C = \(\sqrt{2} F_{\mathrm{A}}, \vec{F}_{\mathrm{R}}\) is along the solid surface, the contact angle is 90° and the liquid surface is horizontal at the edge where it meets the solid, as in figure (a). In general this is not so, and the liquid surface is not horizontal at the edge.

For a liquid which completely wets the solid (e.g., pure water in contact with clean glass), F_C << F_A. For a liquid which partially wets the solid (e.g., kerosine . or impure water in contact with glass), F_C < \(\sqrt{2} F_{\mathrm{A}}\). If F_C << F_A or if F_C < \(\sqrt{2} F_{\mathrm{A}}\), the contact angle is correspondingly zero or acute and the liquid surface curves up and acquires a concave shape until the tangent PT is tangent to \(\vec{F}_{\mathrm{R}}\) fron figure (b).

If F_C > \(\sqrt{2} F_{\mathrm{A}}\), the contact angle is obtuse and the liquid surface curves down and acquires a convex shape until the tangent PT is tangent to \(\vec{F}_{\mathrm{R}}\), from figure (c).

Question 52.
State the conditions for concavity and convexity of a liquid surface where it is in contact with a solid.
Answer:
For a molecule in the liquid surface which is in contact with a solid, the forces on it are largely
(i) the solid-liquid adhesive force \(\vec{F}_{\mathrm{A}}\) normal and into the solid surface and
(ii) the liquid-liquid cohesive force \(\vec{F}_{\mathrm{C}}\) at nearly 45° with the horizontal.

If F_C << F_A or if F_C < \(\sqrt {2}\)F_A , the contact angle is correspondingly zero or acute and the liquid sur-face is concave with the solid.

If F_C > \(\sqrt {2}\)F_A, the contact angle is obtuse and the liquid surface curves down, i.e., convex, with the Solid.

Question 53.
Draw neat labelled diagrams to show angle of contact between (a) pure water and clean glass . (b) mercury and clean glass.
Answer:

Question 54.
Explain the shape of a liquid drop on a solid surface in terms of interfacial tensions.
OR
Account for the angle of contact in terms of interfacial tensions.
OR
Draw diagram showing force due to surface tension at the liquid-solid, air-solid, air-liquid interface, in case of (i) a drop of mercury on a plane solid surface and (ii) a drop of water on a plane solid surface. Discuss the variation of angle of contact.
Answer:
A liquid surface, in general, is curved where it meets a solid. The angle between the solid surface and the tangent to the liquid surface at the extreme edge of the liquid, as measured through the liquid, is called the angle of contact.

Above figure shows the interfacial tensions that act in equilibrium at the common point of the liquid, solid and gas (air + vapour).
T₁ = the liquid-solid interfacial tension
T₂ = the solid-gas interfacial tension
T₃ = the liquid-gas interfacial tension
θ = the angle of contact for the liquid-solid pair is the angle between T₁ and T₃
The equilibrium force equation (along the solid surface) is
T₃ Cos θ + T₁ – T₂ = 0
∴ cos θ = \(\frac{T_{2}-T_{1}}{T_{3}}\) …………… (1)

Case (1) : If T₂ > T₁, cos θ is positive and contact angle θ < 900, so that the liquid wets the surface.
Case (2) : If T₂ < T₁, cos θ is negative and θ is obtuse, so that the liquid is non-wetting.
Case (3): If T₂ – T₁ T₃, cos θ = 1 and θ ≅ 0°.
Case (4) : If T₂ – T₁ ≅ T₃, cos θ will be greater than 1 which is impossible, so that there will be no equilibrium and the liquid will spread over the solid surface.

Question 55.
State the expression for the angle of contact in terms of interfacial tensions?
Answer:
cos θ = \(\frac{T_{2}-T_{1}}{T_{3}}\), where θ is the angle of contact for a liquid-solid pair, T₁ is the liquid-solid interfacial tension, T₂ is the solid-gas (air + vapour) inter-facial tension and T₃ is the liquid-gas interfacial tension.

Question 56.
In terms of interfacial tension, when is the angle of contact acute ?
Answer:
The angle of contact is acute when the solid-gas (air + vapour) interfacial tension is greater than the liquid-solid interfacial tension.

Question 57.
In terms of interfacial tensions, when is the angle of contact obtuse ?
Answer:
The angle of contact is obtuse when the solid-gas (air + vapour) interfacial tension is less than the liquid-solid interfacial tension.

Question 58.
State the factors affecting a liquid-solid angle of contact.
Answer:
Factors affecting a liquid-solid angle of contact:

1. the nature of the liquid and the solid in contact,
2. impurities in the liquid,
3. temperature of the liquid.

Question 59.
Explain the effect of impurity on the angle of contact (or surface tension of a liquid).
Answer:
Effect of impurity :
(i) The angle of contact or the surface tension of a liquid increases with dissolved impurities like common salt. For dissolved impurities, the angle of contact (or surface tension) increases linearly with the concentration of the dissolved materials.

(ii) It decreases with sparingly soluble substances like phenol or alcohol. A detergent is a surfactant whose molecules have hydrophobic and hydrophilic ends; the hydrophobic ends decrease the surface tension of water. With reduced surface tension, the water can penetrate deep into the fibres of a cloth and remove stubborn stains.

(iii) It decreases with insoluble surface impurities like oil, grease or dust. For example, mercury surface contaminated with dust does not form perfect spherical droplets till the dust is removed.

Question 60.
Explain the effect of temperature on the angle of contact (or surface tension of a liquid).
Answer:
Effect of temperature : The surface tension of a liquid decreases with increasing temperature of the liquid. For small temperature differences, the decrease in surface tension is nearly directly proportional to the temperature rise.

If T and T₀ are the surface tensions of a liquid at temperatures θ and 0 °C, respectively, then T = T₀(1 – αθ) where α is a constant for a given liquid. The surface tension of a liquid becomes zero at its critical temperature. The surface tension increases with increasing temperature only in case of molten copper and molten cadmium.

Question 61.
Why cold wash is recommended for new cotton fabrics while hot wash for removing stains?
Answer:
Cold wash is recommended for new/coloured cotton fabrics. Cold water, due to its higher surface tension, does not penetrate deep into the fibres and thus does not fade the colours. Hot water, because of its lower surface tension, can penetrate deep into fabric fibres and remove tough stains.

Question 62.
Explain in brief the pressure difference across a curved liquid surface.
Answer:
Every molecule lying within the surface film of a static liquid is pulled tangentially by forces due to surface tension. The direction of their resultant, \(\vec{F}_{\mathrm{T}}\), on a molecule depends upon the shape of that liquid surface and decides the cohesion pressure at a point just below the liquid surface.

Consider two molecules, A and B, respectively just above and below the free surface of a liquid. So, the level difference between them is negligibly small and the atmospheric pressure on both is the same, p₀, Let \(\vec{F}_{\mathrm{atm}}\) be the downward force on A and B due to the atmospheric pressure.

If the free surface of a liquid is horizontal, the resultant force \(\vec{F}_{\mathrm{T}}\) on molecule B is zero, from figure (a). Then, the cohesion pressure is negligible and the net force on A and B is \(\vec{F}_{\mathrm{atm}}\). The pressure difference on the two sides of the liquid surface is zero.

If the free surface of a liquid is concave, the resultant force \(\vec{F}_{\mathrm{T}}\) on molecule B is outwards (away from the liquid), from figure (b), opposite to \(\vec{F}_{\mathrm{atm}}\). Then, the net force on B is less than \(\vec{F}_{\mathrm{atm}}\) and the cohesion pressure is decreased. The pressure above the concave liquid surface is greater than that just below the liquid surface.

If the free surface of a liquid is convex, the resultant force \(\vec{F}_{\mathrm{T}}\) on molecule B acts inwards (into the liquid), from figure (c), in the direction of \(\vec{F}_{\mathrm{atm}}\). Then, the net force on B is greater than \(\vec{F}_{\mathrm{atm}}\) and the cohesion pressure is increased. The pressure below the convex liquid surface is greater than that just above the liquid surface.

Question 63.
Derive an expression for the excess pressure inside a soap bubble.
OR
Derive Laplace’s law for spherical membrane of a bubble due to surface tension.
Answer:
Consider a small, spherical, thin-filmed soap bubble with a radius R. Let the pressure outside the drop be Po and that inside be p. A soap bubble in air is like a spherical shell and has two gas-liquid interfaces. Hence, the surface area of the bubble is
A = 8πR² ………. (1)
Hence, with a hypothetical increase in radius by an infinitesimal amount dR, the differential increase in surface area and surface energy would be
dA = 16πR ∙ dR and
dW = T ∙ dA = 16πTRdR ………….. (2)
We assume that dR is so small that the pressure inside remains the same, equal to p. All parts of the surface of the bubble experiences an outward force per unit area equal to p – p_o. Therefore, the work done by this outward pressure-developed force against the surface tension force during the increase in radius dR is
dW = (excess pressure × surface area) ∙ dR
= (p – p_o) × 4πR² ∙ dR ………. (3)
From Eqs. (2) and (3),
(p – p_o) × 4πR² ∙ dR = 16πTRdR
∴ p – p_o = \(\frac{4 T}{R}\) …………… (4)
which is the required expression.

[Note : The excess pressure inside a drop or bubble is inversely proportional to its radius : the smaller the bubble radius, the greater the pressure difference across its wall.]

Question 64.
What is the excess of pressure inside a soap bubble of radius 3 cm if the surface tension of the soap solution is 30 dyn/cm ?
Answer:
Excess of pressure, p – p_o = \(\frac{4 T}{R}=\frac{4 \times 30}{3}\)
= 40 dyn/cm²

Question 65.
Two soap bubbles of the same soap solution have radii 3 cm and 1.5 cm. If the excess pressure inside the bigger bubble is 40 dyn/cm², what is the excess pressure inside the smaller bubble ?
Answer:
Excess pressure ∝ T/R. In this case, the surface tension is the same in the two cases. Hence, the excess pressure inside the smaller bubble will be 80 dyn/cm².

Question 66.
Explain : In the absence of gravity or other external forces, a liquid drop assumes a spherical shape.
Answer:
A spherical shape has the minimum surface area-to-volume ratio of all geometric forms. If any . external force distorts the sphere, molecules must be brought from the interior to the surface in order to provide for the increased surface area. This process requires work to be done in order to raise the potential energy of a molecule. The change in free surface energy is equal to the net work done to alter the surface area of the liquid.

However, spontaneous processes are associated with a decrease in free energy. Hence, in the absence of external forces, a liquid drop will spontaneously assume a spherical shape in order to minimize its exposed surface area and thereby its free surface energy.

[ Note : The spontaneous coalescence of two similar liquid droplets into one large drop when brought into contact is a dramatic demonstration of the decrease in free surface energy brought about by the decrease in total surface area by the formation of a single larger drop.]

Question 67.
A small air bubble of radius r in water is at a depth h below the water surface. If p₀ is the atmospheric pressure, ρ is the density of water and T is the surface tension of water, what is the pressure inside the bubble?
Answer:
The absolute pressure within the liquid at a depth h is p = p₀ + ρgh.
Since the excess pressure inside a bubble is \(\frac{2 T}{R}\), the pressure inside the bubble is
P_in = p + \(\frac{2 T}{R}\) = p₀ + ρgh + \(\frac{2 T}{R}\)

68. Solve the following

Question 1.
What is the excess pressure (in atm) inside a soap bubble with a radius of 1.5 cm and surface tension of 3 × 10^-2 N/m? [1 atm = 101.3 kPa]
Solution:
Data : R = 1.5 × 10^-2 m, T = 3 × 10^-2 N/m,
1 atm = 1.013 × 10⁵ Pa
The excess pressure inside a soap bubble is
p – p₀ = \(\frac{4 T}{R}\)
= \(\frac{4 \times 3 \times 10^{-2}}{1.5 \times 10^{-2}}\) = 8Pa
= \(\frac{8}{1.013 \times 10^{5}}\) atm = 7.897 × 10^-5 atm

Question 2.
A raindrop of diameter 4 mm is about to fall on the ground. Calculate the pressure inside the rain drop. [Surface tension of water T = 0.072 N/m, atmospheric pressure = 1.013 × 10⁵ N/m²)
Solution:
Data : D = 4 × 10^-3 m, T = 0.072N/m,
p₀ = 1013 × 10⁵ N/m²
R = \(\frac{D}{2}\) = 2 × 10^-3 m
The excess pressure inside the raindrop is
p – p₀ = \(\frac{2 I}{R}=\frac{2(0.072)}{2 \times 10^{-3}}\) = 72 N/m²
∴ p = 101300 + 72 = 101372 N/m²

Question 3.
What should be the diameter of a soap bubble such that the excess pressure inside it is 51.2 Pa? [Surface tension of soap solution = 3.2 × 10^-2 N/m]
Solution:
Data : p – p₀ = 51.2 Pa, T = 3.2 × 10^-2 N/m
Forasoapbubb1e, p – p₀ = \(\frac{4T}{R}\)
∴ The radius of the soap bubble should be
R = \(\frac{4 T}{p-p_{0}}=\frac{4 \times 3.2 \times 10^{-2}}{51.2}\) = 2.5 × 10^-3 m = 2.5 mm
∴ the diameter of the soap bubble should be 2 × 2.5 = 5 mm.

Question 4.
The lower end of a capillary tube of diameter 1 mm is dipped 10 cm below the water surface in a beaker. What pressure is required to blow a hemispherical air bubble at the lower end of the tube? Present your answer rounded off to 4 significant figures. [Surface tension = 0.072 N/m, density = 10³ kg/m³, atmospheric pressure = 101.3 kPa, g = 9.8 m/s²]
Solution:
Data : r = 0.5 mm = 5 × 10^-4 m, d = 10 cm = 0.1 m,
T = 0.072 N/m, p = 10³ kg/m³, g = 9.8 m/s²,
P = 1 atm = 1.013 × 10⁵ Pa
The pressure outside the bubble at the depth d is
p₀ = P + dρg
= 1.013 × 10⁵ +0.1 × 10³ × 9.8
= (1.013 + 0.0098) × 10⁵ = 1.0228 × 10⁵ Pa
Since a bubble within water has only one, gas-liquid interface, the excess pressure inside the bubble is
p – p₀ = \(\frac{2 T}{r}=\frac{2 \times 0.072}{5 \times 10^{-4}}\) = 0.0288 × 10⁴ Pa
= 0.00288 × 10⁵ Pa
∴ P = (1.0228 + 0.00288) × 10⁵ = 1.02568 × 10⁵ Pa
The pressure inside the air bubble is 1.026 × 10⁵ Pa (or 102.6 kPa), rounded off to four significant figures.

Question 5.
There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.072 N/m and density 10³ kg/m³. The bubble is at a depth of 10 cm below the free surface of the liquid. By what amount is the pressure inside the bubble greater than the’ atmospheric pressure?
Solution:
Data : R = 10^-3 m, T = 0.072 N/m, ρ = 10³ kg/m³, h = 0.1 m
Let the atmospheric pressure be p0. Then, the absolute pressure within the liquid at a depth h is
p = p₀ + ρgh
Hence, the pressure inside the bubble is
p_in = p₀ + \(\frac{2 T}{R}\) = p₀ + ρgh + \(\frac{2 T}{R}\)
The excess pressure inside the bubble over the atmospheric pressure is
p_in – p₀ = ρgh + \(\frac{2 T}{R}\)
= (10³) (9.8) (0.1) + \(\frac{2(0.072)}{10^{-3}}\)
= 980 + 144 = 1124 Pa

Question 6.
Two soap bubbles A and B, of radii 2 cm and 4 cm, respectively, are in a closed chamber where air pressure is maintained at 8 N/m ². If n_A and n_B are the number of moles of air in bubbles A and B, respectively, then find the ratio n_B : n_A. [Surface tension of soap solution = 0.04 N/m. Ignore the effect of gravity.]
Solution:
Data : R_A = 0.02 m, R_B = 0.04 m, p₀ = 8 N/m², T = 0.04 N/m

This is the required ratio.

Question 69.
What is capillary? What is capillarity or capillary action?
Answer:
(1) A tube of narrow bore (i.e. very small diameter) is called a capillary tube. The word capillary is derived from the Latin capillus meaning hair, capillaris in Latin means ‘like a hair’.

(2) If a capillary tube is just partially immersed in a wetting liquid the liquid rises in the capillary tube. This is called capillary rise.

If a capillary tube is just partially immersed in a non-wetting liquid, the liquid falls in the capillary tube. This is called capillary depression.

The rise of a wetting liquid and fall of a non-wetting liquid in a capillary tube is called capillarity.

Question 70.
State any four applications of capillarity.
Answer:
Applications of capillarity:

1. A blotting paper or a cotton cloth absorbs water; ink by capillary action.
2. Oil rises up the wick of an oil lamp and sap rises up xylem tissues of a tree by capillarity.
3. Ground water rises to the open surface through the capillaries formed in the soil. In summer, the farmers plough their fields to break these capillaries and prevent excessive evaporation.
4. Water rises up the crevices in rocks by capillary action. Expansion and contraction of this water due to daily and seasonal temperature variations cause the rocks to crumble.

[Note: The rise of sap is due to the combined action of capilarity and transpiration. The transpiration pull, is considered to be the major driving force for water transport throughout a plant.]

Question 71.
Two capillary tubes have radii in the ratio 1: 2. If they are dipped in the same liquid, what will be the ratio of capillary rise in the two tubes ?
Answer:
T = \(\frac{h r \rho g}{2 \cos \theta}\)
In this case, hr = constant
∴ h₁ : h₂ = r₂ : r₁ = 2 : 1.

Question 72.
The radii of two columns of a U-tube are r₁ and r₂. When a liquid of density ρ and angle of contact θ = 0° is filled in it, the level difference of the liquid in the two columns is h. Find the surface tension of the liquid.
Answer:
Capillary rise, h = \(\frac{2 T \cos \theta}{r \rho g}\), where θ is the angle of contact.
Assuming the two columns of the U-tube to be sufficiently thin,

73. Solve the following

Question 1.
A liquid of density 900 kg/m³ rises to a height of 9 mm in a capillary tube of 2.4 mm diameter. If the angle of contact is 25°, find the surface tension of the liquid.
Solution:
Data : ρ = 900 kg/m³, h = 9 mm = 9 × 10^-3 m,
θ = 25°, g = 9.8 m/s²
r = \(\frac{1}{2}\) × diameter = \(\frac{2.4}{2}\) = 1.2 mm = 1.2 × 10^-3 m
cos θ = cos 25° = 0.9063
The surface tension of the liquid,
T = \(\frac{r h \rho g}{2 \cos \theta}\)
= \(\frac{1.2 \times 10^{-3} \times 9 \times 10^{-3} \times 900 \times 9.8}{2 \times 0.9063}\)
= 5.257 × 10^-2 N/m

Question 2.
A capilary tube of uniform bore is dipped vertically in water which rises by 7 cm in the tube. Find the radius of the capillary tube if the surface tension of water is 70 dyn/cm. [g = 980 cm/s²]
Solution:
Data : h = 7 cm, T = 70 dyn/cm, g = 980 cm/s², ρ = 1 g/cm³ and θ = 0° (for water)
∴ cos θ = 1 .
Surface tension, T = \(\frac{r h \rho g}{2 \cos \theta}\)
∴ The radius of the capillary tube,
r = \(\frac{2 T \cos \theta}{h \rho g}\)
= \(\frac{2 \times 70 \times 1}{7 \times 1 \times 980}\) = 0.02041 cm

Question 3.
A liquid rises to a height of 9 cm in a glass capillary tube of radius 0.02 cm. What will be the height of the liquid column in a glass capillary tube of radius 0.03 cm ?
Solution:
Data : h₁ = 9 cm, r₁ = 0.02 cm, r₂ = 0.03 cm
For the first capillary, T = \(\frac{r_{1} h_{1} \rho g}{2 \cos \theta}\)
For the second capillary, T = \(\frac{r_{2} h_{2} \rho g}{2 \cos \theta}\)
∴ \(\frac{r_{1} h_{1} \rho g}{2 \cos \theta}=\frac{r_{2} h_{2} \rho g}{2 \cos \theta}\)
∴ r₁h₁ = r₂h₂
The height of the liquid column in the second capillary,
h₂ = \(\frac{r_{1} h_{1}}{r_{2}}=\frac{0.02 \times 9}{0.03}\) = 6 cm

Question 4.
Water rises to a height of 5 cm in a certain capillary tube. In the same capillary tube, mercury is depressed by 2.02 cm. Compare the surface tensions of water and mercury.
[Density of water = 1000 kg/m³, density of mercury = 13600 kg/m³, angle of contact for water = 0°, angle of contact for mercury = 148°]
Solution:
Let T_w, θ_w, h_w and ρ_w be the surface
tension, angle of contact, capillary rise and density of water respectively. Let T_m, θ_m, h_m and ρ_m be the corresponding quantities for mercury. The radius (r) of the capillary is the same in both cases.

Data : h_w = 5 cm = 5 × 10^-2 m, θ_w = 0°, ρ_w = 1000 kg/m³, h_m = -2.02 cm
= -2.02 × 10^-2 m,
ρ_m = 13600 kg/m³, θ_m= 148°
[Note : h_m is taken to be negative because for mercury there is capillary depression.]
cos θ_w = cos 0° = 1
cos θ_m = cos 148° = – cos 32° = -0.8480

Question 5.
When a glass capillary tube of radius 0.4 mm is dipped into mercury, the level of mercury inside the capillary stands 1.50 cm lower than that outside. Calculate the surface tension of mercury. [Angle of contact of mercury with glass = 148 °, density of mercury = 13600 kg/m³]
Solution:
Data : r = 0.4 mm = 4 × 10^-4 m, h = -1.50 cm = – 1.50 × 10^-2 m,
ρ = 13.6 × 10³ kg/m³, g = 9.8 m/s², θ = 148°
cos θ = cos 148° = – cos 32 ° = – 0.8480
The surface tension of mercury is
T = \(\frac{r h \rho g}{2 \cos \theta}\)
= \(\frac{\left(4 \times 10^{-4}\right)\left(-1.50 \times 10^{-2}\right)\left(13.6 \times 10^{3}\right)(9.8)}{2(-0.8480)}\)
= 0.4715 N/m

Question 6.
The tube of a mercury barometer is 1 cm in diameter. What correction due to capillarity is to be applied to the barometric reading if the surface tension of mercury is 435.5 dyn/cm and the angle of contact of mercury with glass is 140° ? [Density of mercury = 13600 kg/m³]
Solution:
Data : d = 1 cm, T = 435.5 dyn/cm, θ = 140°, ρ = 13660 kg/m³ = 13.66 g/cm³,
g = 9.8 m/s² = 980 cm/s²

∴ The correction due to capillarity = -0.1001 cm

Question 7.
Calculate the density of paraffin oil, if within a glass capillary of diameter 0.25 mm dipped in paraffin oil of surface tension 0.0245 N/m, the oil rises to a height of 4 cm. [Angle of contact of paraffin oil with glass = 28°, acceleration due to gravity = 9.8 m/s²]
Solution :
Data : d = 0.25 mm, T = 0.0245 N/m, h = 4 cm = 4 × 10^-2 m, θ = 28°, g = 9.8 m/s²

This gives the density of paraffan oil.

Question 8.
A capillary tube of radius r can support a liquid column of weight 6.284 × 10^-4 N. Calculate the radius of the capillary if the surface tension of the liquid is 4 × 10^-2 N/m.
Solution:
Data : mg = 6.284 × 10^-4 N, T = 4 × 10^-2 N/m
Net upward force = weight of liquid column
∴ 2πrT cos θ = mg
Assuming the angle of contact, θ = 0° (∵ data not given), the radius of the capillary is
r = \(\frac{m g}{2 \pi T}=\frac{6.284 \times 10^{-4}}{2(3.142)\left(4 \times 10^{-2}\right)}\) = 2.5 × 10^-3 m

Question 9.
Two vertical glass plates are held parallel 0.5 mm apart, dipped in water. If the surface tension of water is 70 dyn/cm , calculate the height to which water rises between the two plates.
Solution:
Data : x = 0.5 mm = 5 × 10^-4 m, T = 0.07 N/m, ρ(water) = 10³ kg/m³

Let the width of each glass plate be b and the height to which the water rises between the plates be h.

Then, the upward force on the water between the plates due to surface tension = 2Tb cos θ

where θ is the angle of contact of water with glass. The weight of the water between the plates = mg = (bxhρ) = g

where x is the separation between the plates and p is the density of water.
Equating, (bxhρ)g = 2Tb cosθ
∴ The height to which water rises between the two plates,
h = \(\frac{2 T \cos \theta}{x \rho g}=\frac{2\left(7 \times 10^{-2}\right)(1)}{\left(5 \times 10^{-4}\right)\left(10^{3}\right)(9.8)}=\frac{0.2}{7}\)
= 0.02857 m = 2.857 cm

Question 10.
A glass capillary of radius 0.4 mm is inclined at 60° with the vertical in water. Find the length of water column in the capillary tube. [Surface tension of water = 7 × 10^-2 N/m]
Solution:
Data : r = 4 × 10^-4 m, Φ = 60°, T = 7 × 10^-2 N/m
Let h be the capillary rise when the capillary tube is immersed vertically in water. Let l be the length of the water column in the capillary tube above that of the outside level.

Question 74.
What is hydrodynamics?
Answer:
Hydrodynamics is the branch of physics that deals with fluid dynamics, i.e., the study of fluids in motion. Since the most basic fluid motion is highly complex, we consider only ideal fluids-non-viscous and incompressible, i.e., fluids whose internal friction is negligible and density is constant throughout.

Question 75
What is meant by a steady flow ?
Answer:
When a liquid flows slowly over a surface or through a pipe such that its velocity or pressure at any point within the fluid is constant, it is said to be in steady flow.

Question 76.
Explain a streamline and streamline flow.
Answer:
Streamline : Consider point A, from figure, within a fluid. The velocity \(\vec{v}\) at A does not change with time. Hence, every particle passes point A with the same speed and in the same direction. The same is true about the other points such as B and C. A curve which is tangent or parallel to the velocity of the fluid particles at every point will be the path of every particle arriving at A. It is called a streamline. A fluid particle cannot cross a streamline but only flow along it.

Streamline flow : When a liquid flows slowly over a surface or through a pipe with a velocity less than a certain critical velocity, the motion of its molecules is orderly. All molecules passing a given point proceed with the same velocity. This kind of fluid motion is called streamline or steady flow.

Question 77.
Explain a flow tube.
Answer:
A bundle of adjacent streamlines form a tube of flow or flow tube through which the fluid is flowing. In a flow tube, where the streamlines are close together the velocity is high, and where they are widely separated, the fluid is moving slowly. No fluid can cross the boundary of a tube of flow.

Question 78.
Explain turbulent flow.
Answer:
Turbulent flow or turbulence is a non-steady fluid flow in which streamlines and flowtubes change continuously. It has two main causes. First, any obstruction or sharp edge, such as in a tap, creates

turbulence by imparting velocities perpendicular to the flow. Second, if the speed with which a fluid moves relative to a solid body is increased beyond a certain critical velocity the flow becomes unstable or one of extreme disorder. In both cases, the fluid particles still move in general towards the main direction as before. But now all sorts of secondary motions cause them to cross and recross the main direction continuously. The orderly streamlines break up into eddies or vertices and the result is turbulence. In a turbulent flow, regions of fluid move in irregular, colliding paths, resulting in mixing and swirling.

Question 79.
Distinguish between streamline flow and turbulent flow.
Answer:

Streamline flow	Turbulent flow
1. The steady flow of a fluid, with velocity less than certain critical velocity is called streamline or laminar flow.	1. A non-steady irregular fluid flow in which stream­lines and flowtubes change continuously with a veloc­ity greater than certain critical velocity.
2. In a streamline flow, the velocity of a fluid at a given point is always constant.	2. In a turbulent flow, the vel­ocity of a fluid at any point does not remain constant.
3. Streamlines do not change and never intersect.	3. Streamlines and flowtubes change continuously.
4. over a surface, and is in the form of coaxial cylinders through a pipe.	4. Fluid particles still move in general towards the main direction as before. But now all sorts of secondary motions cause eddies or vortices.

Question 80.
Explain the Reynolds number.
OR
What is Reynolds number?
Answer:
Osborne Reynolds found that if the free-stream velocity of a fluid increases when it moves relative to a solid body, a point is reached where the steady flow becomes turbulent. From experiments, he found that the transition from steady to turbulent flow depends on the value of the quantity \(\frac{v_{0} d}{\eta / \rho}\), where v₀ is the free-stream velocity, d is some characteristic dimension of the system, ρ the density of the fluid and η its coefficient of viscosity. For a sphere in a fluid stream, d is its diameter; for water in a pipe, d is the pipe diameter.

This dimensionless number, defined as
Re = \(\frac{v_{0} d \rho}{\eta}\) is called the Reynolds number.
In a system of particular geometry, transition from a steady to turbulent flow is given by a certain value of the Reynolds number called the critical Reynolds number. The free-stream velocity for this critical Reynolds number is called the critical velocity, v_critical = \(\frac{n R_{\mathrm{e}}}{\rho d}\). For a given system geometry, the free stream velocity beyond which a streamline flow becomes turbulent is called critical velocity.

Steady flow takes place for Re up to about 1000. For 1000 < Re < 2000, there is a transition region in which the flow is extremely sensitive to all sorts of small disturbances. For Re > 2000, the flow is completely turbulent.

[Notes : (1) See Q. 95 for “free-stream velocity”. (2) The dimensionless number is named after Osborne Reynolds (1842-1912), British physicist.]

Question 81.
Explain the term viscosity.
Answer:
Suppose a constant tangential force is applied to the surface of a liquid. Under this shearing force, the liquid begins to flow. The motion of a thin layer of the liquid at the surface, relative to a layer below, is opposed by fluid friction. Because of this internal fluid friction, horizontal layers of the liquid flow with varying velocities.

This also happens in a gas. When a solid surface is moved through a gas, a thin layer of the gas moves with the surface. But its motion relative to a layer away is opposed by fluid friction.

The resistance to relative motion between the adjacent layers of a fluid is known as viscosity.

It is a property of the fluid. The resistive force in fluid motion is called the viscous drag.

Question 82.
When a-liquid contained in a bucket is stirred and left alone, it comes to rest after some time. Why?
Answer:
This happens due to the internal friction (viscosity) and friction with the walls and bottom of the bucket.

Question 83.
What do you mean by viscous drag?
Answer:
When a fluid flows past a solid surface, or when a solid body moves through a fluid, there is always a force of fluid friction opposing the motion. This force of fluid friction is called the drag force or viscous drag.

Question 84.
What causes viscous drag in fluids?
Answer:
In liquids, the viscous drag is due to short range molecular cohesive forces while in gases it is due to collisions between fast moving molecules. For laminar flow in both liquids and gases, the viscous drag is proportional to the relative velocity between the layers, provided the relative velocity is small. For turbulent flow, the viscous drag increases rapidly and is proportional to some higher power of the relative velocity.

Question 85.
Define and explain velocity gradient in a steady flow.
Answer:
Definition : In a steady flow of a fluid past a solid surface, the rate at which the velocity changes with distance within a limiting distance from the surface is called the velocity gradient.

When a fluid flows past a surface with a low velocity, within a limiting distance from the surface, its velocity varies with the distance from the surface, from below figure. The layer in contact with the surface is at rest relative to the surface. Starting outwards from the surface, the next layer has an extremely small velocity; each successive layer has a slightly higher velocity than its inner neighbour, as shown. Finally, a layer is reached which has approximately the full, or free-stream, velocity v₀ of the fluid. The situation is reversed if a body is moving in a stationary fluid : the fluid velocity reduces as the distance of a layer from the body increases. Thus, the velocity in each layer increases with its distance from the surface.

Consider the layer of thickness dy at y from the solid surface. Let v and v + dv be the velocities of the fluid at the base and upper edge of this layer. The change in velocity across the layer is dv. Therefore, the rate at which the velocity changes dv between the layers is \(\frac{d v}{d y}\). This is called the velocity gradient.

Question 86.
State and explain Newton’s law of viscosity.
Answer:
Newton’s law of viscosity: In a steady flow of a fluid past a solid surface, a velocity profile is set up such that the viscous drag per unit area on a layer is directly proportional to the velocity gradient.

When a fluid flows past a solid surface in a streamline flow or when a solid body moves through a fluid, the force of fluid friction opposing the motion is called the viscous drag. The magnitude of the viscous drag of a fluid is given by Newton’s law of viscosity.

If \(\frac{d v}{d y} \) is the velocity gradient, the viscous drag per unit area on a layer,
\(\frac{F}{A} \propto \frac{d v}{d y}\)
∴ \(\frac{F}{A}=\eta \frac{d v}{d y}\)
where the constant of proportionality, y, is called the coefficient of viscosity of the fluid.

Question 87.
Define coefficient of viscosity.
Answer:
Coefficient of viscosity : The coefficient of viscosity of a fluid is defined as the viscous drag per unit area acting on a fluid layer per unit velocity gradient established in a steady flow.

Question 88.
Find the dimensions of the coefficient of viscosity. State its SI and CGS units.
Answer:
By Newton’s law of viscosity,
\(\frac{F}{A}=\eta \frac{d v}{d y}\)
where \(\frac{F}{A}\) is the viscous drag per unit area, \(\frac{d v}{d y}\) is the velocity gradient and y is the coefficient of viscosity of the fluid. Rewriting the above equation as
η = \(\frac{(F / A)}{(d v / d y)}\)
[η] = \(\frac{\left[F A^{-1}\right]}{[d v / d y]}\) = [ML^-1T^-2][T¹] = [ML^-1T^-1]
SI unit: the pascal ∙ second (abbreviated Pa ∙ s), 1 Pa ∙ s = 1 N ∙ m^-2 ∙ s

CGS unit: dyne ∙ cm^-2 ∙ s, called the poise [symbol P, named after Jean Louis Marie Poiseuille (1799-1869), French physician].

[Note : The most commonly used submultiples are the millipascal-second (mPa ∙ s) and the centipoise (cP). 1 mPa ∙ s = 1 cP.]

Question 89.
Define the SI and CGS units of coefficient of viscosity.
Answer:
The SI unit of coefficient of viscosity is the pascal- second.

Definition : If a tangential force per unit area of one newton per square metre is required to maintain a difference in velocity of one metre per second between two parallel layers of a fluid in streamline flow separated by one metre, the coefficient of viscosity of the fluid is one pascal-second.

The CGS unit of coefficient of viscosity is the poise.

Definition : If a tangential force per unit area of one dyne per square centimetre is required to maintain a difference in velocity of one centimetre per second between two parallel layers of a fluid in streamline flow separated by one centimetre, the coefficient of viscosity of the fluid is one poise.

Question 90.
Find the conversion factor between the SI and CGS units of coefficient of viscosity using dimensional analysis.
Answer:
The dimensions of the coefficient of viscosity η are
[η] = [ML^-1T^-1]
The SI and CGS units of coefficient of viscosity are the pascal-second and poise, respectively.
1 Pa ∙ s = 1 N ∙ m^-2 ∙ s = 1 kg-m^-1 ∙ s^-1
1 P = 1 dyn ∙ cm^-2 ∙ s = l g-cm^-1 ∙ s^-1
Let 1 Pa ∙ s = xP
∴ 1[M₁L₁^-1T₁^-1] = x[M₂L₂^-1T₂^-1]
where subscripts 1 and 2 pertain to SI and CGS units.

Question 91.
State Stokes’ law. Derive Stokes’ law using dimensional analysis.
Answer:
Stokes’ law : If a fluid flows past a sphere or a sphere moves through a fluid, for small enough
∴ Viscous force = gravitational force-buoyant force
= mg – m_Lg
where m = mass of the sphere = \(\frac{4}{3} \pi r^{3} \rho\) and m_{L</sub. = mass of the liquid displaced = \(\frac{4}{3} \pi r^{3} \rho_{\mathrm{L}}\).}

At its terminal speed v_t, the magnitude of the viscous force by Stokes’ law is

[Note : Theoretically, v → v_t as time t → ∞. In practice, if η is appreciable, then v tends to v_t in a very small time interval.]

[Data: g = 9.8 m/s²]

92. Solve the following

Question 1.
The relative velocity between two layers of a fluid, separately by 0.1 mm, is 2 cm/s. Calculate the velocity gradient.
Solution :
Data : dy = 0.1 mm = 10^-2 cm, dv = 2 cm/s
The velocity gradient, \(\frac{d v}{d y}=\frac{2 \mathrm{~cm} / \mathrm{s}}{10^{-2} \mathrm{~cm}}\) = 200 s^-1

Question 2.
Calculate the force required to move a flat glass plate of area of 10 cm² with a uniform velocity of 1 cm/s over the surface of a liquid 2 mm thick, if the coefficient of viscosity of the liquid is 2 Pa.s.
Solution :
Data : η = 2 Pa.s; A = 10 cm² = 10^-3 m²;
dv = 1 cm/s = 0.01 m/s; dy = 2 mm = 2 × 10^-3 m
According to Newton’s formula,
viscous force f = \(\eta A \frac{d v}{d y}\)
= \(\frac{(2 \mathrm{~Pa} \cdot \mathrm{s})\left(10^{-3} \mathrm{~m}^{2}\right)(0.01 \mathrm{~m} / \mathrm{s})}{2 \times 10^{-3} \mathrm{~m}}\)
= 0.01 N

This force retards the motion of the glass plate. Therefore, in order to keep the plate moving with a uniform velocity, an equal force must be exerted on the plate in the forward direction.

The required force to move the glass plate is 0.01 N

Question 3.
A metal plate of length 10 cm and breadth 5 cm is in contact with a layer of oil 1 mm thick. The horizontal force required to move it with a velocity of 4 cm/s along the surface of the oil is 0.32 N. Find the coefficient of viscosity of the oil. Also express it in poise.
Solution:

Question 4.
A spherical liquid drop of diameter 2 × 10^-4 m is falling with a constant velocity through air, under gravity. If the density of the liquid is 500 kg/m3 and the coefficient of viscosity of air is 2 × 10^-5 Pa.s, determine the terminal velocity of the drop and the viscous force acting on it. Ignore the density of air.
Solution :
Data : r = 1 × 10^-4 m, ρ = 500 kg/m³, ρ_air \(\ll \rho\) η = 2 × 10^-5 Pa.s

(i) The terminal velocity,

Question 5.
If the speed at which water flows through a long cylindrical pipe of radius 8 mm is 10 cm/s, find the Reynolds number. [Density of water = 1 g/cm³, coefficient of viscosity of water = 0.01 poise]
Solution :
Data : v₀ = 10 cm/s, ρ = 1 g/cm³, r = 8 mm
∴ d = 2r = 16 mm = 1.6 cm, η = 0.01 poise

Question 93.
Define volume flow rate or volume flux. Explain how it is related to the velocity of fluid.
OR
What is the difference between flow rate and fluid velocity ? How are they related ?
Answer:
Definition : The volume of fluid passing by a given location per unit time through an area is called the volume flow rate, or simply flow rate, Q.Q = dV/dt

Consider an ideal fluid flowing with velocity v through a uniform flow tube of cross section A. If, as shown in Fig. 2.34, the shaded cylinder of fluid of length x and volume V flows past point P in time t,

which is the required relation.

Question 94.
State the SI unit for volume flow rate.
Answer:
The SI unit for volume flow rate is the cubic metre per second (m³/s).

[Note : Another common unit accepted in SI is the litre per minute (L/min). 1 L = 10^-3 m³ = 10³ cm³. An old non-SI unit from FPS system still used is the cubic feet per second (symbol, cusec).]

Question 95.
Define mass flow rate or mass flux. Explain how it is related to the velocity of fluid.
Answer:
Definition : The mass of fluid passing by a given point per unit time through an area is called the mass flow rate, dmldt.

Consider an ideal fluid of density ρ flowing with velocity v through a uniform flow tube of cross section A. If, as shown in Fig. 2.34, the shaded cylinder of fluid of length x and volume V flows past point P in time t,

which is the required relation.

Question 96.
Explain the continuity condition for a flow tube. Show that the flow speed is inversely proportional to the cross-sectional area of a flow tube.
Answer:
Consider a fluid in steady or streamline flow. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude. Suppose the velocity is \(\overrightarrow{v_{1}}\) at point P and \(\overrightarrow{v_{2}}\) at point Q. If A₁ and A₂ are the cross-sectional areas of the tube and ρ₁ and ρ₂ are the densities of the fluid at these two points, the mass of the fluid passing per unit time across A₁ is A₁ρ₁v₁ and that passing across A₂ is A₂ρ₂v₂. Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires
A₁ρ₁v₁ = A₂ρ₂v₂ …………. (1)

Equation (1) is called the equation of continuity of flow. It holds true for a compressible fluid, (like all gases) for which the density of the fluid may differ from point to point in a tube of flow. For an incompressible fluid (like all liquids), ρ₁ = ρ₂ and Eq. (1) takes the simpler form
A₁v₁ = A₂v₂ ………… (2)
∴ \(\frac{v_{1}}{v_{2}}=\frac{A_{2}}{A_{1}}\) …………… (3)
that is, the flow speed is inversely proportional to the cross-sectional area of a flow tube. Where the area is large, the speed of flow is small, and vice versa.

Equations (2) is the equation of continuity for an incompressible fluid for which density is constant throughout.

Question 97.
Explain why flow speed is greatest where streamlines are closest together.
Answer:
By the equation of continuity, the flow speed is inversely proportional to the area of cross section of a flow tube. Where the area of cross section is small, i. e., streamlines are close, the flow speed is large and vice versa.

Question 98.
You can squirt water a considerably greater distance by placing your thumb over the end of a garden hose. Explain.
Answer:
Placing one’s thumb over the end of a garden hose constricts the open end. By the continuity condition, the speed of water increases as it passes through the constriction. Hence, water squirts out and reaches a longer distance.

99. Solve the following

Question 1.
A liquid is flowing through a horizontal pipe of varying cross section. At a certain point, where the diameter of the pipe is 5 cm, the flow velocity is 0.25 m/s. What is the flow velocity where the diameter is 1 cm ?
Solution:
Data : d₁ = 5 cm, v₁ = 0.25 m/s, d₂ = 1 cm According to the equation of continuity of flow,
A₁ρ₁v₁ = A₂ρ₂v₂
where A₁ and ρ₁ are the cross-sectional area and density of the liquid where the flow velocity is v₁; A₂ and ρ₂ are the corresponding quantities where the flow velocity is v₂.
Assuming the liquid is incompressible,

Question 100.
State Bernoulli’s principle.
Answer:
Where the velocity of an ideal fluid in streamline flow is high, the pressure is low, and where the velocity of a fluid is low, the pressure is high. OR At every point in the streamline flow of an ideal (i.e., nonviscous and incompressible) fluid, the sum of the pressure energy, kinetic energy and potential energy of a given mass of the fluid is constant at every point.

[Note : The above principle is equivalent to a statement of the law of conservation of mechanical energy as applied to fluid mechanics. It was published in 1738 by Daniel Bernoulli (1700 – 82), Swiss mathematician.]

Question 101.
Explain Bernoulli’s equation of fluid flow.
Answer:
Consider an ideal fluid incompressible and nonviscous of density ρ flowing along a flow tube of varying cross section. The system under consideration is the flow tube between points 1 and 2, and the Earth (below figure). From the continuity equation it follows that pressure and speed must be different in regions of different cross section. If the height also changes, there is an additional pressure difference.

The fluid enters the system at point 1 through a. surface of cross section A₁ at speed v₁. The point 1 lies at a height h₁, with respect to an arbitrary reference level y = 0, and the local pressure there is p₁. The fluid leaves the system at point 2 where the corresponding quantities are A₂, v₂, h₂ and p₂.

Consider a small fluid element, of volume ∆V and mass ∆m = ρ∆V, that enters at point 1 and leaves at point 2 during small time interval ∆t. In the absence of internal fluid friction, it can be shown that the work done on the fluid element by the surrounding fluid is
∆W= (p₁ – p₂)∆V
This is sometimes called the pressure energy. During At, the changes in the kinetic energy and potential energy are
∆KE = \(\frac{1}{2}\) ∆m (V₂² – V₁²) = \(\frac{1}{2}\) ρ∆V(v₂² – v₁²)
∆PE = ∆mg(h₂ – h₁) = ρ∆Vg(h₂ – h₁)
Since ∆W is the work done by a non-conservative force,
∆W = ∆KE + ∆PE
∴ (p₁ – p₂)∆V = \(\frac{1}{2}\) ρ∆V(v₂2² – v₁²) + ρ∆Vg(h₂ – h₁) ……….. (1)
∴ p₁ – p₂ = \(\frac{1}{2}\) ρ(v₂² – v₁²) + ρg(h₂ – h₁)
∴ p₁ + \(\frac{1}{2}\) ρv₁² + ρgh₁ = p₂ + \(\frac{1}{2}\) ρv₂² + ρgh₂
or p + \(\frac{1}{2}\) ρv² + ρgh = constant …………. (2)
This is known as Bernoulli’s equation.

[Notes : Equation (1) can be rewitten as
p₁∆V + \(\frac{1}{2}\)ρ∆Vv₁² + ρ∆Vgh₁
p₂∆V + \(\frac{1}{2}\) ρ∆v₂² + ρ∆Vgh₂
or p∆V + \(\frac{1}{2}\)ρ∆v² + ρ∆Vgy = constant ………. (3)
i.e., pressure energy + KE + PE = constant
Dividing Eq. (3) by ∆m = ρ∆V,
\(\frac{p}{\rho}\) + \(\frac{1}{2}\) v² + gy constant

i.e., pressure energy per unit mass + KE per unit mass + PE per unit mass = constant, which is Bernoulli’s principle. Note that in writing

∆W = ∆KE + ∆PE, we have assumed principle of conservation of energy.

Dimensionally, pressure is energy per unit volume. Both terms on the right side of Eq. (2) also have the same dimensions. Hence, the term (p₁ – p₂) is often referred to as pressure energy per unit volume or pressure head. The first term on the right, \(\frac{1}{2}\) p (v₂² – v₁²) head and the second term, pg(h₂ – h₁), is called the potential head.]

Question 102.
State the limitations of Bernoulli’s principle.
Answer:
Limitations: Bernoulli’s principle and his equation for fluid flow is valid only for
(1) an ideal fluid, i.e., one that is incompressible and nonviscous, so that the density remains constant throughout a flow tube and there is no viscous drag which results in energy dissipation or loss,
(2) streamline flow.

Question 103.
State the applications of Bernoulli’s principle.
Answer:
Applications :

1. Venturi meter : It is a horizontal constricted tube that is used to measure flow speed in a gas.
2. Atomizer : It is a hydraulic device used for spraying insecticide, paint, air perfume, etc.
3. Aerofoil : The aerofoil shape of the wings of an aircraft produces aerodynamic lift.
4. Bunsen’s burner : Bernoulli effect is used to admit air into the burner to produce an oxidising flame.

Question 104.
State the law of efflux. Derive an expression for the speed of efflux for a tank discharging through an opening at a depth h below the liquid surface. Hence or otherwise show that the speed of efflux for an open tank is \(\sqrt{2 g h}\).
Answer:
Law of efflux (Torricelli’s theorem) : The speed of efflux for an open tank through an orifice at a depth h below the liquid surface is equal to the speed acquired by a body falling freely through a vertical distance h.

Consider a tank with cross-sectional area A₁ holding a static liquid of density ρ. The tank discharges through an opening (of cross-sectional area A₂) in the side wall at a depth h below the surface of the liquid. The flow speed at which the liquid leaves the tank is called the speed of efflux.

The pressure at point 2 it is the atmospheric pressure p₀. Let the pressure of the air above the liquid at point 1 be ρ. We assume that the tank is large in cross section compared to the opening (A₁ >> A₂), so that the upper surface of the liquid will drop very slowly. That is, we may regard the liquid surface to be approximately at rest v₁ ≈ 0). Bernoulli’s equation, in usual notation, states
p₁ + \(\frac{1}{2}\) ρv₁² + ρgy₁ = p₂ + \(\frac{1}{2}\) ρv₂² + ρgy₂
Substituting p₁ = p, p₂ = p₀, v₁ ≈ 0 and (y₁ – y₂) = h,
v₂² = 2 \(\frac{p-p_{0}}{\rho}\) + 2gh
If the tank is open to the atmosphere, then p = p₀,
v₂ = \(\sqrt{2 g h}\).
which is the law of efflux.

[Note : For an open tank, the speed of the liquid, v₂, leaving a hole a distance h below the surface is equal to that acquired by an object falling freely through a vertical distance h.]

Question 105.
What is a Venturi tube? Explain the working of a Venturi tube.
OR
What is a Venturi meter? Briefly explain its use to determine the flow rate in a pipe.
Answer:
A Venturi meter is a horizontal constricted tube that is used to measure the flow speed through a pipeline. The constricted part of the tube is called the throat. Although a Venturi meter can be used for a gas, they are most commonly used for liquids. As the fluid passes through the throat, the higher speed results in lower pressure at point 2 than at point 1. This pressure difference is measured from the difference in height h of the liquid levels in the U-tube manometer containing a liquid of density ρ_m (from below figure). The following treatment is limited to an incompressible fluid.

Let A₁ and A₂ be the cross-sectional areas at points 1 and 2, respectively. Let v₁ and v₂ be the corresponding flow speeds, p is the density of the fluid in the pipeline. By the equation of continuity,
v₁A₁ = v₂A₂ …………. (1)
Since the meter is assumed to be horizontal, from Bernoulli’s equation we get,
p₁ = \(\frac{1}{2}\) ρv₁² = p₂ + \(\frac{1}{2}\) ρv₂²
∴ p₁ = \(\frac{1}{2}\) ρv₁² = p₂ + \(\frac{1}{2}\) ρv₁² \(\left(\frac{A_{1}}{A_{2}}\right)^{2}\) [from Eq. (1)]
∴ p₁ – p₁ = \(\frac{1}{2}\) ρv₁² [\(\left(\frac{A_{1}}{A_{2}}\right)^{2}\) – 1] …………. (2)
The pressure difference is equal to pm gh, where h is the differences in liquid levels in the manometer. Then,

Equation (3) gives the flow speed of an incom-pressible fluid in the pipeline. The flow rates of practical interest are the mass and volume flow ‘ rates through the meter.
Volume flow rate = A₁v₁
and mass flow rate = density x volume flow rate = ρA₁v₁
[Note ; When a Venturi meter is used in a liquid pipeline, the pressure difference is measured from the difference in height h of the levels of the same liquid in the two vertical tubes, as shown in below figure. Then, the pressure difference is equal to ρgh.

The flow meter is named after Giovanni Battista Venturi (1746-1822), Italian physicist.]

Question 106.
Explain aerodynamic lift on the wings of an aeroplane.
OR
Explain why the upper surface of the wings of an aeroplane is made convex and the lower surface concave.
Answer:
An aeroplane wing has a special characteristic aerodynamic shape called an aerofoil. An aerofoil is convex on the top and slightly concave on the bottom. Its leading edge is well rounded while the trailing edge is sharp. As an aeroplane moves through air, the aerofoil shape makes the air moving over the top and along the bottom of a wing in a certain way.

If the air over the top surface travels faster than the air below the wing, this decreases the air pressure above the wing. The air flowing below the wing moves almost in a straight line, so its speed and air pressure remain the same. The air under the wings therefore pushes upward more than the air on top of the wings pushes downward, thus producing an upward force \(\vec{F}\). It is the pressure difference that generates this force. The component of \(\vec{F}\) perpendicular to the direction of motion is called the aerodynamic lift or, simply, the lift. The component parallel to the direction of flight is the drag. The lift is the force that allows an aeroplane to get off the ground and stay in the air. For an aeroplane to stay in level flight, the lift is equal in magnitude and opposite in direction to the force of gravity.

[Note: For an airborne aeroplane to get to the ground, the direction of \(\vec{F}\) must be reversed. Then, the upper surface should be more concave than the lower surface such that air above the wing travels slower than the air below it, decreasing the air pressure below the wing. This is achieved by small flaps, called ailerons, attached at the trailing end of each wing.]

Question 107.
Explain the working of an atomizer.
OR
A perfume bottle or atomizer sprays a fluid that is in the bottle. How does the fluid rise up in the vertical tube in the bottle?
Answer:
An atomizer is a device which entraps or entrains liquid droplets in a flowing gas. Its working is based on Bernoulli’s principle. A squeeze bulb or a pump is used to create a jet of air over an open tube dipped into a liquid. By Bernoulli’s principle, the high-velocity air stream creates low pressure at the open end of the tube. This causes the liquid to rise in the tube. The liquid is then dispersed into a fine spray of droplets. This type of system is used in a perfume bottle, a paint sprayer, insect and perfume sprays and an automobile carburetor.

[Notes : A Bunsen burner uses an adjustable gas nozzle to entrain air into the gas stream for proper combustion. Aspirators, used as suction pumps, in dental and surgical situations (for draining body fluids) or for draining a flooded basement, is another similar applica-tion. Some chimney pipes have a T-shape, with a cross-piece on top that helps draw up gases whenever there is even a slight breeze.]

Question 108.
Roofs are sometimes blown off vertically during a tropical cyclone, and houses sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.
Answer:
A cyclonic high wind blowing over a roof creates a low pressure above it, in accordance with Bernoulli’s principle. The pressure below the roof is equal to the atmospheric pressure which is now greater than the pressure above the roof. This pressure difference causes an aerodynamic lift that lifts the roof up. Once the roof is lifted up, it blows off in the direction of the wind.

Wind speeds in a tornado may be much higher and thus create much greater pressure differences. Sometimes, wooden houses hit by a tornado explode.

Question 109.
Describe what happens : (1) Hold the short edge of a paper strip (2″ × 6″) up to your lips-the strip slanting downward over your finger-and blow over the top of the strip. (2) Hold two strips of paper up to your lips, separated by your fingers and blow between the strips.
Answer:
(1) The air stream over the top surface travels faster than the air stream below the paper strip. This decreases the air pressure above the strip relative o that below. This produces an aerodynamic lift in accordance with Bernoulli’s principle and the paper strip will lift up.

(2) Air passing between the paper strips flows in a narrower channel and, in accordance with Be rnoulli’s principle, must increase its speed, causing the pressure between them to drop. This will pinch the two strips together.

110. Solve the following

Question 1.
Water is flowing through a horizontal pipe of varying cross section. At a certain point where the velocity is 0.12 m/s, the pressure of water is 0.010 m of mercury. What is the pressure at a point where the velocity is 0.24 m/s?
Solution:
Data : v₁ = 0.12 m/s, p₁ = 0.010 m of Hg, v₂ = 0.24 m/s, y₁ = y₂ (horizontal pipe), ρ_Hg = 13600 kg/m², ρ_w = 1000 kg/m³, g = 9.8 m/s²
p₁ = 0.010 m of Hg .
= (0.010 m) ρ_Hgg
= (0.010 m) (13600 kg/m³) (9.8 m/s²)
= 1332.8 Pa ≅ 1333 Pa
According to Bernoulli’s principle,

This gives the pressure of the water in the pipe where the flow velocity is 0.24 m / s.

Question 2.
A building receives its water supply through an undergound pipe 2 cm in diameter at an absolute pressure of 4 × 10⁵ Pa and flow velocity 4 m/s. The pipe leading to higher floors is 1.5 cm in diameter. Find the flow velocity and pressure at the floor inlet 10 m above.
Solution:
Data : d₁ = 3 cm, p₁ = 4 × 10⁵ Pa, v₁ = 4 m/s,
d₂ = 2 cm, h₂ – h₁ = 10 m
By continuity equation, the flow velocity at the higher floor inlet

Question 3.
Calculate the total energy per unit mass possessed by water at a point where the pressure is 0.1 × 10⁵ N/m², the velocity is 0.02 m/s and the height of the water level from the ground is 10 cm. Density of water = 1000 kg/m³.
Solution:
Data : p = 0.1 × 10⁵ N/m² = 10⁴ Pa, v = 0.02 m/s, y = 10 cm = 0.1 m, ρ = 1000 kg/m³, g = 9.8 m/s²
The total energy per unit mass of water
= \(\frac{p}{\rho}\) + \(\frac{1}{2}\)v² + gy
= \(\frac{10^{4} \mathrm{~Pa}}{10^{3} \mathrm{~kg} / \mathrm{m}^{3}}\) + \(\frac{1}{2}\) (2 × 10^-2 m/s)² + (9.8 m/ s²) (0.1 m)
= 10 + 0.0002 + 0.98 = 10.9802 J/kg

Question 4.
A horizontal wind with a speed of 11 m/s blows past a tall building which has large windowpanes of plate glass of dimensions 4 m × 1.5 m. The air inside the building is at atmospheric pressure. What is the total force exerted by the wind on a window pane ? [Density of air = 1.3 kg/m³]
Solution:
Data : v₁ (inside) = 0 m/s, v₂ (outside) = 11 m/s, ρ = 1.3 kg/m³, p₁ = p₀ = atmospheric pressure,
A = 4m × 1.5 m = 6m
Let p₂ be the air pressure outside a window. At the same height, Bernoulli’s equation gives
∴ p₁ + \(\frac{1}{2}\) ρv₁² = p₂ + \(\frac{1}{2}\)ρv₂²
∴ p₀ + 0 = p₂ + \(\frac{1}{2}\) ρv₂²
∴ The difference in pressure across a windowpane is
p₀ – p₂ = \(\frac{1}{2}\) ρv₂²
Since the right hand side is positive,
p₀ > p₂ and p₀ – p₂ is directed outward.
∴ The total force on a window pane is
F = (p₀ – p₂)A = \(\frac{1}{2}\) ρv₂² A
= \(\frac{1}{2}\) (1.3 kg/m³) (11 m/s)² (6 m²)
= 3.9 × 121 = 471.9 N (outward)

Question 5.
A water tank has a hole at a distance x from the free surface of water in the tank. If the radius of the hole is 2 mm and the velocity of efflux is 11 m/s, find x.
Solution:
Data: r = 2 mm, v = 11 m/s, g = 9.8 m/s²
By Torricelli’s law of efflux, the velocity of efflux,
v = \(\sqrt{2 g x}\)
∴ x = \(\frac{v^{2}}{2 g}=\frac{(11 \mathrm{~m} / \mathrm{s})^{2}}{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)}\)
= \(\frac{121}{19.6}\) = 6.173 m

Multiple Choice Questions

Question 1.
Consider the following statements:
I. A fluid in hydrostatic equilibrium exerts only normal force on any surface within the fluid.
II. A fluid can resist a tangential force. Of these,
(A) only (I) is correct
(B) only (II) is correct
(C) both are correct
(D) both are false.
Answer:
(A) only (I) is correct

Question 2.
Which of the following is correct ?
(A) The free surface of a liquid at rest is horizontal.
(B) The pressure at a point within a liquid at rest is the same in all directions.
(C) The pressure at all points within a liquid at rest is the same.
(D) Both (A) and (B).
Answer:
(D) Both (A) and (B).

Question 3.
The surface of a liquid (of uniform density p) in a container is open to the atmosphere. The atmospheric pressure is ρ₀. The pressure ρgh, at a depth h below the surface of the liquid, is called the
(A) absolute pressure
(B) normal pressure
(C) gauge pressure
(D) none of these.
Answer:
(C) gauge pressure

Question 4.
Three vessels having the same base area are filled with water to the same height, as shown. The force exerted by water on the base is

(A) largest for vessel P
(B) largest for vessel Q
(C) largest for vessel R
(D) the same in all three.
Answer:
(D) the same in all three.

Question 5.
Rain drops or liquid drops are spherical in shape, especially when small, because
(A) cohesive force between the molecules of water have spheres of influence
(B) for a given volume, a spherical drop has the least surface energy
(C) for a given volume, a spherical drop has the maximum surface energy
(D) the pressure inside a drop is many times the atmospheric pressure outside.
Answer:
(B) for a given volume, a spherical drop has the least surface energy

Question 6.
The surface tension acts
(A) perpendicular to the surface and vertically up-wards
(B) perpendicular to the surface and vertically into the liquid
(C) tangential to the surface
(D) only at the liquid-solid interface.
Answer:
(C) tangential to the surface

Question 7.
A thin ring of diameter 8 cm is pulled out of water (surface tension 0.07 N/m). The force required to break free the ring from water is
(A) 0.0088 N
(B) 0.0176 N
(C) 0.0352 N
(D) 3.52 N.
Answer:
(C) 0.0352 N

Question 8.
A matchstick 5 cm long floats on water. The water film has a surface tension of 70 dyn/cm. A little comphor put on one side of stick reduces the surface tension there to 50 dyn/cm. The net force on the matchstick is
(A) 4 dynes
(B) 20 dynes
(C) 100 dynes
(D) 600 dynes.
Answer:
(C) 100 dynes

Question 9.
A big drop of radius R is formed from 1000 droplets of water. The radius of a droplet will be
(A) 10 R
(B) \(\frac{R}{10}\)
(C) \(\frac{R}{100}\)
(D) \(\frac{R}{1000}\)
Answer:
(B) \(\frac{R}{10}\)

Question 10.
The work done in breaking a spherical drop of a liquid (surface tension T) of radius R into 8 equal drops is
(A) πR²T
(B) 2πR²T
(C) 3πR²T
(D) 4πR²T.
Answer:
(D) 4πR²T.

Question 11.
If for a liquid in a vessel, the force of cohesion is more than the force of adhesion,
(A) the liquid does not wet the solid
(B) the liquid wets the solid
(C) the surface of the liquid is plane
(D) the angle of contact is zero.
Answer:
(A) the liquid does not wet the solid

Question 12.
If a liquid does not wet a solid surface, its angle of contact with the solid surface is
(A) zero
(B) acute
(C) 90°
(D) obtuse.
Answer:
(D) obtuse.

Question 13.
The pressure within a bubble is higher than that outside by an amount proportional
(A) directly to both the surface tension and the bubble size
(B) directly to the surface tension and inversely to the bubble size
(C) directly to the bubble size and inversely to the surface tension
(D) inversely to both the surface tension and the bubble size.
Answer:
(B) directly to the surface tension and inversely to the bubble size

Question 14.
The pressure difference across the surface of a spherical water drop of radius 1 mm and surface tension 0.07 N/m is
(A) 28 Pa
(B) 35 Pa
(C) 140 Pa
(D) 280 Pa.
Answer:
(C) 140 Pa

Question 15.
An air bubble just inside a soap solution and a soap bubble blown using the same solution have their radii in the ratio 3 : 2. The ratio of the excess pressure inside them is
(A) 1 : 12
(B) 1 : 6
(C) 1 : 3
(D) 1 : 2.
Answer:
(C) 1 : 3

Question 16.
A liquid rises to a height of 5 cm in a glass capillary tube of radius 0.02 cm. The height to which the liquid would rise in a glass capillary tube of radius 0.04 cm is
(A) 2.5 cm
(B) 5 cm
(C) 7.5 cm
(D) 10 cm.
Answer:
(A) 2.5 cm

Question 17.
In a gravity free space, the liquid in a capillary tube will rise to
(A) the same height as that on the Earth
(B) a lesser height than on the Earth
(C) slightly more height than that on the Earth
(D) the top and overflow.
Answer:
(D) the top and overflow.

Question 18.
In which of the following substances, does the surface tension increase with an increase in tempera¬ture?
(A) Copper
(B) Molten copper
(C) Iron
(D) Molten iron
Answer:
(B) Molten copper

Question 19.
A fluid flows in steady flow through a pipe. The pipe has a circular cross section, but its radius varies along its length. The mass of the fluid passing per second at the entrance point (radius R) of the pipe is Q₁ while that at the exit point (radius R/2) is Q₂. Then, Q₂ is equal to
(A) \(\frac{1}{4}\) Q₁
(B) Q₁
(C) 2Q₁
(D) 4Q₁.
Answer:
(B) Q₁

Question 20.
The dimensions of coefficient of viscosity are
(A) [ML^-1T^-2]
(B) [M^-1LT^-2]
(C) [MLT^-2]
(D) [ML^-1T^-1].
Answer:
(D) [ML^-1T^-1].

Question 21.
The unit of coefficient of viscosity is
(A) the pascal-second
(B) the pascal
(C) the poise-second
(D) both (A) and (C).
Answer:
(A) the pascal-second

Question 22.
A fluid flows past a sphere in streamline flow. The viscous force on the sphere is directly proportional to
(A) the radius of the sphere
(B) the speed of the flow
(C) the coefficient of viscosity of the fluid
(D) all of these.
Answer:
(D) all of these.

Question 23.
Water flows in a streamlined flow through the pipe shown in the following figure. The pressure

(A) is greater at A than at B
(B) at A equals that at B
(C) is less at A than at B
(D) at A is unrelated to that at B.
Answer:
(A) is greater at A than at B

Question 24.
Two steel marbles (of radii R and \(\frac{R}{3}\)) are released in a highly viscous liquid. The ratio of the terminal velocity of the larger marble to that of the smaller is
(A) 9
(B) 3
(C) 1
(D) \(\frac{1}{9}\)
Answer:
(A) 9

Question 25.
A large tank, filled with a liquid, is open to the atmosphere. If the tank discharges through a small hole at its bottom, the speed of efflux does NOT depend on
(A) cross-sectional area of the hole
(B) depth of the hole from the liquid surface
(C) acceleration due to gravity
(D) all of these.
Answer:
(A) cross-sectional area of the hole

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 1 Rotational Dynamics Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 1 Rotational Dynamics

Question 1.
What is circular motion?
Answer:
The motion of a particle along a complete circle or a part of it is called circular motion.

Question 2.
What is a radius vector in circular motion?
Answer:
For a particle performing circular motion, its position vector with respect to the centre of the circle is called the radius vector.
[Note : The radius vector has a constant magnitude, equal to the radius of the circle. However, its direction changes as the position of the particle changes along the circumference.]

Question 3.
What is the difference between rotation and revolution?
Answer:
There is no physical difference between them. It is just a question of usage. Circular motion of a body about an axis passing through the body is called rotation. Circular motion of a body around an axis outside the body is called revolution.

Question 4.
State the characteristics of circular motion.
Answer:

1. It is an accelerated motion : As the direction of velocity changes at every instant, it is an accelerated motion.
2. It is a periodic motion : During the motion, the particle repeats its path along the same trajectory. Thus, the motion is periodic.

Question 5.
Explain angular displacement in circular motion.
Answer:
The change in the angular position of a particle performing circular motion with respect to a reference line in the plane of motion of the particle and passing through the centre of the circle is called the angular displacement.

As the particle moves in its circular path, its angular position changes, say from θ₁ at time t to θ₂ at a short time δt later. In the interval δt, the position vector \(\vec{r}\) sweeps out an angle δθ = θ₂ – θ₁. δθ is the magnitude of the change in the angular position of the particle.

Infinitesimal angular displacement \(\overrightarrow{\delta \theta}\) in an infinitesimal time interval δt → 0, is given a direction perpendicular to the plane of revolution by the right hand thumb rule.

Question 6.
Explain angular velocity. State the right hand thumb rule for the direction of angular velocity.
Answer:
Angular velocity : The time rate of angular displacement of a particle performing circular motion is called the angular velocity.

1. If the particle has an angular displacement \(\delta \vec{\theta}\) in a short time interval δt, its angular velocity
   
2. \(\vec{\omega}\) is a vector along the axis of rotation, in the direction of \(d \vec{\theta}\), given by the right hand thumb rule.

Right hand thumb rule : If the fingers of the right hand are curled in the sense of revolution of the particle, then the outstretched thumb gives the direction of the angular displacement.

[Note : Angular speed, ω = |\(\vec{\omega}\)| = \(\frac{d \theta}{d t}\) is also called angular frequency. ]

Question 7.
Explain the linear velocity of a particle performing circular motion.
OR
Derive the relation between the linear velocity and the angular velocity of a particle performing circular motion.
Answer:
Consider a particle performing circular motion in an anticlockwise sense, along a circle of radius r. In a very small time interval δt, the particle moves from point A to point B through a distance δs and its angular position changes by δθ.

Since is tangential, the instantaneous linear velocity \(\vec{v}\) of a particle performing circular motion is along the tangent to the path, in the sense of motion of the particle.

\(\vec{v}\), \(\vec{\omega}\) and \(\vec{r}\) are mutually perpendicular, so that in magnitude, v = ωr.

Question 8.
State the relation between the linear velocity and the angular velocity of a particle in circular motion.
Answer:
Linear velocity, \(\vec{v}\) = \(\vec{\omega}\) × \(\vec{r}\) where ω is the angular velocity and r is the radius vector.

At every instant, \(\vec{v}\), \(\vec{\omega}\) and \(\vec{r}\) are mutually perpendicular, so that in magnitude v = ωr.

Question 9.
Define uniform circular motion (UCM).
Answer:
A particle is said to perform uniform circular motion if it moves in a circle or a circular arc at constant linear speed or constant angular velocity.

Question 10.
A stone tied to a string is rotated in a horizontal circle (nearly). If the string suddenly breaks, in which direction will the stone fly off ?
Answer:
In a circular motion, the instantaneous velocity \(\vec{v}\) is always tangential, in the sense of the motion. Hence, an inertial observer will see the stone fly off tangentially, in the direction of \(\vec{v}\) at the instant the string breaks.

Question 11.
What is the angular speed of a particle moving in a circle of radius r centimetres with a constant speed of v cm/s ?
Answer:
Angular speed, ω = \(\frac{v \mathrm{~cm} / \mathrm{s}}{r \mathrm{~cm}}\) = \(\frac{v}{r}\) rad/s.

Question 12.
Define the period and frequency of revolution of a particle performing uniform circular motion (UCM) and state expressions for them. Also state their SI units.
Answer:
(1) Period of revolution : The time taken by a particle performing UCM to complete one revolution is called the period of revolution or the period (T) of UCM.
T = \(\frac{2 \pi r}{v}\) = \(\frac{2 \pi}{\omega}\)
where v and ω are the linear and angular speeds, respectively.
SI unit: the second (s)
Dimensions : [M°L°T¹].

(2) Frequency of revolution : The number of revolutions per unit time made by a particle in UCM is called the frequency of revolution (f).

The particle completes 1 revolution in periodic time T. Therefore, it completes 1/T revolutions per unit time.
∴ Frequency f = \(\frac{1}{T}\) = \(\frac{v}{2 \pi r}\) = \(\frac{\omega}{2 \pi}\)
SI unit : the hertz (Hz), 1 Hz = 1 s^-1
Dimensions : [M°L°T^-1]

Question 13.
If the angular speed of a particle in UCM is 20π rad/s, what is the period of UCM of the particle?
Answer:
The period of UCM of the particle,
T = \(\frac{2 \pi}{\omega}\) = \(\frac{2 \pi}{20 \pi}\) = 0.1 s

Question 14.
Why is UCM called a periodic motion?
Answer:
In a uniform motion, a particle covers equal distances in equal intervals of time. Any motion which repeats itself in equal intervals of time is called a periodic motion. In a uniform circular motion (UCM), the particle takes the same time to complete each revolution, a distance equal to the circumference of the circle. Therefore, it is a periodic motion.

Question 15.
Give one example of uniform circular motion.
Answer:

1. Circular motion of every particle of the blades of a fan or the dryer drum of a washing machine when the fan or the drum is rotating with a constant angular speed.
2. Motion of the hands of a clock.
3. Motion of an Earth-satellite in a circular orbit.

Question 16.
What can you say about the angular speed of an hour hand as compared to that of the Earth’s rotation about its axis ?
Answer:
The periods of rotation of an hour hand and the Earth are T_h = 12 h and T_E = 24 h, respectively, so that their angular speeds are ω_h = \(\frac{2 \pi}{12}\) rad/h and ω_E = \(\frac{2 \pi}{24}\) rad/h.
∴ ω_h = 2ω_E

Question 17.
Explain the acceleration of a particle in UCM. State an expression for the acceleration.
Answer:
A particle in uniform circular motion (UCM) moves in a circle or circular arc at constant linear speed v. The instantaneous linear velocity \(\vec{v}\) of the particle is along the tangent to the path in the sense of motion of the particle. Since \(\vec{v}\) changes in direction, without change in its magnitude, there must be an acceleration that must be

1. perpendicular to \(\vec{v}\)
2. constant in magnitude
3. at every instant directed radially inward, i.e., towards the centre of the circular path.

Such a radially inward acceleration is called a centripetal acceleration.
∴ \(\vec{a}\) = \(\frac{d \vec{v}}{d t}\) = \(\overrightarrow{a_{\mathrm{r}}}\)

If \(\vec{\omega}\) is the constant angular velocity of the particle and r is the radius of the circle,
\(\overrightarrow{a_{\mathrm{r}}}\) = –\(\omega^{2} \vec{r}\)
where ω = |\(\vec{\omega}\)| and the minus sign shows that the direction of \(\vec{a}_{\mathrm{r}}\) is at every instant opposite to that of the radius vector \(\vec{r}\). In magnitude,
a_r = ω²r = \(\frac{v^{2}}{r}\) = ωv

[Note : The word centripetal comes from Latin for ‘centre-seeking’.]

Question 18.
Draw a diagram showing the linear velocity, angular velocity and radial acceleration of a particle performing circular motion with radius r.
Answer:

Question 19.
If a particle in UCM has linear speed 2 m/s and angular speed 5 rad/s, what is the magnitude of the centripetal acceleration of the particle ?
Answer:
The magnitude of the centripetal acceleration of the particle is a_r = ωv = (5)(2) = 10 m/s²

Question 20.
State any two quantities that are uniform in UCM.
Answer:
Linear speed and angular speed. (Also, kinetic energy, angular speed and angular momentum.)

Question 21.
State any two quantities that are nonuniform in UCM.
Answer:
Velocity and acceleration are nonuniform in UCM.
(Also, centripetal force.)

Question 22.
What is a nonuniform circular motion?
Answer:
Consider a particle moving in a plane along a circular path of constant radius. If the particle is speeding up or slowing down, its angular speed ω and linear speed v both change with time. Then, the particle is said to be in a non uniform circular motion.

Question 23.
Explain angular acceleration.
Answer:
Angular acceleration : The time rate of change of angular velocity of a particle performing circular motion is called the angular acceleration.

(i) If \(\delta \vec{\omega}\) is the change in angular Velocity in a short time interval St, the angular acceleration

(ii) The direction of \(\vec{\alpha}\) is the same as that of \(d \vec{\omega}\). We consider the case where a change in \(\vec{\omega}\) arises due to a change in its magnitude only. If the particle is speeding up, i.e., ω is increasing with time, then \(\vec{\alpha}\) is in the direction of \(\vec{\omega}\). If the particle is slowing down, i.e., ω is decreasing with time, then \(\vec{\alpha}\) is directed opposite to \(\vec{\omega}\).

(iii) If the angular speed changes from ω₁ to ω₂ in time f, the magnitude (α) of the average angular acceleration is
α = \(\frac{\omega_{2}-\omega_{1}}{t}\)

Question 24.
Explain the tangential acceleration of a particle in non uniform circular motion.
Answer:
Tangential acceleration : For a particle performing circular motion, the linear acceleration tangential to the path that produces a change in the linear speed of the particle is called the tangential acceleration.
Explanation :
(i) If a particle performing circular motion is speeding up or slowing down, its angular speed co and linear speed v both change with time.

The linear acceleration that produces a change only in the linear speed must be along \(\vec{v}\). Hence, it is called the tangential acceleration, \(\overrightarrow{a_{\mathrm{t}}}\). In magnitude, a_t = dv/dt

(ii) If the linear speed v of the particle is increasing, \(\overrightarrow{a_{\mathrm{t}}}\) is in the direction of \(\vec{v}\). If v is decreasing, \(\overrightarrow{a_{\mathrm{t}}}\) is directed opposite to \(\vec{v}\).

Question 25.
Obtain the relation between the magnitudes of the linear (tangential) acceleration and angular acceleration in non uniform circular motion.
Answer:
Consider a particle moving along a circular path of constant radius r. If the particle is speeding up or slowing down, its motion is nonuniform, and its angular speed ω and linear speed v both change with time. At any instant, v, ω and r are related by v = ωr
The angular acceleration of the particle is
α = \(\frac{d \omega}{d t}\)

The tangential acceleration \(\overrightarrow{a_{\mathrm{t}}}\) is the linear acceleration that produces a change in the linear speed of the particle and is tangent to the circle. In magnitude,

This is the required relation.

Question 26.
Obtain an expression for the acceleration of a particle performing circular motion. Explain its two components.
OR
For a particle performing uniform circular motion, \(\vec{v}\) = \(\vec{\omega}\) × \(\vec{r}\). Obtain an expression for the linear acceleration of a particle performing non-uniform circular motion.
OR
In circular motion, assuming \(\vec{v}\) = \(\vec{\omega}\) × \(\vec{r}\), obtain an expression for the resultant acceleration of a particle in terms of tangential and radial components.
Answer:
Consider a particle moving along a circular path of constant radius r. If its motion is nonuniform, then its angular speed ω and linear speed v both change with time.

acceleration is called the radial or centripetal acceleration \(\overrightarrow{a_{\mathrm{r}}}\).
\(\overrightarrow{a_{\mathrm{r}}}\) = \(\vec{\omega}\) × \(\vec{v}\) … (5)
In magnitude, a_r = ωv
since \(\vec{\omega}\) is perpendicular to \(\vec{v}\).
∴ \(\vec{a}\) = \(\overrightarrow{a_{\mathrm{t}}}\) + \(\overrightarrow{a_{\mathrm{r}}}\) …. (6)
This is the required expression.

Question 27.
What is the angle between linear acceleration and angular acceleration of a particle in nonuniform circular motion ?
Answer:
In a nonu niform circular motion, the angular acceleration is an axial vector, perpendicular to the plane of the motion. The linear acceleration is in the plane of the motion. Hence, the angle between them is 90°.

Question 28.
What are the differences between a nonuniform circular motion and a uniform circular motion? (Two points of distinction) Give examples.
Answer:
(i) Nonuniform circular motion :

1. The angular and tangential accelerations are non-zero, so that linear and angular speeds both change with time.
   
2. The net linear acceleration, being the resultant of the radial and tangential accelerations, is not radial, \(\vec{a}\) = \(\overrightarrow{a_{\mathrm{c}}}\) + \(\overrightarrow{a_{t}}\),
3. The magnitudes of the centripetal acceleration and the centripetal force are not constant.
4. Example : Motion of the tip of a fan blade when the fan is speeding up or slowing down.

(ii) Uniform circular motion :

1. The angular and tangential accelerations are zero, so that linear speed and angular velocity are constant.
2. The net linear acceleration is radially inward, i.e., centripetal.
3. The magnitudes of the centripetal acceleration and the centripetal force are also constant.
4. Example : Motion of the tips of the hands of a mechanical clock.

Question 29.
Write the kinematical equations for circular motion in analogy with linear motion.
Answer:
For circular motion of a particle with constant angular acceleration α,

where ω₀ and ω are the initial and final angular speeds, t is the time, ω_av the average angular speed and θ_o and θ the initial and final angular positions of the particle.

Then, the angular kinematical equations for the circular motion are (in analogy with linear kinematical equations for constant linear acceleration)
ω = ω₀ + αt
θ – θ₀ = ω₀t + \(\frac{1}{2} \alpha t^{2}\)
ω² = \(\omega_{0}^{2}\) + 2α (θ – θ₀)

Question 30.
Solve the following :

Question 1.
Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?
Solution :
Data : d = 40 km, /= 1 rot/s
∴ r = \(\frac{d}{2}\) = \(\frac{40 \mathrm{~km}}{2}\) = 20 km = 2 × 10⁴ m
Linear speed, v = ωr = (2πf)r
= (2 × 3.142 × 1)(2 × 10⁴)
= 6.284 × 2 × 10⁴
= 1.257 × 10⁵ m/s (or 125.6 km/s)

Question 2.
A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is 50 cm. If the body revolves with a constant angular speed of 20 rad/s, find the

1. period of revolution
2. linear speed
3. centripetal acceleration of the body.

Solution :
Data : m = 100 g = 0.1 kg, r = 50 cm = 0.5 m, ω = 20 rad/s

1. The period of revolution of the body,
   
2. Linear speed, v = ωr = 20 × 0.5 = 10 m/s
3. Centripetal acceleration,
   a_c = w²r = (20)² × 0.5 = 200 m/s²

Question 3.
Calculate the angular speed of the Earth due to its spin (rotational motion).
Solution :
Data : T = 24 hours = 24 × 60 × 60 s

The angular speed of the Earth due to its spin (rotational motion) is 7.273 × 10^-5 rad/s.

Question 4.
Find the angular speed of rotation of the Earth so that bodies on the equator would feel no weight. [Radius of the Earth = 6400 km, g = 9.8 m/s²]
Solution :
Data : Radius of the Earth = r = 6400 km = 6.4 × 10⁶ m, g = 9.8 m/s²

As the Earth rotates, the bodies on the equator revolve in circles of radius r.

Question 31.
Write the kinematical equations for circular motion in analogy with linear motion.
Answer:
For circular motion of a particle with constant angular acceleration α,

where ω₀ and ω are the initial and final angular speeds, t is the time, ω_av the average angular speed and 0o and 0 the initial and final angular positions of the particle.

Then, the angular kinematical equations for the circular motion are (in analogy with linear kinematical equations for constant linear acceleration)
ω = ω₀ + αt
θ – θ_o = ω₀t + \(\frac{1}{2} \alpha t^{2}\)
ω² = \(\omega_{0}^{2}\) + 2α (θ – θ_o)

Question 32.
Solve the following :

Question 1.
Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?
Solution :
Data : d = 40 km, f= 1 rot/s
∴ r = \(\frac{d}{2}\) = \(\frac{40 \mathrm{~km}}{2}\) = 20 km = 2 × 10⁴ m
Linear speed, v = ωr = (2πf)r
= (2 × 3.142 × 1)(2 × 10⁴)
= 6.284 × 2 × 10⁴
= 1.257 × 10⁵ m/s (or 125.6 km/s)

Question 2.
A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is 50 cm. If the body revolves with a constant angular speed of 20 rad/s, find the

1. period of revolution
2. linear speed
3. centripetal acceleration of the body.

Solution :
Data : m = 100 g = 0.1 kg, r = 50 cm = 0.5 m, ω = 20 rad/s

1. The period of revolution of the body,
   
2. Linear speed, v = ωr = 20 × 0.5 = 10 m/s
3. Centripetal acceleration,
   a_c = ω² = (20)² × 0.5 = 200 m/s²

Question 3.
Calculate the angular speed of the Earth due to its spin (rotational motion).
Solution :
Data : T = 24 hours = 24 × 60 × 60 s

The angular speed of the Earth due to its spin (rotational motion) is 7.273 × 10^-5 rad/s.

Question 4.
Find the angular speed of rotation of the Earth so that bodies on the equator would feel no weight. (Radius of the Earth = 6400 km, g = 9.8 m/s²]
Solution :
Data : Radius of the Earth = r = 6400 km
= 6.4 × 10⁶ m, g = 9.8 m/s²
As the Earth rotates, the bodies on the equator revolve in circles of radius r.
These bodies would not feel any weight if their centripetal acceleration (ωr) is equal to the acceleration due to gravity (g).
∴ ω²r = g
The angular speed of the Earth’s rotation,

Question 5.
To simulate the acceleration of large rockets, astronauts are seated in a chamber and revolved in a circle of radius 9.8 m. What angular speed is required to generate a centripetal acceleration 8 times the acceleration due to gravity? [g = 9.8 m/s²]
Solution :
Data : r = 9.8 m, g = 9.8 m/s², a = 8g
Centripetal acceleration = ω²r
∴ ω²r = 8g
∴ 9.8 ω² = 8(9.8)
∴ ω² = 8
The required angular speed, ω = \(\sqrt{8}\) = 2\(\sqrt{2}\) = 2.828 rad/s

Question 6.
The angular position of a rotating object is given by θ(t) = (1.55t² – 7.75 t + 2.87) rad, where t is measured in second.
(i) When is the object momentarily at rest ?
(ii) What is the magnitude of its angular acceleration at that time ?
Solution :

Question 7.
A motor part at a distance of 1.5 m from the motor’s axis of rotation has a constant angular acceleration of 0.25 rad/s². Find the magnitude of its linear acceleration at the instant when its angular speed is 0.5 rad/s.
Solution :
Data : r = 1.5 m, α = 0.25 rad/s², ω = 0.5 rad/s² a_r = ω²r = (0.5)²(1.5) = 0.25 × 1.5 = 0.375 m/s² a_t = αr = 0.25 × 1.5 = 0.375 m/s²
The linear acceleration,

Question 8.
A coin is placed on a stationary disc at a distance of 1 m from the disc’s centre. At time t = 0 s, the disc begins to rotate with a constant angular acceleration of 2 rad/s² around a fixed vertical axis through its centre and perpendicular to its plane.
Find the magnitude of the linear acceleration of the coin at t = 1.5 s. Assume the coin does not slip.
Solution :

Question 9.
A railway locomotive enters a stretch of track, which is in the form of a circular arc of radius 280 m, at 10 m/s and with its speed increasing uniformly. Ten seconds into the stretch its speed is 14m/s and at 18s its speed is 19 m/s. Find
(i) the magnitude of the locomotive’s linear acceleration when its speed is 14 m/s
(ii) the direction of this acceleration at that point with respect to the locomotive’s radial acceleration
(iii) the angular acceleration of the locomotive.
Answer:
Data : r = 280 m, v₁ = 10 m/s at t₁ = 0, v₂ = 14 m/s at t₂ = 10 s, v₃ = 19 m/s at t₃ = 18 s
(i) At t = t₂, the radial acceleration is

Since the tangential acceleration is constant, it may be found from the data for any two times.

(ii) If θ is the angle between the resultant linear acceleration and the radial acceleration,
tan θ = \(\frac{a_{\mathrm{t}}}{a_{\mathrm{r}}}\) = \(\frac{0.5}{0.7}\) = 0.7142
∴ θ = tan^-1 0.7142 = 35°32′

(iii) a_t = αr
The angular acceleration,
α = \(\frac{a_{\mathrm{t}}}{r}\) = \(\frac{0.5}{280}\)
= 1.785 × 10^-3 rad/s²
= 1.785 mrad/s²

Question 10.
The frequency of revolution of a particle performing circular motion changes from 60 rpm to 180 rpm in 20 seconds. Calculate the angular acceleration of the particle.
Solution :
Data : f₁ =60 rpm = \(\frac{60}{60}\) rev/s = 1 rev/s, f₂ = 180 rpm = \(\frac{180}{60}\) rev/s = 3 rev/s, t = 20 s
The angular acceleration in SI units,

OR
Using non SI units, the angular frequencies are ω₁ = 60 rpm = 1 rps and ω₂ = 180 rpm = 3 rps.

Question 11.
The frequency of rotation of a spinning top is 10 Hz. If it is brought to rest in 6.28 s, find the angular acceleration of a particle on its surface.
Solution:
Data: f₁ = 10Hz, f₂ = 0 Hz, t = 6.28s
The angular acceleration,

Question 12.
A wheel of diameter 40 cm starts from rest and attains a speed of 240 rpm in 4 minutes. Calculate its angular displacement in this time interval.
Solution:

Question 13.
A flywheel slows down uniformly from 1200 rpm to 600 rpm in 5 s. Find the number of revolutions made by the wheel in 5 s.
Solution :
Data : ω₀ = 1200 rpm, ω = 600 rpm, f = 5 s
Since the flywheel slows down uniformly, its angular acceleration is constant. Then, its average angular speed,

Its angular displacement in time t,
θ = ω_av.t = 15 × 5 = 75 revolutions

Question 33.
Define and explain centripetal force.
Answer:
Definition : In the uniform circular motion of a particle, the centripetal force is the force on the particle which at every instant points radially towards the centre of the circle and produces the centripetal acceleration to move the particle in its circular path.

Explanation : A uniform circular motion is an accelerated motion, with a radially inward (i.e., centripetal) acceleration –\(\frac{v^{2}}{r} \hat{\mathrm{r}}\) or \(-\frac{v^{2}}{r} \hat{\mathrm{r}}\), where \(\vec{r}\) is the radius vector and \(\hat{\mathbf{r}}\) is a unit vector in the direction of \(\vec{r}\). Hence, a net real force must act on the particle to produce this acceleration. This force, which at every instant must point radially towards the centre of the circle, is called the centripetal force. If m is the mass of the particle, the centripetal force is

Notes :

1. As viewed from an inertial frame of reference, the centripetal force is necessary and sufficient for the particle to perform UCM. At any instant, if the centripetal force suddenly vanishes, the particle would fly off in the direction of its linear velocity at that instant.
2. In case the angular or linear speed changes with time, as in nonuniform circular motion, the force is not purely centripetal but has a tangential component which accounts for the tangential acceleration.

Question 34.
Give any two examples of centripetal force.
Answer:
Examples of centripetal force :

1. For an Earth-satellite in a circular orbit, the centripetal force is the gravitational force exerted by the Earth on the satellite.
2. In the Bohr atom, the centripetal force on an electron in circular orbit around the nucleus is the attractive Coulomb force of the nucleus.
3. When an object tied at the end of a string is revolved in a horizontal circle, the centripetal force is the tension in the string.
4. When a car takes a turn in a circular arc on a horizontal road with constant speed, the force of static friction between the car tyres and road surfaces is the centripetal force.

Note : The tension in a string or the force of friction is electromagnetic in origin.

Question 35.
Define and explain centrifugal force.
Answer:
Definition : In the reference frame of a particle performing circular motion, centrifugal force is defined as a fictitious, radially outward force on the particle and is equal in magnitude to the particle’s mass times the centripetal acceleration of the reference frame, as measured from an inertial frame of reference.

Explanation : A uniform circular motion is an accelerated motion, with a centripetal acceleration of magnitude v²/r or ω²r. A frame of reference attached to the particle also has this acceleration and, therefore, is an accelerated or noninertial reference frame. The changing direction of the linear velocity appears in this reference frame as a tendency to move radially outward. This is explained by assuming a fictitious centrifugal, i.e., radially outward, force acting on the particle. Since the particle is stationary in its reference frame, the magnitude of the centrifugal force is mv²/r or mω²r, the same as that of the centripetal force on the particle.

Note : The word ‘centrifugal’ comes from the Latin for ‘fleeing from the centre’. The word has the same root fuge from the Latin ‘to flee’ as does refugee.

Question 36.
Give any two examples of centrifugal force.
Answer:
Examples of centrifugal force :

1. A person in a merry-go-round experiences a radially outward force.
2. Passengers of a car taking a turn on a level road experience a force radially away from the centre of the circular road.
3. A coin on a rotating turntable flies off for some high enough angular speed of the turntable.
4. As the Earth rotates about its axis, the centrifugal force on its particles is directed away from the axis. The force increases as one goes from the poles towards the equator. This leads to the bulging of the Earth at the equator.

Question 37.
Explain why centrifugal force is called a pseudo force.
Answer:
A force which arises from gravitational, electromagnetic or nuclear interaction between matter is called a real force. The centrifugal force does not arise due to any of these interactions. Therefore, it is not a real force.

The centrifugal force in the noninertial frame of reference of a particle in circular motion is the effect of the acceleration of the frame of reference. Therefore, it is called a pseudo or fictitious force.

Question 38.
Distinguish between centripetal force and centrifugal force. State any two points of distinction.
Answer:

Centripetal force	Centrifugal force
1. Centripetal force is the force required to provide centri­petal acceleration to a par­ticle to move it in a circular path.	1. The centrifugal tendency of the particle, in its acceler­ated, i.e., non-inertial, frame of reference, is explained by assuming a centrifugal force acting on it.
2. At every instant, it is directed radially towards the centre of the circular path.	2. At every instant, it is directed radially away from the centre of the circular path.
3. It is a real force arising from gravitational or electromag­netic interaction between matter.	3. It is a pseudo force since it is the effect of the acceleration of the reference frame of the revolving particle.

Question 39.
Solve the following :

Question 1.
An object of mass 0.5 kg is tied to a string and revolved in a horizontal circle of radius 1 m. If the breaking tension of the string is 50 N, what is the maximum speed the object can have?
Solution :
Data : m = 0.5 kg, r = 1 m, F = 50 N
The maximum centripetal force that can be applied is equal to the breaking tension.

This is the maximum speed the object can have.

Question 2.
A certain string 500 cm long breaks under a tension of 45 kg wt. An object of mass 100 g is attached to this string and whirled in a horizontal circle. Find the maximum number of revolutions that the object can make per second without breaking the string, [g = 9.8 m/s²]
Solution :
Data : m = 100 g = 0.1 kg, r = 500 cm = 5 m, g = 9.8 m/s², F = 45 kg wt = 45 × 9.8 N
The breaking tension is equal to the maximum centripetal force that can be applied.
∴ F = mω²r ,
But ω = 2πf, where/is the corresponding frequency of revolution.

The maximum number of revolutions per second, f = 4.726 Hz

Question 3.
A disc of radius 15 cm rotates with a speed of 33\(\frac{1}{3}\) rpm. Two coins are placed on it at 4 cm and 14 cm from its centre. If the coefficient of friction between the coins and the disc is 0.15, which of the two coins will revolve with the disc ?
Solution :
Data : r = 15 cm = 0.15 m,
f = 33\(\frac{1}{3}\) rpm = \(\frac{100}{3 \times 60}\)rev/s = \(\frac{5}{9}\) Hz, µ_s = 0.15, r₁ = 4 cm = 0.04 m, r₂ = 14 cm = 0.14 m

To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.

Limiting force of static friction, f_s = µ_s N = µ_s (mg) where m is the mass of the coin and N = mg is the normal force on the coin.
∴ mω²r ≤ µ_s(mg) or ω²r ≤ µ_sg
µ_sg = 0.15 × 9.8 = 1.47 m/s²
For the first coin, r₁ = 0.04 m.
∴ ω²r₁ = (3.491)² × 0.04 = 12.19 × 0.04 = 0.4876 m/s²
Since, ω²r₁ < µ_sg, this coin will revolve with the disc. For the second coin, r₂ = 0.14 m.
∴ ω²r₂ = (3.491)² × 0.14 = 12.19 × 0.14 = 1.707 m/s²
Since, ω²r₂ > µ_sg, this coin will not revolve with the disc.
Thus, only the coin placed at 4 cm from the centre will revolve with the disc.

Question 40.
Derive an expression for the maximum safe speed for a vehicle on a horizontal circular road without skidding off. State its significance.
Answer:
Consider a car of mass m taking a turn of radius r along a level road. If µ_s is the coefficient of static friction between the car tyres and the road surface, the limiting force of friction is f_s = µ_sN = µ_smg where N = mg is the normal reaction. The forces on the car, as seen from an inertial frame of reference are shown in below figure.

Then, the maximum safe speed v_max with which the car can take the turn without skidding off is set by maximum centripetal force = limiting force of static friction

This is the required expression.
Significance : The above expression shows that the maximum safe speed depends critically upon friction which changes with circumstances, e.g., the nature of the surfaces and presence of oil or water on the road. If the friction is not sufficient to provide the necessary centripetal force, the vehicle is likely to skid off the road.

[Note : At a circular bend on a level railway track, the centrifugal tendency of the railway carriages causes the flange of the outer wheels to brush against the outer rail and exert an outward thrust on the rail. Then, the reaction of the outer rail on the wheel flange provides the necessary centripetal force.]

Question 41.
Derive an expression for the maximum safe speed for a vehicle on a circular horizontal road without toppling/overturning/rollover.
Answer:
Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

1. the lateral limiting force of static friction \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels-acting along the axis of the wheels and towards the centre of the circular path which provides the necessary centripetal force.
2. the weight \(m \vec{g}\) acting vertically downwards at the centre of gravity (C.G.)
3. the normal reaction \(\vec{N}\) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
   

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance b between the left and right wheels called the track width.

The friction force \(\overrightarrow{f_{\mathrm{s}}}\) on the wheels produces a torque T, that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll

When the inner wheel just gets lifted above the ground, the normal reaction \(\vec{N}\) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \(\tau_{\mathrm{r}}\) which tends to restore the car back on all four wheels
\(\tau_{\mathrm{r}}\) = mg.\(\frac{b}{2}\) … (3)

The car does not topple as long as the restoring torque \(\tau_{\mathrm{r}}\) counterbalances the toppling torque \(\tau_{\mathrm{t}}\). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force \(m \vec{g}\) passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

Question 42.
A carnival event known as a “well of death” consists of a large vertical cylinder inside which usually a stunt motorcyclist rides in horizontal circles. Show that the minimum speed necessary to keep the rider from falling is given by v = \(\sqrt{r g / \mu_{s}}\), in usual notations.
Answer:
The forces exerted on the rider are

1. the normal force \(\vec{N}\) exerted by the wall, directed radially inward, is the centripetal force,
2. the upward frictional force \(\overrightarrow{f_{\mathrm{s}}}\) exerted by the wall, since the motorcycle has a tendency to slide down,
3. the downward gravitational force \(m \vec{g}\).

which is the required expression.

Question 43.
A road at a bend should be banked for an optimum or most safe speed v₀. Derive an expression for the required angle of banking.
OR
Obtain an expression for the optimum or most safe speed with which a vehicle can be driven along a curved banked road. Hence show that the angle of banking is independent of the mass of a vehicle.
Answer:
Consider a car taking a left turn along a road of radius r banked at an angle θ for a designed optimum or most safe speed v₀. Let m be the mass of the car. In general, the forces acting on the car are
(a) its weight \(m \vec{g}\), acting vertically down
(b) the normal reaction of the road \(\vec{N}\), perpendicular to the road surface
(c) the frictional force \(\overrightarrow{f_{s}}\) along the inclined surface of the road.

At the optimum speed, frictional force is not relied upon to contribute to the necessary lateral centripetal force. Thus, ignoring \(\overrightarrow{f_{\mathrm{s}}}\), resolve \(\vec{N}\) into two perpendicular components : N cos θ vertically up and N sin θ horizontally towards the centre of the circular path. Since there is no acceleration in the vertical direction, N cos θ balances mg and N sin θ provides the necessary centripetal force.

Equation (3) gives the expression for the required angle of banking. From EQ. (3), we can see that θ depends upon v₀, r and g. The angle of banking is independent of the mass of a vehicle negotiating the curve. Also, for a given r and θ, the recommended optimum speed is
v₀ = \(\sqrt{r g \tan \theta}\) … (4)

Question 44.
State any two factors on which the most safe speed of a car in motion along a banked road depends.
Answer:
The angle of banking of the road and the radius of the curved path.

Question 45.
A curved horizontal road must be banked at an angle θ’ for an optimum speed v. What will happen to a vehicle moving with a speed v along this road if the road is banked at an angle θ such that
(i) θ < θ’
(ii) θ > θ’?
Answer:
(i) For θ < θ’, the horizontal component of the normal reaction would be less than the optimum value and will not be able to provide the necessary centripetal force. Then, the vehicle will tend to skid outward, up the inclined road surface.

(ii) For θ > θ’, the horizontal component of the normal reaction would be more than the necessary centripetal force. Then, the vehicle will tend to skid down the banked road.

Question 46.
A banked circular road is designed for traffic moving at an optimum or most safe speed v₀. Obtain an expression for
(a) the minimum safe speed
(b) the maximum safe speed with which a vehicle can negotiate the curve without skidding.
Answer:
Consider a car taking a left turn along a road of radius r banked at an angle θ for a designed optimum speed v. Let m be the mass of the car. In general, the forces acting on the car are
(a) its weight \(m \vec{g}\), acting vertically down
(b) the normal reaction of the road \(\vec{N}\), perpendicular to the road surface
(c) the frictional force \(\overrightarrow{f_{\mathrm{s}}}\) along the inclined surface of the road.
If µ_s is the coefficient of static friction between the tyres and road, f_s = µ_sN.

(a) For minimum safe speed : If the car is driven at a speed less than the optimum speed v₀, it may tend to slide down the inclined surface of the road so that \(\overrightarrow{f_{\mathrm{s}}}\) is up the incline.

Resolve \(\vec{N}\) and \(\overrightarrow{f_{\mathrm{s}}}\) into two perpendicular components : Ncos θ and f_s sin θ vertically up; N sin θ horizontally towards centre of the circular path. So long as the car takes the turn without skidding off, the horizontal components N sin θ and f_s cos θ together provide the necessary centripetal force, and N cos θ balances the sum mg + f_s sin θ. If v_max is the maximum safe speed without skidding,

(b) For maximum safe speed : If the car is driven fast enough, at a speed greater than the optimum speed v, it may skid off up the incline so that \(\overrightarrow{f_{\mathrm{s}}}\) is down the incline.

Resolve \(\vec{N}\) and \(\overrightarrow{f_{\mathrm{s}}}\) into two perpendicular components : N cos θ vertically up and f_s sin θ vertically down; N sin θ and f_s cos θ horizontally towards the centre of the circular path. So long as the car takes the turn without skidding off, the horizontal components N sin θ and f cos θ together provide the necessary centripetal force, and N cos θ balances the sum mg + f sin θ. If v is the maximum safe
speed without skidding,

Ignoring few special cases, the maximum value of µ_s = 1. Thus, for θ ≥ 45°, v_max = ∞, i.e., on a heavily banked road a car is unlikely to skid up the incline and the minimun limit is more important.

Question 47.
Solve the following :

Question 1.
Find the maximum speed with which a car can be safely driven along a curve of radius 100 m, if the coefficient of friction between its tyres and the road is 0.2 [g = 9.8 m/s²].
Solution :
Data : r = 100 m, µ_s = 0.2, g = 9.8 m/s²
The maximum speed, v = \(\sqrt{r \mu_{s} g}\)
= \(\sqrt{100 \times 0.2 \times 9.8}\) = 14 m/s

Question 2.
A flat curve on a highway has a radius of curvature 400 m. A car goes around the curve at a speed of 32 m/s. What is the minimum value of the coefficient of friction that will prevent the car from sliding?
Solution:
Data : r = 400 m, v = 32 m/s, g = 10 m/s²

Question 3.
A car can be driven on a flat circular road of radius r at a maximum speed v without skidding. The same car is now driven on another flat circular road of radius 2r on which the coefficient of friction between its tyres and the road is the same as on the first road. What is the maximum speed of the car on the second road such that it does not skid?
Solution:
Data: v₁ = v, r₁ = r, r₂ = 2r
On a flat circular road, the maximum safe speed is

Question 4.
On a dry day, the maximum safe speed at which a car can be driven on a curved horizontal road without skidding is 7 m/s. When the road is wet, the frictional force between the tyres and road reduces by 25%. How fast can the car safely take the turn on the wet road ?
Solution:
Let subscripts 1 and 2 denote the values of a quantity under dry and wet conditions, respectively.
Data : v₁ = 7 m/s, f₂ = f₁, – 0.25f₁ = 0.75f₁

On a dry horizontal curved road, the frictional force between the tyres and road is f₁ = µ₁mg, where m is the mass of the car and g is the gravitational acceleration.

The maximum safe speed for taking a turn of radius r on a dry horizontal curved road is

Question 5.
A coin kept at a distance of 5 cm from the centre of a turntable of radius 1.5 m just begins to slip when the turntable rotates at a speed of 90 rpm. Calculate the coefficient of static friction between the coin and the turntable. [g = π² m/s²]
Solution:
Data: r = 5 cm = 0.05 m, f = 90 rpm = \(\frac{90}{60}\) rps = 1.5 rps, g = π² m/s²

The centripetal force for the circular motion of the coin is provided by the friction between the coin and the turntable. The coin is just about to slip off the turntable when the limiting force of friction is equal to the centripetal force.

Question 6.
A thin cylindrical shell of inner radius 1.5 m rotates horizontally, about a vertical axis, at an angular speed ω. A wooden block rests against the inner surface and rotates with it. If the coefficient of static friction between block and surface is 0.3, how fast must the shell be rotating if the block is not to slip and fall ?

Solution :
Data : r = 1.5 m, µ_s = 0.3
The normal force \(\vec{N}\) of the shell on the block is the centripetal force which holds the block in place. \(\vec{N}\) determines the friction on the block, which in turn keeps it from sliding downward. If the block is not to slip, the friction force \(\overrightarrow{f_{\mathrm{s}}}\) must balance the weight \(m \vec{g}\) of the block.
∴ N = mω²r and f_s = μ_sN = mg
∴ μ_s(ω²r) = mg

This gives the required angular speed.

Question 7.
A motorcyclist rounds a curve of radius 25 m at 36 km/h. The combined mass of the motorcycle and the man is 150 kg.

1. What is the centripetal force exerted on the motorcyclist ?
2. What is the upward force exerted on the motorcyclist?

Solution :
Data : r = 25m, v = 36 km/h = 36 × \(\frac{5}{18}\)m/s = 10 m/s, m = 150 kg, g = 10 m/s²

1. Centripetal force, F = \(\frac{m v^{2}}{r}\) = \(\frac{150 \times(10)^{2}}{25}\) = 600 N
2. Upward force = normal reaction of the road surface = mg = 150 × 10 = 1500 N

Question 8.
A motorcyclist is describing a circle of radius 25 m at a speed of 5 m/s. Find his inclination with the vertical. What is the value of the coefficient of friction between the tyres and ground ?
Solution :
Data : v = 5 m/s, r = 25 m, g = 10 m/s²

Question 9.
A motor van weighing 4400 kg (i.e., a motor van of mass 4400 kg) rounds a level curve of radius 200 m on an unbanked road at 60 km/h. What should be the minimum value of the coefficient of friction to prevent skidding ? At what angle should the road be banked for this velocity?
Solution :
Data : m = 4400 kg, r = 200 m,
v = 60 km/h = 60 × \(\frac{5}{18}\)m/s = \(\frac{50}{3}\) m/s, g = 10 m/s²

[Note : In part (ii), v is to be taken as the optimum speed.]

Question 10.
An amusement park ride (known variously as the Rotor, the Turkish Twist and the Gravitron) consists of a large vertical cylinder that is spun about it axis fast enough such that the riders remain pinned against its inner wall. The floor drops away once the cylinder has attained its full rotational speed. The radius of the cylinder is R and the coefficient of static friction between a rider and the wall is μ_s.
(i) Show that the minimum angular speed necessary to keep a rider from falling is given by ω = \(\sqrt{g / \mu_{s} R} \text {. }\).
(ii) Obtain a numerical value for the frequency of rotation of the cylinder in rotations per minute if R =4 m and
µ_s = 0.4.
Solution:
Data: R = 4 m, µ_s = 0.4, g = 10 m/s²
The forces exerted on the rider, when the floor
drops away, are

1. the normal force \(\vec{N}\) exerted by the wall, directed radially inward, is the centripetal force
2. the upward frictional force \(\overrightarrow{f_{\mathrm{s}}}\) exerted by the wall
3. the downward gravitational force mg .

∴ N = mω²R and f_s = µ_sN = µ_s (mω²R) where m is the mass of the rider and ω is the angular speed of the Rotor cylinder. For the rider not to fall, \(\overrightarrow{f_{\mathrm{s}}}\) must balances \(m \vec{g}\).

This is the minimum angular speed necessary. Since ω = 2πf, the corresponding frequency of rotation of the cylinder is

Question 11.
The two rails of a broad-gauge railway track are 1.68 m apart. At a circular curve of radius 1.6 km, the outer rail is raised relative to the inner rail by 8.4 cm. Find the angle of banking of the track and the optimum speed of a train rounding the curve.
Solution :
Data : l = 1.68 m = 168 cm, r = 1.6 km = 1600 m, h = 8.4 cm, g = 10 m/s²

Question 12.
A metre gauge train is moving at 72 kmph along a curved railway track of radius of curvature 500 m. Find the elevation of the outer rail above the inner rail so that there is no side thrust on the outer rail.
Solution :
Data : r = 500 m, v = 72 kmph = 72 × \(\frac{5}{18}\) m/s = 20 m/s, g = 10 m/s², l = 1 m
tan θ = \(\frac{v^{2}}{r g}\)
= \(\frac{(20)^{2}}{500 \times 10}\) = 0.08
The required angle of banking,
θ = tan^-1 (0.08) = 4°4′
The elevation of the outer rail relative to the inner rail,
h = l sin θ
= (1)(sin 4°4′) = 0.0709 m = 7.09 cm

Question 13.
A circular race course track has a radius of 500 m and is banked at 10°. The coefficient of static friction between the tyres of a vehicle and the road surface is 0.25. Compute
(i) the maximum speed to avoid slipping
(ii) the optimum speed to avoid wear and tear of the tyres.
Solution :
Data : r = 500 m, θ = 10°, µ_s = 0.25, g = 9.8 m/s², tan 10° = 0.1763

(i) On the banked track, the maximum speed of the vehicle without slipping (skidding) is

(ii) The optimum speed of the vehicle on the track is

Question 48.
Define a conical pendulum.
Answer:
A conical pendulum is a small bob suspended from a string and set in UCM in a horizontal plane with the centre of its circular path below the point of suspension such that the string makes a constant angle θ with the vertical.
OR
A conical pendulum is a simple pendulum whose bob revolves in a horizontal circle with constant speed such that the string describes the surface of an imaginary right circular cone.

Question 49.
Derive an expression for the angular speed of the bob of a conical pendulum.
OR
Derive an expression for the frequency of revolution of the bob of a corical pendulum.
Answer:
Consider a conical pendulum of string length L with its bob of mass m performing UCM along a circular path of radius r.
At every instant of its motion, the bob is acted upon by its weight \(m \vec{g}\) and the tension \(\vec{F}\) in the string. If the constant angular speed of the bob is ω, the necessary horizontal centripetal force is F_c = mω²r

F_c is the resultant of the tension in the string and the weight. Resolve \(\vec{F}\) into components F cos θ vertically opposite to the weight of the bob and F sin θ horizontal. F cos θ balances the weight. F sin θ is the necessary centripetal force.
∴ F sin θ = mω²r … (1)
and F cos θ = mg … (2)

is the required expression for ω.
[Note : From Eq. (4), cos θ = g/ω²L. Therefore, as ω increases, cos θ decreases and θ increases.
If n is the frequency of revolution of the bob,

is the required expression for the frequency.

Question 50.
What will happen to the angular speed of a conical pendulum if its length is increased from 0.5 m to 2 m, keeping other conditions the same?
Answer:
The angular speed of the conical pendulum will become half the original angular speed.

Question 51.
Write an expression for the time period of a conical pendulum. State how the period depends on the various factors.
Answer:
If T is the time period of a conical pendulum of string length L which makes a constant angle θ with the vertical,
T = 2π\(\sqrt{\frac{L \cos \theta}{g}}\)
is the required expression
(Note: L cos θ = OC = h, where h is the axial height of the cone.
∴ T = 2π\(\sqrt{\frac{h}{g}}\)

where g is the acceleration due to gravity at the place.

From the above expression, we can see that

1. T ∝ \(\sqrt{L}\)
2. T ∝ \(\sqrt{\cos \theta}\) if θ increases, cos θ and T decrease
3. T ∝ \(\frac{1}{\sqrt{g}}/latex]
4. The period is independent of the mass of the bob.

Question 52.
Solve the following :

Question 1.
A stone of mass 2 kg is whirled in a horizontal circle attached at the end of a 1.5 m long string. If the string makes an angle of 30° with the vertical, compute its period.
Solution :
Data : L = 1.5 m, θ = 30°, g = 10 m/s²
The period of the conical pendulum,

Question 2.
A string of length 0.5 m carries a bob of mass 0. 1 kg at its end. If this is to be used as a conical pendulum of period 0.4πs, calculate the angle of inclination of the string with the vertical and the tension in the string.
Solution :
Data : L = 0.5m, m = 0.1 kg, T = 0.4πs, g = 10 m/s²

Question 3.
In a conical pendulum, a string of length 120 cm is fixed at a rigid support and carries a bob of mass 150 g at its free end. If the bob is revolved in a horizontal circle of radius 0.2m around a vertical axis, calculate the tension in the string. [g = 9.8 m/s²]
Solution:
Data : L = 120 cm = 1.2 m, m = 150 g = 0.15 kg,
r = 0.2 m, g = 9.8 m/s²

Question 4.
A stone of mass 1 kg, attached at the end of a 1 m long string, is whirled in a horizontal circle. If the string makes an angle of 30° with the vertical, calculate the centripetal force acting on the stone.
Solution :
Data : m = 1 kg, L = 1 m, θ = 30°, g = 10 m/s²

Question 53.
What is vertical circular motion? Comment on its two types.
Answer:
A body revolving in a vertical circle in the gravitational field of the Earth is said to perform vertical circular motion.

A vertical circular motion controlled only by gravity is a nonuniform circular motion because the linear speed of the body does not remain constant although the motion can be periodic.

In a controlled vertical circular motion, such as that a body attached to a rod, the linear speed of the body can be constant (including zero) so that such a motion can be uniform and periodic.

Question 54.
A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the linear speed of the body at the

1. lowest position
2. highest position ?

Answer:

1. [latex]\sqrt{5 r g}\)
2. \(\sqrt{r g}\) in the usual notation.

Question 55.
A body, tied to a string, performs circular motion in a vertical plane such that the tension in the string is zero at the highest point. What is the angular speed of the body at the

1. highest position
2. lowest position ?

Answer:

1. \(\sqrt{g / r}\)
2. \(\sqrt{5 g / r}\) in the usual notation.

Question 56.
In a vertical circular motion controlled by gravity, derive an expression for the speed at an arbitrary position. Hence, show that the speed decreases while going up and increases while coming down.
OR
In a nonuniform vertical circular motion, derive expressions for the speed and tension/normal force at an arbitrary position.
OR
Show that a vertical circular motion controlled by gravity is a non uniform circular motion.
Answer:
Consider a small body of mass m tied to a string and revolved in a vertical circle of radius r. At every instant of its motion, the body is acted upon by its weight \(m \vec{g}\) and the tension \(\vec{T}\) in the string. At any instant, when the body is at the position P, let the string make an angle θ with the vertical, \(m \vec{g}\) is resolved into components, mg cos θ (radial) and mg sin θ (tangential).

At point P shown, the net force on the body towards the centre, T-mg cos θ, is the necessary centripetal force on the body. If v is its speed at P,


Let v₂ be the speed of the body at the lowest point B, which is the reference level for zero potential energy. Then, the body has only kinetic energy

As the body goes from B to P, it rises through a height h = r – r cos θ = r(1 – cos θ).
Total energy at P = KE + PE

Assuming that the total energy of the body is conserved, total energy at any point = total energy at the bottom.
Then, from Eqs. (2) and (3),

From the above expression, it can be seen that the linear speed v changes with θ. Thus, as θ increases, (while going up) cos θ decreases, 1 – cos θ increases, and v decreases. While coming down, θ decreases and v increases. Hence, a vertical circular motion controlled by gravity is a nonuniform circular motion.

Substituting for v² from Eq. (4) in Eq. (1),

Equation (6) is the expression for the tension in the string at any instant in terms of the speed at the lowest point.

Question 57.
A body at the end of a rod is revolved in a non-uniform vertical circular motion. Show that
(i) it must have a minimum speed 2\(\sqrt{g r}\) at the bottom
(ii) the difference in tensions in the rod at the highest and lowest positions is 6 mg.
Answer:
Consider a body of mass m attached to a rod and revolved in a vertical circle of radius r at a place where the acceleration due to gravity is g. We shall assume that the rod is not rigid so that the tension in the rod changes. As the rod is rotated in a nonuniform circular motion, the tension in the rod changes from a minimum value T₁ when the body is at the highest point to a maximum value T₂ when the body is at the bottom of the circle. At every instant, the body is acted upon by two forces, namely/its weight \(m \vec{g}\) and the tension \(\vec{T}\) in the string.

Question 58.
You may have seen in a circus a motorcyclist driving in vertical loops inside a hollow globe (sphere of death). Explain clearly why the motor-cyclist does not fall down when at the highest point of the chamber.
Answer:
A motorcyclist driving in vertical loops inside a hollow globe performs vertical circular motion. Suppose the mass of the motorcycle and motorcyclist is m and the radius of the chamber is r. At every instant of the motion, the motorcyclist is acted upon by the weight \(m \vec{g}\) and the normal reaction \(\vec{N}\).

At the highest point, let v₁ be the speed and \(\vec{N}_{1}\) the normal reaction. Here, both \(\vec{N}_{1}\) and \(m \vec{g}\) are parallel, vertically downward. Hence, the net force on the motorcyclist towards the centre O is N₁ + mg. If this force is able to provide the necessary centripetal force at the highest point, the motorcycle does not lose contact with the globe and fall down.

The minimum value of this force is found from the limiting case when N, just becomes zero and the weight alone provides the necessary centripetal force :
\(\frac{m v_{1}^{2}}{r}\) = mg
This requires that the motorcycle has a minimum speed at the highest point given by \(v_{1}^{2}\) = gr or v₁ = \(\sqrt{g r}\)

[Note : The ‘globe of death’ is a circus stunt in which stunt drivers ride motorcycles inside a mesh globe. Starting from small horizontal circles, they eventually perform revolutions along vertical circles. The linear speed is more for larger circles but angular speed is more for smaller circles as in conical pendulum.]

Question 59.
A car crosses over a bridge which is in the form of a convex arc with a uniform speed,
(i) State the expression for the normal reaction on the car.
OR
How does the normal reaction on the car vary with speed?
(ii) Hence show that the maximum speed with which the car can cross the bridge without losing contact with the road is equal to \(\sqrt{r g}\).
Answer:
Suppose a car of mass m, travelling with a uniform speed v, crosses over a bridge which is in the form of a convex arc of radius r.
(i) The forces acting on it at the highest point are as shown in below figure. Their resultant mg-N provides the centripetal force.

is the required expression. It shows that as v increases, N decreases.

(ii) Equation (1) shows that for g – \(\frac{v^{2}}{r}\) = 0, i.e., for centripetal acceleration equalling the gravitational acceleration, N = 0. That is, for \(\frac{v^{2}}{r}\) = g or v = \(\sqrt{r g}\), the
car just loses contact with the road. Therefore, this is the maximum speed with which a car can cross the bridge, irrespective of its mass.

[Data : Take g = 10 m/s² unless specified otherwise]

Question 60.
Solve the following :

Question 1.
An object of mass 1 kg tied to one end of a string of length 9 m is whirled in a vertical circle. What is the minimum speed required at the lowest position to complete the circle ? [g = 9.8 m/s²]
Solution :
Data : m = 1 kg, r = 9 m, g = 9.8 m/s²
The minimum speed of the object at the lowest position is

Question 2.
A stone of mass 5 kg, tied at one end of a rope of length 0.8 m, is whirled in a vertical circle. Find the minimum velocity at the highest point and at the midway point, [g = 9.8 m/s²]
Solution:
Data : m = 5 kg, r = l = 0.8 m, g = 9.8 m/s²

1. The minimum velocity of the stone at the highest point in its path,
   v = \(\sqrt{r g}\) = \(\sqrt{0.8 \times 9.8}\) = 2.8 m/s
2. The minimum velocity of the stone at the midway point in its path,
   v = \(\sqrt{3 r g}\) = \(\sqrt{3 \times 0.8 \times 9.8}\) = 4.85 m/s

Question 3.
A small body of mass 0.3 kg oscillates in a vertical plane with the help of a string 0.5 m long with a constant speed of 2 m/s. It makes an angle of 60° with the vertical. Calculate the tension in the string.
Solution :
Data : m = 0.3 kg, r = 0.5 m, v = 2 m/s, θ = 60°, g = 10 m/s²

Question 4.
A bucket of water is whirled in a vertical circle at an arm’s length. Find the minimum speed at the top so that no water spills out. Also find the corresponding angular speed. [Assume r = 0.75 m]
Solution :
Data : r = 0.75 m, g = 10 m/s²
At the highest point the minimum speed required is v = \(\sqrt{r g}\) = \(\sqrt{0.75 \times 10}\) = 2.738 m/s
The corresponding angular speed is 2.738
ω = \(\frac{v}{r}\) = \(\frac{2.738}{0.75}\) = 3.651 rad/s

Question 5.
A pendulum, with a bob of mass m and string length l, is held in the horizontal position and then released into a vertical circle. Show that at the lowest position the velocity of the bob is \(\sqrt{2 g l}\) and the tension in the string is 3 mg.
Solution :
Taking the reference level for zero potential energy to be the bottom of the vertical circle, the initial potential energy of the bob at the horizontal position = mgh = mgl.

Hence, at the bottom where the speed of the bob is v, it has only kinetic energy = \(\frac{1}{2}\)mv².

This gives the required velocity at the lowest position.
Also, at the bottom, the tension (T) and the centripetal acceleration are upward while the force of gravity is downward.

Equations (1) and (2) give the required expressions for the velocity and tension at the lowest position.

Question 6.
A stone of mass 100 g attached to a string of length 50 cm is whirled in a vertical circle by giving it a velocity of 7 m/s at the lowest point. Find the velocity at the highest point.
Solution :
Data : m = 0.1 kg, r = l = 0.5 m, v₂ = 7 m/s, g = 10 m/s²
The total energy at the bottom, E_bot
= KE + PE = \(\frac{1}{2} m v_{2}^{2}\) + 0 = \(\frac{1}{2}\)(0.1) (7)² = 2.45 J
The total energy at the top, E_top = KE + PE = \(\frac{1}{2} m v_{1}^{2}\) + mg (2r)
= \(\frac{1}{2}\)(0.1)\(v_{1}^{2}\) + (0.1) (10) (2 × 0.5)
= 0.05\(v_{1}^{2}\) + 1
By the principle of conservation of energy,

Question 7.
A pilot of mass 50 kg in a jet aircraft executes a “loop-the-loop” manoeuvre at a constant speed of 250 m/s. If the radius of the vertical circle is 5 km, compute the force exerted by the seat on the pilot at
(i) the top of the loop
(ii) the bottom of the loop.
Solution :
Data: m = 50 kg, v = 250m/s, r = 5 km = 5 × 10³ m, g = 10 m/s²

(i) At the top of the loop : The forces on the pilot are the gravitational force \(m \vec{g}\) and the normal force \(\vec{N}_{1}\), exerted by the seat, both acting downward. So the net force downward that causes the centripetal acceleration has a magnitude

(ii) At the bottom of the loop : The forces on the pilot are the downward gravitational force \(m \vec{g}\) and the upward normal force \(\vec{N}_{2}\) exerted by the seat. So the net upward force that causes the centripetal acceleration has a magnitude N₂ – mg.

The forces exerted by the seat on the pilot at the top and bottom of the loop are 125 N and 1125 N, respectively.

Question 8.
A ball released from a height h along an incline, slides along a circular track of radius R (at the end of the incline) without falling vertically downwards. Show that h_min = \(\frac{5}{2}\) R.
Solution:
To just loop-the-loop, the ball must have a speed v₂ = \(\sqrt{5 R g}\) at the bottom of the circular track.

If h_min is the minimum height above the bottom of the circular track from which the ball must be released, by the principle of conservation of energy, we have,
mgh_min = \(\frac{1}{2} m v_{2}^{2}\) = \(\frac{1}{2} m(5 R g)\)

Note : 1f the ball rolls all along the track without slipping, its total energy at the top of the circular track should take into account the rotational kinetic energy of the ball.

Question 9.
A block of mass 1 kg is released from P on a frictionless track which ends with a vertical quarter circular turn.

What are the magnitudes of the radial acceleration and total acceleration of the block when it arrives at Q ?
Solution :
Data : m = 1 kg, h = 6 m, r = 2 m, g = 10 m/s²
Let v be the speed of the block at Question Then, the total energy of the block at Q is
E = KE + PE = \(\frac{1}{2} m v^{2}\) + mgr
By the principle of conservation of energy,

The radial acceleration has a magnitude 40 m/s². The total acceleration has a magnitude 41.23 m/s² and makes an angle of 14°2′ with the radial acceleration.

Question 10.
A loop-the-loop cart runs down an incline into a vertical circular track of radius 3 m and then describes a complete circle. Find the minimum height above the top of the circular track from which the cart must be released.
Solution :

Data : r = 3 m
To just loop-the-loop, the cart must have a speed V₁ = \(\sqrt{r g}\) at the top of the loop.

If h is the minimum height above the top of the loop from which the cart must be released, by the principle of conservation of energy, we have, mgh = \(\frac{1}{2} m v_{1}^{2}\) = \(\frac{1}{2} m g r\)
∴ h = \(\frac{r}{2}\) = \(\frac{3}{2}\) = 1.5 m

Question 11.
A motorcyclist rides in vertical circles in a hollow sphere of radius 5 m. Find the required minimum speed and minimum angular speed, so that he does not lose contact with the sphere at the highest point. [g = 9.8 m/s²]
Solution :
Data : r = 5 m, g = 9.8 m/s²
Let v and ω be respectively the required minimum speed and angular speed at the highest point.

Question 12.
The vertical section of a road over a bridge in the direction of its length is in the form of an arc of a circle of radius 4.4 m. Find the maximum speed with which a vehicle can cross the bridge without losing contact with the road at the highest point, if the centre of gravity of the vehicle is 0.5 m from the ground.
Solution :
Data: While travelling along the bridge, the vehicle moves along a vertical circle of radius r = 4.4 + 0.5 = 4.9 m, g = 10 m/s².
If m is the mass and v is the maximum speed of the vehicle, then at the highest point,

Question 13.
A small body tied to a string is revolved in a vertical circle of radius r such that its speed at the top of the circle is \(\sqrt{2 g r}\). Find
(i) the angular position of the string when the tension in the string is numerically equal to 5 times the weight of the body.
(ii) the KE of the body at this position
(iii) the minimum and maximum KEs of the body.
[Take m = 0.1 kg, r = 1.2 m, g = 10 m/s²]
Solution :
Data : v_top = \(\sqrt{2 g r}\), T = 5 mg, m = 0.1 kg, r = 1.2 m, g = 10 m/s²

Let the angular position of the string, θ = 0° when the body is at the bottom of the circle.

We assume total energy to be conserved and take the reference level for zero potential energy to be the bottom of the circle.

Total energy at the top, E

At P, the vertical displacement of the body from the bottom is r(1 – cos θ). Its total energy there is also E.

Question 14.
An object of mass 0.5 kg attached to a rod of length 0.5 m is whirled in a vertical circle at a constant angular speed. If the maximum tension in the rod is 5 kg wt, calculate the linear speed of the object and the maximum number of revolutions it can complete in a minute.
Solution :
Data : m = 0.5 kg, r = l = 0.5 m, g = 10 m/s²,
T₂ = 5 kg wt = 5 × 10 N

As the rod is rotated in a vertical circle at a constant angular speed, the linear speed of the object at the end of the rod is constant, say v. However, the tension in the rod changes from a minimum value T₁ when the object is at the highest point to a maximum value T₂ when the object is at the bottom of the circle.

At the bottom of the circle, the tension and acceleration are upward while the force of gravity is downward.

∴ The maximum number of revolutions the object can complete in a minute is 127.98.

Question 15.
A small body of mass m = 0.1 kg at the end of a cord 1 m long swings in a vertical circle. Its speed is 2 mIs when the cord makes an angle θ = 30° with the vertical. Find the tension in the cord.
Solution:
Data: m = 0.1 kg, r = 1 m, y = 2 m/s, θ = 30°,
g = 9.8 m/s²
When the cord makes an angle θ with the vertical, the centripetal force on the body is

Question 16.
A bucket of water is tied to one end of a rope 8 m long and rotated about the other end in a vertical circle. Find the number of revolutions per minute such that water does not spill.
Solution:
[Important note : The circular motion of the bucket in a vertical plane under gravity is not a uniform circular motion. Assuming the critical case of the motion such that the bucket has the minimum speed at the highest point required for the water to stay put in the bucket, we can find the minimum frequency of revolution. ]
Data :r = 8m, g = 9.8 m/s², π = 3.142
Assuming the bucket has a minimum speed v = \(\sqrt{r g}\) at the highest point, the corresponding angular speed is

Question 61.
Derive an expression for the kinetic energy of a body rotating with constant angular velocity.
Answer:
Consider a rigid body rotating with a constant angular velocity \(\vec{\omega}\) about an axis passing through the point O and perpendicular to the plane of the figure. Suppose that the body is made up of N particles of masses m₁, m₂, …, m_N situated at perpendicular distances r₁, r₂, , r_N, respectively, from the axis of rotation as shown in below figure.

As the body rotates, all the particles perform uniform circular motion with the same angular velocity \(\vec{\omega}\). However, they have different linear speeds depending upon their distances from the axis of rotation.

The linear speed of the particle with mass ml is v₁ = r₁ω. Therefore, its kinetic energy is

where I = \(\sum_{i=1}^{N} m_{i} r_{i}^{2}\) is the moment of inertia of the body about the axis of rotation.
Equation (2) gives the required expression.

Question 62.
Define moment of inertia. State the factors which it depends on. Obtain its dimensions and state its SI unit.
OR
Define moment of inertia. State its dimensions and SI units.
Answer:
(1) Moment of inertia : The moment of inertia of a body about a given axis of rotation is defined as the sum of the products of the masses of the particles of the body and the squares of their respective distances from the axis of rotation.

If the body is made up of N discrete particles of masses m₁, m₂, …,m_N situated at respective distances r₁, r₂, …, r_N from the axis of rotation, the moment of inertia of the body is

For a rigid body, having a continuous and uniform distribution of mass, the moment of inertia is
I = \(\int r^{2} d m\) …(2)
where dm is the mass of an infinitesimal element, situated at distance r from the axis of rotation.

(2) The moment of inertia of a rigid body depends on

1. the mass and shape of the body
2. orientation and position of the rotation axis
3. distribution of the mass about the rotation axis.

(3) Dimensions :
[Moment of inertia] = [mass] [distance]²
= [M] [L²] = [M¹L²T⁰]

(4) SI unit : The kilogram-metre² (kg.m²).

Question 63.
Explain the physical significance of moment of inertia.
Answer:
(1) The physical significance of moment of inertia can be understood by comparing the formulae in the following table.

Linear motion	Rotational motion
1. Momentum = mass × velocity	1. Angular momentum = moment of inertia × angular velocity
2. Force = mass × acceleration	2. Torque = moment of inertia × angular acceleration
3. Kinetic energy = \(\frac{1}{2} M v^{2}\)	3. Kinetic energy = \(\frac{1}{2} I \omega^{2}\)

(2) Force produces acceleration, while torque produces angular acceleration. Force and torque are analogous quantities. Also, momentum and angular momentum are analogous quantities.
(3) By comparing the above formulae, we find that moment of inertia plays the same role in rotational motion as that played by mass in linear motion. The moment of inertia of a body is its rotational inertia, that which opposes any tendency to change its angular velocity. In the absence of a net torque, the body continues to rotate with a uniform angular velocity.

Question 64.
Three point masses M₁, M₂ and M₃ are located at the vertices of an equilateral triangle of side a. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through M₁ ?
Answer:

The moment of inertia of the system about the altitude passing through M₁ is

Question 65.
Find the moment of inertia of a hydrogen molecule about its centre of mass if the mass of each hydrogen atom is m and the distance between them is R.
Answer:
We assume the rotation axis to be a transverse axis through the centre of mass of the linear molecule H₂. Then, each of the hydrogen atom is a distance \(\frac{1}{2}\)R from the CM. Therefore, the MI of the molecule about this axis,

Notes :

1. For a H₂ molecule, mH = 1.674 × 10^-27 kg and bond length = 7.774 × 10^-11 m, so that I = 5.065 × 10^-48 kg.m².
2. As atoms are treated as particles, we do not consider rotation about the line passing through the atoms.

Question 66.
Solve the following :

Question 1.
Four particles of masses 0.2 kg, 0.3 kg, 0.4 kg and 0.5 kg respectively are kept at comers A, B, C and D of a square ABCD of side 1 m. Find the moment of inertia of the system about an axis passing through point A and perpendicular to the plane of the square.
Solution :
Data : m₁ = 0.2 kg, m₂ = 0.3 kg, m₃ = 0.4 kg, m₄ = 0.5 kg

The axis of rotation passes through point A and is perpendicular to the plane of the square. Hence the distance (r₁) of mass ml from the axis is r₁ =0, that of mass m₂ is r₂ = AB = 1 m, that of mass m₃ is r₃ = \(\sqrt{2}\)AC m and that of mass m₄ is r₄ = AD = 1 m.
The moment of inertia,

Question 2.
The moment of inertia of the Earth about its axis of rotation is 9.83 × 10 kg.m² and its angular speed is 7.27 × 10^-5 rad/s. Calculate its
(i) kinetic energy of rotation
(ii) radius of gyration. [ Mass of the Earth = 6 × 10²⁴ kg]
Solution :
Data : I = 9.83 × 10³⁷ kg.m², ω = 7.27 × 10^-5 rad/s, M = 6 × 10²⁴ kg

(i) The kinetic energy of rotation of the Earth,

Question 67.
State an expression for the moment of inertia of a thin ring about its transverse symmetry axis.
Answer:
A thin uniform ring (or hoop) has all its mass uniformly distributed along the circumference of a circle. It is taken to be a two-dimensional body. It is also assumed that the radial thickness of the ring is so small as to be completely negligible in comparison to its radius.

Consider a thin ring (or hoop) of radius R and mass M. The axis of rotation through its centre C is perpendicular to its plane. C is also its centre of mass (CM).

The MI of the ring about the transverse symmetry axis is
I_CM = MR²

Question 68.
Derive an expression for the moment of inertia of a thin uniform disc about its transverse symmetry axis.
Answer:
A thin uniform disc has all its mass homogeneously distributed over its circular surface area. It is taken to be a two-dimensional body, i.e., its axial thickness is small as to be completely negligible in comparison to its radius. Consider a thin disc of radius R and mass M. Its mass per unit area is

The axis of rotation is the transverse symmetry axis, through its centre of mass (CM) and perpendicular to its plane. For rotation about this axis, we consider the disc to consist of a large number of thin concentric rings, having the same rotation axis as the transverse symmetry axis of the disc. One such elemental ring at a distance r from the rotation axis shown in below figure, has mass dm and radial width dr.

Since the disc is uniform, the area and mass of this elemental ring are

and its moment of inertia (MI) about the given axis is dm.r².
Therefore, the MI of the disc is

This gives the required expression.

Question 69.
Is radius of gyration of a rigid body a constant quantity?
Answer:
Radius of gyration of a rigid body depends on the distribution of mass of the body about a rotation axis and, therefore, changes with the choice of the rotation axis. Hence, unlike the mass of the body which is constant, radius of gyration and moment of inertia of the body are not constant.

Question 70.
State an expression for the radius of gyration of
(i) a thin ring
(ii) a thin disc, about respective transverse symmetry axis.
OR
Show that for rotation about respective transverse symmetry axis, the radius of gyration of a thin disc is less than that of a thin ring.
Answer:
(i) The MI of the ring about the transverse symmetry axis is
I_CM = MR² … (1)
Radius of gyration : The radius of gyration of the ring about the transverse symmetry axis is
K = \(\sqrt{I_{\mathrm{CM}} / M}\) = \(\sqrt{R^{2}}\) = R …… (2)

(ii) The MI of the disc about the transverse symmetry axis is
I_CM = \(\frac{1}{2}\)MR<² … (3)
Radius of gyration : The radius of gyration of the disc for the given rotation axis is

Question 71.
State and prove the theorem of parallel axis.
Answer:
Theorem of parallel axis : The moment of inertia of a body about an axis is equal to the sum of

1. its moment of inertia about a parallel axis through its centre of mass and
2. the product of the mass of the body and the square of the distance between the two axes.

Proof : Let I_CM be the moment of inertia (MI) of a body of mass M about an axis through its centre of mass C, and I be its MI about a parallel axis through any point O. Let h be the distance between the two axes.

Consider an infinitesimal mass element dm of the body at a point P. It is at a perpendicular distance CP from the rotation axis through C and a perpendicular distance OP from the parallel axis through O. The MI of the element about the axis through C is CP²dm. Therefore, the MI of the body about the axis through the CM is I_CM = \(\int \mathrm{CP}^{2} d m\). Similarly, the MI of the body about the parallel axis through O is I = \(\int \mathrm{OP}^{2} d m\).

∴ I = I_CM + Mh²
This proves the theorem of parallel axis.

Question 72.
State and prove the theorem of perpendicular axes about moment of inertia.
Answer:
Theorem of perpendicular axes : The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes in its plane and through the point of intersection of the perpendicular axis and the lamina.

Proof : Let Ox and Oy be two perpendicular axes in the plane of the lamina and Oz, an axis perpendicular to its plane. Consider an infinitesimal mass element dm of the lamina at the point P(Y, y). MI of the lamina about the z-axis, I_z = \(\int \mathrm{OP}^{2} d m\)

The element is at perpendicular distance y and x from the x- and y- axes respectively. Hence, the moments of inertia of the lamina about the x- and y-axes are, respectively,

This proves the theorem of perpendicular axes.

Question 73.
About which axis of rotation is the radius of gyration of a body the least ?
Answer:
The radius of gyration of a body is the least about an axis through the centre of mass (CM) of the body.

From the parallel axis theorem, we know that a given body has the smallest possible moment of inertia about an axis through its CM. The radius of gyration of a body about a given axis is directly proportional to the square root of its moment of inertia about that axis. Hence, the conclusion.
{OR I = I_CM + Mh². ∴ Mk² = \([/latexM k_{\mathrm{CM}}^{2}] + Mh².
∴ k² = [latex]k_{\mathrm{CM}}^{2}\) + h², which shows that k is minimum, equal to k_CM when h = 0.}

Question 74.
State an expression for the moment of inertia of a thin uniform rod about an axis through its centre and perpendicular to its length. Hence deduce the expression for its moment of inertia about an axis through its one end and perpendicular to its length.
OR
State an expression for the moment of inertia of a thin uniform rod about its transverse symmetry axis. Hence, deduce the expression for its moment of inertia about a parallel axis through one end. Also deduce the expressions for the corresponding radii of gyration.
Answer:
(1) MI about a transverse axis through centre : Consider a thin uniform rod AB of mass M and length L, rotating about a transverse axis through its centre C. C is also its centre of mass (CM).

(2) MI about a transverse axis through one end : Let I be its MI about a transverse axis through its end A. By the theorem of parallel axis,
I = I_CM + Mh² … (2)
In this case,

(3) Radii of gyration : The radius of gyration of the rod about its transverse symmetry axis is

The radius of gyration of the rod about the transverse axis through an end is

Question 75.
State the expression for the MI of a thin spherical shell (i.e., a thin-walled hollow sphere) about its diameter. Hence obtain the expression for its MI about a tangent.
Answer:
Consider a uniform, thin-walled hollow sphere radius R and mass M. An axis along its diameter is an axis of spherical symmetry through its centre of mass. The MI of the thin spherical shell about its diameter is
I_CM = \(\frac{2}{3}\)MR²

Let I be its MI about a tangent parallel to the diameter. Here, h = R = distance between the two axes. Then, according to the theorem of parallel axis,

Question 76.
Calculate the moment of inertia by direct integration of a thin uniform rod of mass M and length L about an axis perpendicular to the rod and passing through the ród at L/3, as shown below.
Check your answer with the parallel-axis theorem.

Answer:
Method of direct integration : Consider a thin uniform rod of mass M and length L. The axis of rotation is perpendicular to the rod and passing through the rod at L/3. We consider the origin of coordinates to be at this point and the x-axis to be along the rod,

Since the mass density is constant, the linear mass density is
λ = M/L
An element of the rod has mass dm and length dl = dx.

If the distance of each mass element from the axis is given by the variable x, the moment of inertia of an element about the axis of rotation is dI = x²dm
Since the rod extends from x= – L/3 to x = 2L/3, the MI of the rod about the axis is

Method of parallel-axis : The MI of the thin rod about a transverse axis through its CM is
I_CM = \(\frac{1}{12} M L^{2}\)
The given axis of rotation is at a distance h = \(\frac{L}{2}\) – \(\frac{L}{3}\) = \(\frac{L}{6}\) from the transverse symmetry axis.
Therefore, the MI of the rod about the given axis is

the same as arrived at by direct integration method.

Question 77.
State an expression for the moment of inertia of a thin ring about its transverse symmetry axis. Hence deduce the expression for its moment of inertia about a tangential axis perpendicular to its plane. Also deduce the expressions for the corresponding radius of gyration.
Answer:
(1) MI about the transverse symmetry axis : Consider a thin ring (or hoop) of radius R and mass M. The axis of rotation through its centre C is perpendicular to its plane. C is also its centre of mass (CM). It is assumed that the radial thickness of the ring is so small as to be completely negligible in comparison to radius R.

The MI of the ring about the transverse symmetry axis is
I_CM = MR² …(1)
Radius of gyration : The radius of gyration of the ring about the transverse symmetry axis is
k = \(\sqrt{I_{\mathrm{CM}} / M}\) = \(\sqrt{R^{2}}\) = R …(2)

(2) MI about a tangent perpendicular to its plane : Let I be its MI about a parallel axis, tangent to the ring. Here, h = R = distance between the two axes.
By the theorem of parallel axis,
I = I_CM + Mh² … (3)
= MR² + MR² = 2MR² …(4)
Radius of gyration : The radius of gyration of the ring about a transverse tangent is
k = \(\sqrt{I / M}\) = \(\sqrt{2 R^{2}}\) = \(\sqrt{2} R\) …(5)

Question 78.
Assuming the expression for the moment of inertia of a ring about its transverse symmetry axis, obtain the expression for its moment of inertia about
(1) a diameter
(2) a tangential axis in its plane. Also deduce the expressions for the corresponding radii of gyration.
Answer:
Let M be the mass of a thin ring of radius R. Let /CM be the moment of inertia (MI) of the ring about its transverse symmetry axis. Then,
I_CM = MR … (1)

(1) MI about a diameter : Let x- and y-axes be along two perpendicular diameters of the ring as shown in below figure. Let I_x, I_y and I_z be the moments of inertia of the ring about the x, y and z axes, respectively.

Both I_x and I_y represent the moment of inertia of the ring about its diameter and, by symmetry, the MI of the ring about any diameter is the same.
∴ I_x = I_y ….. (2)
Also, I_z being the MI of the ring about its transverse symmetry axis,

(2) MI about a tangent in its plane: Let I be its MI about an axis in plane of the ring, i.e., parallel to a diameter, and tangent to it. Here, h = R and
I_CM = I_x = \(\frac{1}{2}\)MR².
By the theorem of parallel axis,
= \(\frac{1}{2} M R^{2}\) + MR² = \(\frac{3}{2} M R^{2}\) … (7)

Question 79.
State an expression for the MI of a thin uniform disc about a transverse axis through its centre. Hence, derive an expression for the MI of the disc about its tangent perpendicular to the plane. Deduce the expressions for the corresponding radii of gyration.
Answer:
(1) MI about the transverse symmetry axis : Consider a thin uniform disc of radius R and mass M. The axis of rotation through its centre C is perpendicular to its plane. C is also its centre of mass (CM).

Radius of gyration : The radius of gyration of the disc for the given rotation axis is

(2) MI about a tangent perpendicular to its plane : Let I be the MI of the disc about a tangent perpendicular to its plane.

According to the theorem of parallel axis,

Question 80.
Assuming the expression for the moment of inertia of a thin uniform disc about a transverse axis through its centre, obtain an expression for its moment of inertia about any diameter. Hence, write the expression for the corresponding radius of gyration.
Answer:
Consider a thin uniform disc of mass M and radius R in the xy plane, as shown in below figure. Let I_x, I_y and I_z be the moments of inertia of the disc about the x, y and z axes respectively. But, I_x = I_y, since each

represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the MI of the disc about any diameter is the same.
As I_z is the MI of the disc about the z-axis through its centre and perpendicular to its plane,
I_z = \(\frac{1}{2}\)MR² … (1)

According to the theorem of perpendicular axes,

Question 81.
Given the moment of inertia of a thin uniform disc about its diameter to be \(\frac{1}{4}\)MR², where M and R are respectively the mass and radius of the disc, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Answer:
Consider a thin uniform disc of mass M and radius R in the xy plane. Let Ix, ly and Iz be the moments of inertia of the disc about the x, y and z axes respectively.
Now, I_x = I_y
since each represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the MI of the disc about any diameter is the same.
∴ I_x = I_y = \(\frac{1}{4}\)MR² (Given)
According to the theorem of perpendicular axes,
I_z = I_x + I_y = 2(\(\frac{1}{4}\)MR²) = \(\frac{1}{2}\)MR²
Let I be the MI of the disc about a tangent normal to the disc and passing through a point on its edge (i.e., a tangent perpendicular to its plane). According to the theorem of parallel axis,
I = I_CM + Mh²
Here, I_CM = I_z = \(\frac{1}{2}\)MR² and h = R.
∴ I = \(\frac{1}{2}\)MR² + MR² = \(\frac{3}{2}\)MR²
which is the required expression.

Question 82.
Assuming the expression for the moment of inertia of a thin uniform disc about its diameter, show that the moment of inertia of the disc about a tangent in its plane is \(\frac{5}{4}\)MR². Write the expression for the corresponding radius of gyration.
Answer:
Let M be the mass and R be the radius of a thin uniform disc. Let I_CM be the moment of inertia (MI) of the disc about a diameter. Then,

Question 83.
State the expressions for the moment of inertia of a solid cylinder of uniform cross section about
(1) an axis through its centre and perpendicular to its length
(2) its own axis of symmetry.
OR
State the expressions for the MI of a solid cylinder about
(1) a transverse symmetry axis
(2) its cylindrical symmetry axis. Also deduce the expressions for the corresponding radii of gyration.
Answer:
Consider a solid cylinder of uniform density, length L, radius R and total mass M.

Notes :

1. For R « L, a solid cylinder can be approximated as a thin rod, and the expression for the MI about its transverse symmetry axis reduces to the corresponding expression for a thin rod, viz., ML²/12.
2. The MI of a solid cylinder about its cylindrical symmetry axis is the same as that of a disc about its transverse symmetry axis and having the same mass and radius.

Question 84.
Assuming the expression for the moment of inertia of a uniform solid cylinder about a transverse symmetry axis, obtain the expression for its moment of inertia about a transverse axis through its one end.
Answer:
Let M be the mass, L the length and R the radius of a uniform solid cylinder. Let I_CM be the moment of inertia (MI) of the cylinder about a transverse symmetry axis. Then,

Question 85.
State an expression for the moment of inertia of a solid sphere about its diameter. Write the expression for the corresponding radius of gyration.
Answer:
Consider a solid sphere of uniform density, radius R and mass M. An axis along its diameter is an axis of spherical symmetry through its centre of mass.

The MI of the solid sphere about its diameter is
I_CM = \(\frac{2}{5}\)MR²
The corresponding radius of gyration is

Question 86.
A uniform solid sphere of mass 15 kg has radius 0.1 m. What is its moment of inertia about a diameter?
Answer:
Moment of inertia of the sphere about a diameter
= \(\frac{2}{5}\)MR² = \(\frac{2}{5}\) × 15 × (0.1)² = 6 × 10^-2 kg.m²

Question 87.
Assuming the expression for the MI of a uniform solid sphere about its diameter, obtain the expression for its moment of inertia about a tangent.
Answer:
Let M be the mass of a uniform solid sphere of radius R. Let I_CM be its MI about any diameter.

Let I be its MI about a parallel axis, tangent to the sphere. Here, h = R = distance between the two axis.
By the theorem of parallel axis, I = I_CM + Mh²
= \(\frac{2}{5}\)MR² + MR² = \(\frac{7}{5}\)MR²

Question 88.
The moment of inertia of a uniform solid sphere about a diameter is 2 kg m². What is its moment of inertia about a tangent ?
Answer:
Moment of inertia of a solid sphere about its 2
diameter, I_CM = \(\frac{2}{5}\)MR².

Question 89.
The radius of gyration of a uniform solid sphere of radius R is \(\sqrt{\frac{2}{5}}\)R for rotation about its diameter. Show that its radius of gyration for rotation about a tangential axis of rotation is \(\sqrt{\frac{7}{5}}\)R.
Answer:
Let the mass of the uniform solid sphere of radius R be M. Let I_CM and k_d be its MI about any diameter and the corresponding radius of gyration, respectively. Then,

Let I and k_t be its MI about a parallel tangential axis and the corresponding radius of gyration, respectively. Here, h = R = distance between the two axis.
∴ I = \(M k_{\mathrm{t}}^{2}\)
By the theorem of parallel axis,
I = I_CM + Mh²

Question 90.
State the expression for the MI of a thin spherical shell (i.e., a thin-walled hollow sphere) about its diameter. Hence obtain the expression for its MI about a tangent.
Answer:
Consider a uniform, thin-walled hollow sphere radius R and mass M. An axis along its diameter is an axis of spherical symmetry through its centre of mass. The MI of the thin spherical shell about its diameter is
I_CM = \(\frac{2}{3} M R^{2}\)

Let I be its MI about a tangent parallel to the diameter. Here, h = R = distance between the two axes. Then, according to the theorem of parallel axis,

Question 91.
Find the ratio of the radius of gyration of a solid sphere about its diameter to the radius of gyration of a hollow sphere about its tangent, given that both the spheres have the same radius.
Answer:
The radius of gyration of a body about a given axis, k = \(\sqrt{I / M}\), where M and I are respectively the mass of the body and its moment of inertia (MI) about the axis.

For a solid sphere rotating about its diameter,

Question 92.
Calculate the moment of inertia by direct integration of a thin uniform rectangular plate of mass M, length l and breadth b about an axis passing through its centre and parallel to its breadth.
Answer:
Consider a thin uniform rectangular plate of mass M, length l and breadth b. The axis of rotation passes through its centre and is parallel to its breadth. We consider the origin of coordinates to be at the centre of the plate and orient the axes as shown in below figure. Since the plate is thin, we can take the mass as distributed entirely in the xy-plane. Then, the surface mass density is constant and equal to

A rectangular element of the plate, shown shaded, has mass dm, length b and breadth dy.
∴ dm = σdA = σ(b dy)

If the distance of each element from the rotation axis is given by the variable y, the moment of inertia of :
an element about the axis of rotation is
dI_x = y²dm
Since the rod extends from y = -1/2 to y = 1/2, the MI of the thin plate about the axis is

Notes:

(1) The MI of a thin rectangular plate about an axis passing through its centre and parallel to its length (i.e., about the y-axis) is I_y = \(\frac{1}{12}\)Mb². Then, by the theorem of perpendicular axes, the MI of a thin plate about its transverse symmetry axis (i.e., about the z-axis) is I_z = I_x + I_y = \(\frac{1}{12} M\left(l^{2}+b^{2}\right)\)

(2) Suppose, for a rectangular bar of sides l, b and w, we take the origin of coordinates at the centre of mass of the bar and the x, y and z axes parallel to the respective sides. Then, I_x = \(\frac{1}{12}\)M(b² + w²), I_y = \(\frac{1}{12}\)M(w² + l²) and I_z = \(\frac{1}{2} M\left(l^{2}+b^{2}\right)\)

Question 93.
State the MI of a thin rectangular plate-of mass M, length l and breadth b- about an axis passing through its centre and parallel to its length. Hence find its MI about a parallel axis along one edge.
Answer:
Consider a thin rectangular plate of mass M, length l and breadth b. The MI of the plate about an axis passing through its centre and parallel to its edge of length l is

For a parallel axis along its one edge, h = \(\frac{1}{2} b\).
Therefore, by the theorem of parallel axis, the MI about this axis is

Question 94.
State the MI of a thin rectangular plate-of mass M, length l and breadth b about its transverse axis passing through its centre. Hence find its MI about a parallel axis through the midpoint of edge of length b.
Answer:
Consider a thin rectangular plate of mass M, length l and breadth b. The MI of the plate about its transverse axis passing through its centre is
I_CM = M(l² + b²)
For a parallel axis through the midpoint of its breadth, h = \(\frac{1}{2} l\). Therefore, by the theorem of parallel axis, the MI about this axis is

Question 95.
A uniform solid right circular cone of base radius R has mass M. Prove that the moment of inertia of the cone about its central symmetry axis is \(\frac{3}{10} M R^{2}\).
Answer:
Consider a uniform solid right circular cone of mass M, base radius R and height h. The axis of rotation passes through its centre and the vertex, Its constant mass density is

We consider an elemental disc of mass dm, radius r and thickness dz. If the distance of each mass element from the axis is given by the variable z,

Question 96.
Solve the following :

Question 1.
Calculate the moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation passing through
(i) its diameter
(ii) a tangent perpendicular to its plane.
Solution :
Data : M = 500 g 0.5 kg, R = 0.5 m

(i) The moment of inertia of the ring about its

(ii) The moment of inertia of the ring about a tangent perpendicular to its plane
= 2MR² = 2 × 0.5 × (0.5)² = 0.25 kg.m²

Question 2.
A thin uniform rod 1 m long has mass 1 kg. Find its moment of inertia and radius of gyration for rotation about a transverse axis through a point midway between its centre and one end.
Solution :
Data : M = 1 kg, L = 1 m
Let I_CM and I be the moments of inertia of the rod about a transverse axis through its centre, and a parallel axis midway between its centre and one end.

Question 3.
The moment of inertia of a disc about an axis through its centre and perpendicular to its plane is 10 kg.m². Find its MI about a diameter.
Solution :
Data : I_z = 10 kg.m²
If the disc lies in the xy plane with its centre at the origin then, according to the theorem of perpendicular axes,
I_x + I_y = I_z
Since, I_x = I_y, 2I_x = I_z
∴ Its MI about a diameter,
I_x = \(\frac{I_{z}}{2}\) = \(\frac{10}{2}\) = 5 kg.m²

Question 4.
A solid cylinder of uniform density and radius 2 cm has a mass of 50 g. If its length is 12 cm, calculate its moment of inertia about an axis passing through its centre and perpendicular to its length.
Solution :
Data : M = 50 g, R = 2 cm, L = 12 cm

Question 5.
A compound object is formed of a thin rod and a disc attached at the end of the rod. The rod is 0.5 m long and has mass 2 kg. The disc has mass of 1 kg and its radius is 20 cm. Find the moment of inertia of the compound object about an axis passing through the free end of the rod and perpendicular to its length.
Solution :
Data : L = 0.5 m, R = 0.2 m, M_rod = 2 kg, M_disk = 1 kg About a transverse axis through CM,

Question 6.
The radius of gyration of a body about an axis at 6 cm from its centre of mass is 10 cm. Find its radius of gyration about a parallel axis through its centre of mass.
Solution :
Let O be a point at 6 cm from the centre of mass of the body.
Let I = MI about an axis through O,
k = radius of gyration about the axis through O,
I_CM = MI about a parallel axis through the centre of mass of the body,
k_CM = radius of gyration about a parallel axis through the centre of mass,
M = mass of the body,
h = distance between the two axes.
Data : h = 6 cm, k = 10 cm
By the theorem of parallel axis,
I = I_CM + Mh²
Also, I = Mk² and I_CM = \(M k_{\mathrm{CM}}^{2}\)

The radius of gyration about a parallel axis through its centre of mass is 8 cm.

Question 7.
The radius of gyration of a disc about its transverse symmetry axis is 2 cm. Determine its radius of gyration about a diameter.
Solution :
Data : k_CM = 2 cm
Let M and R be the mass and radius of the disc. Let I_CM and k_CM be the MI and radius of gyration of the disc about its transverse symmetry axis. Let I and k be the MI and radius of gyration of the disc about its diameter. Then

Question 8.
Calculate the MI and rotational kinetic energy of a thin uniform rod of mass 10 g and length 60 cm when it rotates about a transverse axis through its centre at 90 rpm.
Solution :
Data : M = 10 g = 10^-2 kg, L = 60 cm = 0.6 m,
f = 90 rpm = 90/60 Hz = 1.5 Hz
The MI of the rod about a transverse axis through its centre is

Angular speed, ω = 2πf = 2 × 3.142 × 1.5 = 9.426 rad/s
Rotational KE = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\left(3 \times 10^{-4}\right)(9.426)^{2}\)
= 0.01333 J

Question 9.
A thin rod of uniform cross section is made up of two sections made of wood and steel. The wooden section has length 50 cm and mass 0.6 kg. The steel section has length 30 cm and mass 3 kg. Find the moment of inertia of the rod about a transverse axis passing through the junction of the two sections.
Solution:
Data : L₁ =0.5m, M₁ =0.6 kg, L₂ = 0.3m,
The moment of inertia of a thin rod about a transverse axis through its end is \(\frac{M L^{2}}{3}\).

Question 10.
The mass and the radius of the Moon are, respectively, about \(\frac{1}{81}\) time and about \(\frac{1}{3.7}\) time those of the Earth. Given that the rotational period of the Moon is 27.3 days, compare the rotational kinetic energy of the Earth with that of the Moon.
Solution :
Data : M_M = \(\frac{1}{81}\)M_E, R_M = \(\frac{1}{3.7}\)R_E, T_M = 27.3 days, T_E = 1 day

Let I_E and I_M be the moments of inertia of the Earth and the Moon about their respective axes of rotation, and ω_E and ω_M be their respective rotational angular speeds. Assuming the Earth and the Moon to be solid spheres of uniform densities,

Question 11.
A solid sphere of radius R, rotating with an angular velocity ω about its diameter, suddenly stops rotating and 75% of its KE is converted into heat. If c is the specific heat capacity of the material in SI units, show that the temperature of 3R²CO²
the sphere rises by \(\frac{3 R^{2} \omega^{2}}{20 c}\).
Answer:
The MI of a solid sphere about its diameters, I = \(\frac{2}{5}\)MR²
where M is its mass.
The rotational KE of the sphere,

If ∆θ is the rise in temperature,

Question 97.
Define the angular momentum of a particle.
Answer:
Definition : The angular momentum of a particle is defined as the moment of the linear momentum of the particle. If a particle of mass m has linear momentum \(\vec{p}(=m \vec{v})\)), then the angular momentum of this particle with respect to a point O is a vector quantity defined as \(\vec{l}=\vec{r} \times \vec{p}=m(\vec{r} \times \vec{v})\)), where \(\vec{r}\) is the position vector of the particle with respect to O.

It is the angular analogue of linear momentum.

[Note : As the particle moves relative to O in the direction of its momentum \(\vec{p}(=m \vec{v})\), position vector \(\vec{r}\) rotates around O. However, to have angular momentum about O, the particle does not itself have to rotate around O.]

Question 98.
State the dimensions and SI unit of angular momentum.
Answer:

1. Dimensions : [Angular momentum] = [M¹L²T^-1]
2. SI unit: The kilogram.metre²/second (kg.m²/s).

Question 99.
Express the kinetic energy of a rotating body in terms of its angular momentum.
Answer:
The kinetic energy of a body of moment of inertia I and rotating with a constant angular velocity ω is
E = \(\frac{1}{2} I \omega^{2}\)
The angular momentum of the body, L = Iω.
∴E = \(\frac{1}{2}(I \omega) \omega=\frac{1}{2} L \omega\)
This is the required relation.

Question 100.
Why do grinding wheels have large mass and moderate diameter?
Answer:
A grinding wheel, used for abrasive machining operations (e.g., sharpening), is typically in the form of a heavy disc of moderate diameter. A grinding machine needs to have a high frequency of revolution but the machining operations exert braking torques on its wheel.

Angular momentum is directly proportional to mass. Hence, heavier the wheel, the greater is its angular momentum and lesser is the decelerating effect of the braking torques. Also, angular acceleration is inversely proportional to the moment of inertia. Since the wheel is made heavy, its diameter is kept moderate so that a large angular acceleration and high angular velocity can be achieved with a motor of given power.

Question 101.
Solve the following :

Question 1.
The angular momentum of a body changes by 80 kg.m²/s when its angular velocity changes from 20 rad/s to 40 rad/s. Find the change in its kinetic energy of rotation.
Solution :
Data : ω₁ = 20 rad/s, ω₂ = 40 rad/s
If I is the MI of the body, its initial angular momentum is Iω₁, and final angular momentum is Iω₂.
Change in angular momentum
= Iω₂ – Iω₁(ω₂ – ω₁)
∴ 80 = I(40 – 20)
∴ I = 4 kg.m²

Question 2.
A wheel of moment of inertia 1 kg.m² is rotating at a speed of 40 rad/s. Due to the friction on the axis, the wheel comes to rest in 10 minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.
Solution :
Data : I = 1 kg.m², ω₁ = 40 rad/s, ω₂ = 0 at
t = 10 minutes = 60 × 10 s = 600 s,
t’ = 8 minutes = 60 × 8 s = 480 s

This is the required angular momentum of the wheel.

Question 3.
A flywheel rotating about an axis through its centre and perpendicular to its plane loses 100 J of energy on slowing down from 60 rpm to 30 rpm. Find its moment of inertia about the given axis and the change in its angular momentum.
Solution :
Data : f₁ = 60 rpm = 60/60 rot/s = 1 rot/s, f₂ = 30 rpm = 30/60 rot/s = \(\frac{1}{2}\) rot/s, ∆E = – 100 J

This gives the MI of the flywheel about the given axis.

(ii) Angular momentum, L = Iω = I(2πf) 2πIf
The change in angular momentum, ∆L
= L₂ – L₁ = 2πI(f₂ – f₁)
= 2 × 3.142 × 6.753\(\left(\frac{1}{2}-1\right)\)
= -3.142 × 6.753= -21.22 kg.m²/s

Question 102.
A torque of 4 N-m acting on a body of mass 1 kg produces an angular acceleration of 2 rad/s². What is the moment of inertia of the body?
Answer:
The moment of inertia of the body.

Question 103.
Two identical rings are to be rotated about different axes of rotation as shown by applying torques so as to produce the same angular acceleration in both. How is it possible ?

Answer:
The MI of ring 1 about a transverse tangent is I₁ = 2MR²
The MI of ring 2 about its diameter is
I₂ = \(\frac{1}{2}\)MR²
Since, torque T = \(\tau=I \alpha\), to produce the same angular acceleration in both,

∴ It will be possible to produce the same angular acceleration in both the rings only if \(\tau_{1}=4 \tau_{2}\).

Question 104.
Two wheels have the same mass. First wheel is in the form of a solid disc of radius R while the second is a disc with inner radius r and outer radius R. Both are rotating with same angular velocity ω₀ about transverse axes through their centres. If the first wheel comes to rest in time t₁ while the second comes to rest in time t₂, are t₁ and t₂ different? Why?
Answer:
The moments of inertia of the two wheels about transverse axes through their centres are

Question 105.
Solve the following :

Question 1.
A torque of magnitude 400 N-m, acting on a body of mass 40 kg, produces an angular acceleration of 20 rad/s². Calculate the moment of inertia and radius of gyration of the body.
Solution :
Data : T = 400 N.m, M = 40 kg, α = 20 rad/s²

Question 2.
A body starts rotating from rest. Due to a couple of 20 N.m, it completes 60 revolutions in one minute. Find the moment of inertia of the body.
Solution :

Question 3.
A wheel of moment of inertia 2 kg-m2 rotates at 50 rpm about its transverse axis. Find the torque that can stop the wheel in one minute.
Solution :
Data : I = 2 kg.m², f₀ = 50 rpm = \(\frac{50}{60}\) = \(\frac{5}{6}\) rev/s, t = 60 s

Question 4.
A circular disc of moment of inertia 10 kg.m² is rotated about its transverse symmetry axis at a constant frequency of 60 rpm by an electric motor of power 31.42 watts. When the motor is switched off, how many rotations does it complete before coming to rest?
Solution :
Data : I = 10 kg.m², P = 31.42 watts,
f = 60 rpm = 60/60 Hz = 1 Hz
In rotational motion,
power = torque × angular velocity

This torque provided by the motor overcomes the torque of the frictional forces and maintains a constant frequency of rotation.

When the motor is switched off, the disc slows down due to the retarding torque of the frictional forces.

Question 5.
A flywheel of mass 4 kg and radius 10 cm, rotating with a uniform angular velocity of 5 rad/s, is subjected to a torque of 0.01 N.m for 10 seconds.
If the torque increases the speed of rotation, find
(i) the final angular velocity of the flywheel
(ii) the change in its angular velocity
(iii) the change in its angular momentum
(iv) the change in its kinetic energy.
Solution :
Data : M = 4 kg, R = 10 cm = 0.1 m, ω₁ = 5 rad/s, \(\tau\) = 0.01 N.m, t = 10 s

(i) The final angular velocity of the flywheel,
ω₂ = ω₁ + αt
= 5 + 0.5 × 10 = 10 rad/s

(ii) The change in its angular velocity
= ω₂ – ω₁ = 5 rad/s

(iii) The change in its angular momentum
= Iω₂ – Iω₁ = I (ω₂ – ω₁)
= 0.02 × 5 = 0.1 kg.m²/s

(iv) The change in its kinetic energy

Question 6.
A torque of 20 N.m sets a stationary circular disc into rotation about a transverse axis through its centre and acts for 2π seconds. If the disc has a mass 10 kg and radius 0.2 m, what is its frequency of rotation after 2π seconds ?
Solution :
Data : \(\tau\) = 20 N.m, t = 2π s, M = 10 kg, R = 0.1 m Let f₁ and f₂ be the initial and final frequencies of rotation of the disc, and ω₁ and ω₂ be its initial and final angular speeds. Since the disc was initially stationary, f₁ = ω₁, = 0 and ω₂ = 2πf₂.
The MI of the disc about the given axis is

Now, ω₂ = ω₁ + αt = 0 + α t
∴ 2πf₂ = αt
∴ f₂ = \(\frac{\alpha t}{2 \pi}\) = \(\frac{100(2 \pi)}{2 \pi}\) = 100 Hz

Question 7.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. If the rope is pulled downwards with a force of 30 N, find
(i) the angular acceleration of the cylinder
(ii) the linear acceleration of the rope.
Solution :
Data : M = 3 kg, R = 0.4 m, F = 30 N
(i) The MI of a hollow cylinder about its cylinder axis, I = MR²

(ii) The linear acceleration of the rope = the tangential acceleration at = αR = 25 × 0.4 = 10 m/s²

Question 106.
State and prove the principle (or law) of conservation of angular momentum.
Answer:
Principle (or law) of conservation of angular momentum : The angular momentum of a body is conserved if the resultant external torque on the body is zero.
Proof : Consider a moving particle of mass m whose position vector with respect to the origin at any instant is \(\vec{r}\).
Then, at this instant, the linear velocity of this particle is \(\vec{v}=\frac{\overrightarrow{d r}}{d t}\), its linear momentum is \(\vec{p}=m \vec{v}\) and its angular momentum about an axis through the origin is \(\vec{l}=\vec{r} \times \vec{p}\).

Suppose its angular momentum \(\vec{l}\) changes with time due to a torque \(\vec{\tau}\) exerted on the particle.

∴ \(\vec{l}\) = constant, i.e., \(\vec{l}\) is conserved. This proves the principle (or law) of conservation of angular momentum.

Alternate Proof : Consider a rigid body rotating with angular acceleration \(\vec{\alpha}\) about the axis of rotation. If I is the moment of inertia of the body about the axis of rotation, \(\vec{\omega}\) the angular velocity of the body at time t and \(\vec{L}\) the corresponding angular momentum of the body, then

∴ \(\vec{L}\) = constant, i.e., \(\vec{L}\) is conserved. This proves the principle (or law) of conservation of angular momentum.

Question 107.
What happens when a ballet dancer stretches her arms while taking turns?
Answer:
When a ballet dancer stretches her arms while pirouetting, her moment of inertia increases, and consequently her angular speed decreases to conserve angular momentum.

Question 108.
If the Earth suddenly shrinks so as to reduce its volume, mass remaining unchanged, what will be the effect on the duration of the day?
Answer:
If the Earth suddenly shrinks, mass remaining constant, the moment of inertia of the Earth will decrease, and consequently the angular velocity of rotation ω about its axis will increase. Since period \(T \propto \frac{1}{\omega}\), the duration of the day T will decrease.

Question 109.
Two discs of moments of inertia I₁ and I₂ about their transverse symmetry axes, respectively rotating with angular velocities to ω₁ and ω₂, are brought into contact with their rotation axes coincident. Find the angular velocity of the composite disc.
Answer:
We assume that the initial angular momenta (\(\vec{L}_{1}\) and \(\vec{L}_{2}\)) of the discs are either in the same direction or in opposite directions. Then,
the total initial angular momentum = \(\vec{L}_{1}+\vec{L}_{2}=I_{1} \overrightarrow{\omega_{1}}+I_{2} \overrightarrow{\omega_{2}}\)

After they are coupled, the total moment of inertia, i.e., the moment of inertia of the composite disc is I = I₁ + I₂ and the common angular velocity is \(\vec{\omega}\). Assuming conservation of angular momentum,

Question 110.
A boy standing at the centre of a turntable with his arms outstretched is set into rotation with angular speed ω rev/min. When the boy folds his arms back, his moment of inertia reduces to \(\frac{2}{5}\)th its initial value. Find the ratio of his final kinetic energy of rotation to his initial kinetic energy.
Answer:
Data : I₂ = \(\frac{2}{5}\)I₁
L = Iω
Assuming the angular momentum \(\vec{L}\) is conserved, in magnitude,

This gives the required ratio.

Question 111.
Name the quantity that is conserved when

1. \(\vec{F}_{\text {external }}\) is zero
2. \(\vec{\tau}_{\text {external }}\) is zero.

Answer:

1. Total linear momentum is conserved when \(\vec{F}_{\text {external }}\) is Zer0
2. Angular momentum is conserved when \(\vec{\tau}_{\text {external }}\) is zero.

Question 112.
What is the rotational analogue of the equation \(\vec{F}_{\text {external }}\) = \(\frac{d \vec{p}}{d t}\)?
Answer:
\(\vec{F}_{\text {external }}\) = \(\frac{d \vec{L}}{d t}\)

Question 113.
Fly wheels used in automobiles and steam engines producing rotational motion have discs with a large moment of inertia. Explain why?
Answer:
A flywheel is used as

(i) a mechanical energy storage, the energy being stored in the form of rotational kinetic energy

(ii) a direction and speed stabilizer. A flywheel rotor is typically in the form
of a disc. Rotational kinetic energy, \(E_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}\), where I is the moment of inertia and ω is the angular speed. That is, E_rot ∝ I. Therefore, higher the moment of inertia, the higher is the rotational kinetic energy that can be stored or recovered.

Also, angular momentum, \(\vec{L}=I \vec{\omega}\), i.e., \(|\vec{L}| \propto I\). A torque aligned with the symmetry axis of a flywheel can change its angular velocity and thereby its angular momentum. A flywheel with a large angular momentum will require a greater torque to change its angular velocity. Thus, a flywheel can be used to stabilize direction and magnitude of its angular velocity by undesired torques.

Question 114.
Solve the following :

Question 1.
A uniform horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 180 rpm. A blob of wax of mass 1.9 g falls on it and sticks to it at 25 cm from the axis. If the frequency of rotation is reduced by 60 rpm, calculate the moment of inertia of the disc.
Solution :
Data : f₁ = 180 rpm = 180/60 rot/s = 3 rot/s, f₂ = (180 – 60) rpm = 120/60 rot/s = 2 rot/s, m = 1.9 g = 1.9 × 10^-3 kg, r = 25 cm = 0.25 m
Let I₁ be the MI of the disc. Let I₂ be the MI of the disc and the blob.

Question 2.
A horizontal disc is rotating about a transverse axis through its centre at 100 rpm. A 20 gram blob of wax falls on the disc and sticks to it at 5 cm from its axis. The moment of inertia of the disc about its axis passing through its centre is 2 × 10^-4 kg.m². Calculate the new frequency of rotation of the disc.
Solution :
Data : f₁ = 100 rpm, m = 20 g = 20 × 10^-3 kg, r = 5 cm = 5 × 10^-2 m, I₁ = I_disc = 2 × 10^-4 kg.m²
The MI of the disc and blob of wax is
I₂ = I₁ + mr²

This is the new frequency of rotation.

Question 3.
A ballet dancer spins about a vertical axis at 2.5 π rad/s with his arms outstretched. With the arms folded, the MI about the same axis of rotation changes by 25%. Calculate the new speed of rotation in rpm.
Solution:
Let I₁, ω₁, and f₁, be the moment of inertia, angular velocity and frequency of rotation of the ballet dancer with arms outstretched, and I₂, ω₂ and f₂ be the corresponding quantities with arms folded.
Data : ω₁ = 2.5 π rad /s
Since moment of inertia with arms folded is less than that with arms outstretched,

Question 4.
Two wheels, each of moment of inertia 4 kg.m², rotate side by side at the rate of 120 rpm and 240 rpm in opposite directions. If both the wheels are coupled by a weightless shaft so that they now rotate with a common angular speed, find this new rate of rotation.
Solution :
Data : I = 4 kg.m², f₁ = 120 rpm, f₂ 240 rpm
Initially, the angular velocities of the two wheels \(\overrightarrow{\omega_{1}}\), and \(\overrightarrow{\omega_{2}}\)) and, therefore, their angular momenta (\(\vec{L}_{1}\) and \(\vec{L}_{2}\)) are in opposite directions.

The magnitude of the total initial angular momentum
= – L₁ + L₂ = -Iω₁ + Iω₂ (∵ I₁ = I₂ = I)
= 2πl(f₂ – f₁) … (1)

After coupling onto the same shaft, the total moment of inertia is 21. Let ω = 2πf be the common angular speed.

The magnitude of the total final angular momentum = 2I.ω = 4πl.f … (2)

From Eqs. (1) and (2), by the principle of conservation of angular momentum,
4πf = 2πl(f₂ – f₁)
∴ f = \(\frac{f_{2}-f_{1}}{2}\) = \(\frac{240-120}{2}\) = 60 rpm
This gives their new rate of rotation.

Question 5.
A homogeneous (uniform) rod XY of length L and mass M is pivoted at the centre C such that it can rotate freely in a vertical plane. Initially, the rod is horizontal. A blob of wax of the same mass M as that of the rod falls vertically with speed V and sticks to the rod midway between points C and Y. As a result, the rod rotates with angular speed ω. What will be the angular speed in terms of V and L?
Solution :
The initial angular momentum of the rod is zero.
The initial angular momentum of the falling blob of wax about the point C is (in magnitude)
= mass × speed × perpendicular distance between its direction of motion and point C
= MV.\(\frac{L}{4}\)
The total initial angular momentum of the rod and blob of wax = \(\frac{M V L}{4}\) … (1)

After the blob of wax sticks to the rod, and the system rotates with an angular speed ω about the horizontal axis through point C perpendicular to the plane of the figure, the total final angular momentum of the system about this axis

This gives the required angular speed.

Question 6.
A satellite moves around the Earth in an elliptical orbit such that at perigee (closest approach) it is two Earth radii above the Earth’s surface. At apogee (farthest position), it travels with one-fourth the speed it has at perigee. In terms of the Earth’s radius R, what is the maximum distance of the satellite from the Earth’s surface ?
Solution:
Let r_p and r_a be the distances of the satellite from the centre of the Earth at perigee and apogee, respectively. Let v_p and v_a be its linear (tangential) velocities at perigee and apogee.
Data : r_p = 2R + R = 3R, v_a = \(\frac{1}{4}\)v_p

Let L_p and L_a be the angular momenta of the satellite about the Earth’s centre. Because the gravitational force (\(\vec{F}\)) on the satellite due to the Earth is always radially towards the centre of the Earth, its direction is opposite to that of the position vector (\(\vec{r}\)) of the satellite relative to the centre of the Earth, so that the torque \(\vec{\tau}=\vec{r} \times \vec{F}=0\). Hence, the angular momentum of the satellite about the Earth’s centre is constant in time.

At apogee, the distance of the satellite from the Earth’s surface is 12R – R = 11R.

Question 7.
A torque of 100 N.m is applied to a body capable of rotating about a given axis. If the body starts from rest and acquires kinetic energy of 10000 J in 10 seconds, find
(i) its moment of inertia about the given axis
(ii) its angular momentum at the end of 10 seconds.
Solution :
Data : \(\tau\) = 100 N.m, ω_i = 0, E_i = 0, E_f = 10⁴J, t = 10 s

Since the body starts from rest, its initial angular momentum, L_i = 0.
The final angular momentum,
L_f = τ∆t = (100)(10) = 10³ kg.m²/s
The final rotational kinetic energy, E_f = \(\frac{1}{2} L_{\mathrm{f}} \omega_{\mathrm{f}}\)

Question 8.
Two identical metal beads, each of mass M but negligible width, can slide along a thin smooth uniform horizontal rod of mass M and length L. The rod is capable of rotating about a vertical axis passing through its centre. Initially, the beads are almost touching the axis of rotation and the rod is rotating at speed of 14 rad/s. Find the angular speed of the system when the beads have moved up to the ends of the rod. (Assume that no external torque acts on the system.)
Solution :
Data : ω₁ = 14 rad/s
The MI of the rod about a transverse axis through its CM,
I_rod = \(\frac{M L^{2}}{12}\)
Since the beads are almost particle-like, and initially touching the rotation axis, their MI about the vertical axis is taken to be zero.
When the beads move upto the ends of the rod, r = L/2, their MI about the vertical axis is

Question 115.
Discuss how pure rolling (i.e., rolling without slipping) on a plane surface is a combined translational and rotational motion.
Answer:
Rolling motion (without slipping) is an important case of combined translation and rotation. Consider a circularly symmetric rigid body, like a wheel or a disc, rolling on a plane surface with friction along a straight path.

The centre of mass of the wheel is at its geometric centre O. For purely translational motion (the wheel sliding smoothly along the surface without rotating at all), every point on the wheel has the same linear velocity \(\vec{v}_{\mathrm{CM}}\) = \(\vec{v}_{\mathrm{O}}\) as the centre O. For purely rotational motion (as if the horizontal rotation axis through O were stationary), every point on the wheel rotates about the axis with angular velocity \(\vec{\omega}\); in this case, every point on the rim has the same linear speed ωR.

We view the combined motion in the inertial frame of reference in which the surface is at rest. In this frame, since there is no slipping, the point of contact of the wheel with the surface is instantaneously stationary, v_A = 0, so that the wheel is turning about an instantaneous axis through the point of contact A. The instantaneous linear speed of point C (at the top) is V_C = ω(2R) – faster than any other point of the wheel.

Question 116.
Deduce an expression for the kinetic energy of a body rolling on a plane surface without slipping.
OR
Obtain an expression for the total kinetic energy of a rolling body in the form \(\frac{1}{2} M v^{2}\left[1+\frac{k^{2}}{R^{2}}\right]\)
OR
Derive an expression for the kinetic energy when a rigid body is rolling on a horizontal surface without slipping. Hence, find the kinetic energy of a solid sphere.
Answer:
Consider a symmetric rigid body, like a sphere or a wheel or a disc, rolling on a plane surface with friction along a straight path. Its centre of mass (CM) moves in a straight line and, if the frictional force on the body is large enough, the body rolls without slipping. Thus, the rolling motion of the body can be treated as translation of the CM and rotation about an axis through the CM. Hence, the kinetic energy of a rolling body is
E = E_tran + E_rot ….. (1)
where E_tran and E_rot are the kinetic energies associated with translation of the CM and rotation about an axis through the CM, respectively.

Let M and R be the mass and radius of the body. Let ω, k and I be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its centre, and v be the translational speed of the centre of mass.

Equation (4) or (5) or (6) gives the required expression.

Question 117.
A uniform solid sphere of mass 10 kg rolls on a horizontal surface. If its linear speed is 2 m/s, what is its total kinetic energy?
Answer:
Total kinetic energy of the sphere
= \(\frac{7}{10}\)Mv² = \(\frac{7}{10}\) × 10 × (2)² = 28 J

Question 118.
A disc of mass 4 kg rolls on a horizontal surface. If its linear speed is 3 m/s, what is its total kinetic energy?
Answer:
Total kinetic energy of the disc
= \(\frac{3}{4}\)Mv² = \(\frac{3}{4}\) × 4 × (3)² = 27 J

Question 119.
Assuming the expression for the kinetic energy of a body rolling on a plane surface without slipping, deduce the expression for the total kinetic energy of rolling motion for
(i) a ring
(ii) a disk
(iii) a hollow sphere. Also, find the ratio of rotational kinetic energy to total kinetic energy for each body.
Answer:
For a body of mass M and radius of gyration k, rolling on a plane surface without slipping with speed v, its total KE and rotational KE are respectively

[Note : The moment of inertia of all the round bodies above can be expressed as I = βMR², where β is a pure number less than or equal to 1. β is equal to 1 for a ring or a thin-walled hollow cylinder, \(\frac{1}{2}\) for a disc or solid cylinder, \(\frac{2}{3}\) for a hollow sphere and \(\frac{2}{5}\) for a solid sphere.
All uniform rings or hollow cylinders of the same mass and moving with the same speed have the same total kinetic energy, even if their radii are different. All discs or solid cylinders of the same mass and moving with the same speed have the same total kinetic energy; all solid spheres of the same mass and moving with the same speed have the same total kinetic energy. Also, for the same mass and speed, bodies with small c have less total kinetic energy.

Question 120.
State the expression for the speed of a circularly symmetric body rolling without slipping down an inclined plane. Hence deduce the expressions for the speed of
(i) a ring
(ii) a solid cylinder
(iii) a hollow sphere
(iv) a solid sphere, having the same radii.
Answer:
Consider a circularly symmetric body, of mass M and radius of gyration k, starting from rest on an inclined plane and rolling down without slipping. Its speed after rolling down through a height h is

[Note : If the inclined plane is ‘smooth’, i.e., there is no friction, the bodies will slide along the plane without any rotation. They will then have only translational kinetic energy, undergo equal acceleration and all three would arrive at the bottom at the same time with the same speed.]

Question 121.
State with reason if the statement is true or false : A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
Answer:
The statement is true.
Explanation : Rolling on a surface (horizontal or inclined) without slipping may be viewed as pure rotation about an horizontal axis through the point of contact, when viewed in the inertial frame of reference in which the surface is at rest. The point of contact of the wheel with the surface will be instantaneously at rest, resulting in a rolling motion, provided the wheel is able to ‘grip’ the surface, i.e., friction is necessary. With little or no friction, the wheel will slip at the point of contact. On an inclined plane, this will result in pure translation along the plane. On a horizontal surface, the wheel will simply rotate about its axis through the centre without translation.

Question 122.
A ring and a disc roll down an inclined plane through the same height. Compare their speeds at the bottom of the plane.
Answer:

Question 123.
State the expression for the acceleration of a circularly symmetric rigid body rolling without slipping down an inclined plane. Hence, deduce the acceleration of
(i) a ring
(ii) a solid cylinder
(iii) a hollow cylinder
(iv) a solid sphere, rolling without slipping down an inclined plane.
Answer:
A circularly symmetric rigid body, of radius R and radius of gyration k, on rolling down an inclined plane of inclination θ has an acceleration

Question 124.
A spherical shell rolls down a plane inclined at 30° to the horizontal. What is its acceleration ?
Answer:
The acceleration of the spherical shell,
a = \(\frac{3}{5} g \sin \theta\) = 0.6g sin 30° = 0.6g × \(\frac{1}{2}\) = 0.3g m

Question 125.
A spherical shell and a uniform solid sphere roll down the same inclined plane. Compare their accelerations.
Answer:
The ratio of the accelerations, in the usual notation,

Question 126.
A solid sphere, starting from rest, rolls down two different inclined planes from the same height but with different angles of inclination θ₁ > θ₂. On which plane will the sphere take longer time to roll down?
Answer:
Let L₁ and L₂ be the distances rolled down by the sphere along the corresponding inclines from the same height h.
∴ L₁ sin θ₁ = L₂ sin θ₂ = h

The sphere will take longer time to roll down from the same height on the plane with smaller inclination.

Question 127.
Two circular discs A and B, having the same mass, have four identical small circular discs placed on them, as shown in the diagram. They are simultaneously released from rest at the top of an inclined plane. If the discs roll down without slipping, which disc will reach the bottom first?

Answer:
The disc A has the smaller discs closer to the centre than disc B. Hence, the moment of inertia of disc A (I_A) is less than that of disc B (I_B).

[Suppose the larger discs have radius R, the smaller discs have mass m and radius r, and the centre of each smaller disc on disc A is at a distance x from the centre. Then, x = \(\sqrt{2} r\)r and, it can be shown that, I_B – I_A = 4m[R² – (x – r)²] > 0.]

Each composite disc is equivalent to a disc of the same radius R and mass M’ = M + 4m, where m is the mass of each smaller disc, but of different thicknesses.

Suppose, starting from rest, the composite discs roll down the same distance L along a plane inclined at an angle θ, their respective accelerations will be

i.e., the disc A will reach the bottom first.

Question 128.
Solve the following :

Question 1.
A lawn roller of mass 80 kg, radius 0.3 m and moment of inertia 3.6 kg.m², is drawn along a level surface at a constant speed of 1.8 m/s. Find
(i) the translational kinetic energy
(ii) the rotational kinetic energy
(iii) the total kinetic energy of the roller.
Answer:
Data : M = 80 kg, R = 0.3 m, I = 3.6 kg.m², v = 1.8 m/s
(i) The translational kinetic energy of the centre of mass of the roller,

Question 2.
A solid sphere of mass 1 kg rolls on a table with linear speed 2 m/s, find its total kinetic energy.
Solution :
Data : M = 1 kg, v = 2 m/s
The total kinetic energy of a rolling body,

Question 3.
A ring and a disc having the same mass roll on a horizontal surface without slipping with the same linear velocity. If the total KE of the ring is 8 J, what is the total KE of the disc?
Solution :
Data : M_ring = M_disc = M, v_ring = v_disc = v, E_ring = 8J
The total kinetic energies of rolling without slipping on a horizontal surface,
E_ring = Mv² and E_disc = \(\frac{3}{4} M v^{2}\)
since they have the same mass and linear velocity.
∴ E_disc = \(\frac{3}{4}\)E_ring = \(\frac{3}{4}\) × 8 = 6J

Question 4.
A solid cylinder, of mass 2 kg and radius 0.1 m, rolls down an inclined plane of height 3 m. Calculate its rotational energy when it reaches the foot of the plane.
Solution :
Data : M = 2 kg, R = 0.1 m, h = 3 m, g = 10 m/s²

Question 5.
A solid sphere rolls up a plane inclined at 45° to the horizontal. If the speed of its centre of mass at the bottom of the plane is 5 m/s, find how far the sphere travels up the plane.
Solution :
Data : v = 5 m/s, θ = 45°, g = 9.8 m/s²
The total energy of the sphere at the bottom of the plane is
E = \(\frac{7}{10} M v^{2}\)
where M is the mass of the sphere.
In rolling up the incline through a vertical height h, it travels a distance L along the plane. Then,

The sphere travels 2.526 m up the plane.

Question 129.
Choose the correct option:

Question 1.
The bulging of the Earth at the equator and flattening at the poles is due to
(A) centripetal force
(B) centrifugal force
(C) gravitational force
(D) electrostatic force.
Answer:
(B) centrifugal force

Question 2.
A body of mass 0.4 kg is revolved in a horizontal circle of radius 5 m. If it performs 120 rpm, the centripetal force acting on it is
(A) 4π² N
(B) 8π² N
(C) 16π² N
(D) 32π² N.
Answer:
(D) 32π² N.

Question 3.
Two particles with their masses in the ratio 2 : 3 perform uniform circular motion with orbital radii in the ratio 3 : 2. If the centripetal force acting on them is the same, the ratio of their speeds is
(A) 4 : 9
(B) 1 : 1
(C) 3 : 2
(D) 9 : 4.
Answer:
(C) 3 : 2

Question 4.
When a motorcyclist takes a circular turn on a level race track, the centripetal force is
(A) the resultant of the normal reaction and frictional force
(B) the horizontal component of the normal reaction
(C) the frictional force between the tyres and road
(D) the vertical component of the normal reaction.
Answer:
(C) the frictional force between the tyres and road

Question 5.
The maximum speed with which a car can be driven safely along a curved road of radius 17.32 m and banked at 30° with the horizontal is [g = 10 m/s²]
(A) 5 m/s
(B) 10 m/s
(C) 15 m/s
(D) 20 m/s.
Answer:
(B) 10 m/s

Question 6.
A track for a certain motor sport event is in the form of a circle and banked at an angle 6. For a car driven in a circle of radius r along the track at the optimum speed, the periodic time is

Answer:
(C) \(2 \pi \sqrt{\frac{r}{g \tan \theta}}\)

Question 7.
The period of a conical pendulum in terms of its length (l), semivertical angle (θ) and acceleration due to gravity (g), is

Answer:
(C) \(4 \pi \sqrt{\frac{l \cos \theta}{4 g}}\)

Question 8.
A conical pendulum of string length L and bob of mass m performs UCM along a circular path of radius r. The tension in the string is

Answer:
(A) \(\frac{m g L}{\sqrt{L^{2}-r^{2}}}\)

Question 9.
The centripetal acceleration of the bob of a conical pendulum is

Answer:
(D) \(\frac{r g}{L \cos \theta}\)

Question 10.
A small object tied at the end of a string is to be whirled in a vertical circle of radius r. If its speed at the lowest point is \(2 \sqrt{g r}\), then
(A) the string will be slack at the lowest point
(B) it will not reach the midway point
(C) its speed at the highest point will be \(\sqrt{g r}\)
(D) it will just reach the highest point with zero speed.
Answer:
(D) it will just reach the highest point with zero speed.

Question 11.
A small bob of mass m is tied to a string and revolved in a vertical circle of radius r. If its speed at the highest point is \(\sqrt{3 r g}\), the tension in the string at the lowest point is
(A) 5 mg
(B) 6 mg
(C) 7 mg
(D) 8 mg.
Answer:
(D) 8 mg.

Question 12.
A small object, tied at the end of a string of length r, is launched into a vertical circle with a speed \(2 \sqrt{g r}\) at the lowest point. Its speed when the string is horizontal is
(A) > \(3 \sqrt{g r}\)
(B) = \(3 \sqrt{g r}\)
(C) = \(2 \sqrt{g r}\)
(D) 0.
Answer:
(C) = \(2 \sqrt{g r}\)

Question 13.
Two bodies with moments of inertia I₁ and I₂ (I₁ > I₂) rotate with the same angular momentum. If E₁ and E₂ are their rotational kinetic energies, then
(A) E₂ > E₁
(B) E₂ = E₁
(C) E₂ < E₁
(D) E₂ ≤ E₁
Answer:
(A) E₂ > E₁

Question 14.
The radius of gyration k for a rigid body about a given rotation axis is given by

Answer:
(B) \(k^{2}=\frac{1}{M} \int r^{2} d m\)

Question 15.
Three point masses m, 2m and 3m are located at the three vertices of an equilateral triangle of side l. The moment of inertia of the system of particles about an axis perpendicular to their plane and equidistant from the vertices is
(A) 2ml²
(B) 3ml²
(C) \(2 \sqrt{3}\) ml²
(D) 6ml²
Answer:
(A) 2ml²

Question 16.
The moment of inertia of a thin uniform rod of mass M and length L, about an axis passing through a point midway between the centre and one end, and perpendicular to its length, is
(A) \(\frac{48}{7}\)ML²
(B) \(\frac{7}{48}\)ML²
(C) \(\frac{1}{48}\)ML²
(D) \(\frac{1}{16}\)ML²
Answer:
(B) \(\frac{7}{48}\)ML²

Question 17.
A thin uniform rod of mass M and length L has a small block of mass M attached at one end. The moment of inertia of the system about an axis through its CM and perpendicular to the length of the rod is
(A) \(\frac{13}{12}\) ML²
(B) \(\frac{1}{3}\) ML²
(C) \(\frac{5}{24}\) ML²
(D) \(\frac{7}{48}\) ML²
Answer:
(C) \(\frac{5}{24}\) ML²

Question 18.
A thin wire of length L and uniform linear mass density λ is bent into a circular ring. The MI of the ring about a tangential axis in its plane is

Answer:
(C) \(\frac{3 \lambda L^{3}}{8 \pi^{2}}\)

Question 19.
When a planet in its orbit changes its distance from the Sun, which of the following remains constant ?
(A) The moment of inertia of the planet about the Sun
(B) The gravitational force exerted by the Sun on the planet
(C) The planet’s speed
(D) The planet’s angular momentum about the Sun
Answer:
(D) The planet’s angular momentum about the Sun

Question 20.
If L is the angular momentum and I is the moment of inertia of a rotating body, then \(\frac{L^{2}}{2 I}\) represents its
(A) rotational PE
(B) total energy
(C) rotational KE
(D) translational KE.
Answer:
(C) rotational KE

Question 21.
A thin uniform rod of mass 3 kg and length 2 m rotates about an axis through its CM and perpendicular to its length. An external torque changes its frequency by 15 Hz in 10 s. The magnitude of the torque is
(A) 3.14 N.m
(B) 6.28 N.m
(C) 9.42 N.m
(D) 12.56 N.m.
Answer:
(C) 9.42 N.m

Question 22.
The flywheel of a motor has mass 300 kg and radius of gyration 1.5 m. The motor develops a constant torque of 2000 N.m and the flywheel starts from rest. The work done by the motor during the first 4 revolutions is
(A) 2 kJ
(B) 8 kJ
(C) 8n kJ
(D) 16π kJ.
Answer:
(D) 16π kJ.

Question 23.
Two uniform solid spheres, of the same mass but radii in the ratio R₁ : R₂ = 1 : 2, roll without slipping on a plane surface with the same total kinetic energy. The ratio ω₁ : ω₂ of their angular speed is
(A) 2 : 1
(B) \(\sqrt{2}\) : 1
(C) 1 : 1
(D) 1 : 2.
Answer:
(A) 2 : 1

Question 24.
A circularly symmetric body of radius R and radius of gyration k rolls without slipping along a flat surface. Then, the fraction of its total energy associated with rotation is [c = k²/R²]
(A) c
(B) \(\frac{c}{1+c}\)
(C) \(\frac{1}{c}\)
(D) \(\frac{1}{1+c}\)
Answer:
(B) \(\frac{c}{1+c}\)