CBSE Sample Papers for Class 11 Physics Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 3 with Solutions

Time Allowed : 3 hours
Maximum Marks : 70

General Instructions:

1. 1. All questions are compulsory. There are 40 questions.
   2. This Question paper has four sections : Section A, Section B, Section C, Section D
   3. Section A contains twenty five questions of one mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains three question of five marks each.
   4. There is no overall choice. However, internal choices have been provided in seven questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks weightage. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:
      c = 3 × 10⁸ m/s,
      h = 6.63 × 10^-34 Js
      e = 1.6 × 10^-19 C,
      Radius of Earth, R_e = 6.4 × 10⁶ m
      Universal Gravitational constant. G = 6.67 × 10^-11 Nm2kg-2
      mass of electron, m_e = 9.1 × 10^-31 kg,
      mass of neutron, m_n = 1.675 × 10^-27 kg
      mass of proton, m_p = 1.673 × 10^-27 kg
      Avogadro’s number = 6.023 × 10²³ atom per gram
      Boltzmann constant = 1.38 × 10^-23 JK^-1

Section – A (25 Marks)

Question numbers 1 to 25 carry 1 mark each.

Question 1.
Which of the following measurements is most precise ? [1]
(A) 5.00 mm
(b) 5.00 cm
(C) 5.00 m
(D) 5.00 km
Answer:
Option (A) is correct.

Question 2.
The displacement of a particle is given by x = (t – 2)² where x is in metres and t in seconds. The distance covered by the particle in first 4 seconds is : [1]
(A) 4 m
(B) 8 m
(C) 12 m
(D) 16 m
Answer:
Option (B) is correct.

Explanation:
At t = 0, x = x₀ = (0 – 2)² = 4 m
t = 1 s, x = x₁ = (1 – 2)² = 1 m
t = 2 s, x = x₂ = (2 – 2)² = 0 m
t = 3 s, x = x₃ = (3 – 2)² = 1 m
t = 4 s, x = x₄ = (4 – 2)² = 4 m
The distance covered by the particle in 1^st second is D₁ = x₀ – x₁ = 3 m
Similarly, D₂ = 1 m, D₃ = 1 m, D₄ = 3 m
The distance covered by the particle in first 4 seconds is
D = D₁ + D₂ + D₃ + D₄
= 3 m + 1 m + 1 m + 3 m = 8 m

Question 3.
At a metro station, a girl walks up a stationary escalator in time t₁. If she remains stationary on the escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be : [1]
(A) (t₁ + t₂)/2
(B) t₁t₂(t₂ – t₁)
(C) t₁t₂/(t₂ + t₁)
(D) t₁ – t₂
OR
Young’s modulus of steel is 1.9 × 10¹¹ N/m². When expresses in CGS units of dyne/cm², it will be equal to (1 N = 10⁵ dyne, 1 m² = 10⁴ cm²) : [1]
(A) 1.9 × 10¹⁰
(B) 1.9 × 10¹¹
(C) 1.9 × 10¹²
(D) 1.9 × 10¹³
Answer:
Option (C) is correct.

Explanation:
Let L be the length of escalator. Velocity of girl w.r.t. escalator,
υ_ge = \(\frac{\mathrm{L}}{t_1}\)
∴ Velocity of escalator,
υ_e = \(\frac{\mathrm{L}}{t_2}\)
∴ Velocity of girl w.r.t. ground would be
υ_g = υ_ge + υ_e
= L \(\left(\frac{1}{t_1}+\frac{1}{t_2}\right)\)
The desired time is, t = \(\frac{\mathrm{L}}{v_g}=\frac{\mathrm{L}}{\mathrm{L}\left(\frac{1}{t_1}+\frac{1}{t_2}\right)}=\frac{t_1 t_2}{t_1+t_2}\)

OR

Option (C) is correct.
Explanation:
1.9 × 10¹¹ \(\left(\frac{10^5 \mathrm{dyne}}{10^4 \mathrm{~cm}^2}\right)\) = 1.9 × 10¹² dyne cm^-2

Question 4.
Which of the diagrams shown in Fig. most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit? [1]

Answer:
Option (D) is correct.

Explanation:
As Earth moves once around the sun in its elliptical orbit, its kinetic energy is maximum when it is farthest from the sun. As kinetic energy is never zero during its motion.

Question 5.
Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods. [1]
(A) Both the rods will elongate but there shall be no perceptible change in shape.
(B) The steel rod will elongate and change shape but the rubber rod will only elongate.
(C) The steel rod will elongate without any perceptible change in shape, but the rubber road will elongate and the shape of the bottom edge will change to an ellipse.
(D) The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
Answer:
Option (D) is correct.

Question 6.
As the temperature is increased, the time period of a pendulum [1]
(A) increases as its effective length increases even though its centre of mass still remains at the centre of the bob.
(B) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob.
(C) increases as its effective length increases due to shifting of centre of mass below the centre of the bob.
(D) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.
OR
The equation of motion of a particle is x = a cos (αt)² The motion is :
(A) Periodic but not oscillatory.
(B) Periodic and oscillatory.
(C) Oscillatory but not periodic.
(D) Neither periodic nor oscillatory.
Answer:
Option (A) is correct.
OR
Option (C) is correct.

Question 7.
Change in temperature of the medium changes : [1]
(A) frequency of sound waves.
(B) amplitude of sound waves
(C) wavelength of sound waves.
(D) loudness of sound waves
A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in :
(A) one second
(B) 0.5 second
(C) 2 seconds
(D) data given is insufficient.
Answer:
Option (C) is correct.
OR
Option (B) is correct.

Explanation :
μ = \(\frac{\mathrm{M}}{\mathrm{L}}=\frac{2.5}{20 \mathrm{~m}}\) kg = 0.125 kg/m
\(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{200 \mathrm{~N}}{0.125 \mathrm{~kg} / \mathrm{m}}}\)
υ = 40 m/s
t = \(\frac{\mathrm{L}}{v}=\frac{20 \mathrm{~m}}{40 \mathrm{~m} / \mathrm{s}}\) = 0.5 s

Question 8.
Gravitational forces are : [1]
(A) always attractive
(B) always repulsive
(C) sometimes attractive and some time repulsive
(D) None
Answer:
Option (A) is correct.

Question 9.
Unit of longitudinal strain is : [1]
(A) cm
(B) cm^-2
(C) g cm^-1
(D) None
Answer:
Option (D) is correct.

Question 10.
For an adiabatic change in a gas [1]
(A) TVT^γ-1 = constant
(B) TV^γ = constant
(C) TV^γ+1 = constant
(D) TV^-γ = constant
Answer:
Option (A) is correct.

Question 11.
Velocity-time plot of two particles make angles of 60° and 30° with the axis on which time is plotted. What is the ratio of their acceleration ? [1]
Answer:
Acceleration is given by the slope of velocity-time graph of an object.
Ratio of Acceleration = \(\frac{\text { Slope of graph of particle I }}{\text { Slope of graph of particle II }}\)
\(\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}\) = 3

Commonly Made Error
Students normally use here the angle directly without the use of tangent.

Answering Tip
Students should learn the formula and then apply the tangent of the angles given to find the ratio of the acceleration.

Question 12.
Give example of a motion where x > 0, υ < 0, a > 0 at a particular instant. [1]
Answer:
Example : x(t) = A + Be^-γ
Here, A > B and γ > 0 and all are positive constants.

Question 13.
Give a relation for impulse in scalar form. [1]
OR
Why does a swimmer push the water backwards ?
Answer:
I = F × t = p₂ – p₁
where notations have their usual meanings.

OR

So as to get forward push according to Newton’s third law of motion.

Question 14.
What should be the angle between the force and the displacement for maximum and minimum work? [1]
Answer:
For maximum work θ = 0° and minimum work θ = 180°

Question 15.
It is correct to say that the C.M. of system of n-particles is always given by average position vectors of the constituent particles ? If not, when the statement is true ? [1]
Answer:
No, this statement is true when all the particles of the system are of same mass.

Commonly Made Error
Students fail to give the reason.

Answering Tip
Mathematical treatment of centre of mass will help in giving a justification to the statement given.

Question 16.
Potential energy of a galaxy is taken as positive or negative [1]
OR
Give an expression for gravitational potential at the surface of earth.
Answer:
Galaxy has gravitational forces which are attractive so potential energy of a galaxy is taken as negative.

Commonly Made Error
Confusion in sign of potential energy of galaxy.

Answering Tip
Gravitational forces being attractive forces so potential energy associated with them has a negative value.

OR

Using V = – \(\frac{\mathrm{GM}}{\mathrm{r}}\)
when r = R, i.e., radius of earth
We have V = – \(\frac{\mathrm{GM}}{\mathrm{R}}\)

Question 17.
What is Young’s Modulus? [1]
Answer:
It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality. Normal stress
Y = \(\frac{\text { Normal stress }}{\text { Longitudinal strain }}\)

Question 18.
Where and why high viscosity liquids are used in trains ? [1]
Answer:
High viscosity liquids are used as buffers in trains so that the shocks may be absorbed easily.

Question 19.
Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1Å and 2Å the gases may be considered under identical conditions of temperature pressure and volume. [1]
Answer:
We have l ∝ \(\frac{1}{d^2}\)
(Here d is diameter of molecule of gas)
So, l₁ ∝ \(\frac{1}{d_1^2}\) ……………. (i)
l₂ ∝ \(\frac{1}{d_2^2}\) …………. (ii)
[Given d₁ = 1Å, d₂ = 2Å]
By eqn, (i) and (ii)
\(\)
\(\)
l₁ : l₂ = 4 : 1

Question 20.
What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator? [1]
OR
Is it possible to monitor the temperature of a wire by measuring its vibrational frequency ?
Read the following text and answer any 4 of the following questions on the basis of the same:
A gas which obeys the ideal gas equation PV = nRT at all temperature and pressure values is called an ideal gas equation. Molecules of such a gas are free from inter molecular attraction and have negligible volume. The gas particles are equally sized and motion of the particles follows Newton’s laws of motion. These particles do not undergo any energy loss as they have perfect elastic collisions. Figure shows plot of \(\frac{P V}{T}\) versus P for 1.00 × 10^-3 kg of oxygen gas at two different temperatures.

Answer:
\(\frac{a_{\max }}{v_{\max }}=\frac{\omega^2 \mathrm{~A}}{\omega \mathrm{A}}\) = ω
OR

Yes, v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{m}}\)
As temperature increases, length increases Therefore, frequency of vibration v decreases. Hence, changes in temperature can be monitored.

Question 21.
What does the dotted line signify in the above diagram? [1]
(A) Ideal gas behaviour
(B) Real gas behaviour
(C) Oxygen gas at temperature T₁
(D) Oxygen gas at temperature T₂
Answer:
Option (A) is correct.

Explanation:
The dotted line is a straight line parallel to P axis, so it signifies the ideal gas behaviour.

Question 22.
Which is of the following is correct ? [1]
(A) T₁ < T₂
(B) T₁ > T₂
(C) T₁ = T₂
(D) T₁ ≤ T₂
Answer:
Option (B) is correct.

Explanation:
The real gas approaches the ideal behaviour and the temperature increases. Since the plot of temperature Tj is close to the ideal behaviour, it can be concluded that T₁ > T₂.

Question 23.
The value of \(\frac{P V}{T}\) where curves meet on y-axis is [1]
(A) 8.31 JK^-1
(B) 0.26 JK^-1
(C) 0.52 JK^-1
(D) 4.15 JK^-1
Answer:
Option (B) is correct.

Explanation:
The ideal gas equation is given by
\(\frac{\mathrm{PV}}{\mathrm{T}}\) = nR
M_O2 = 32 g
nR = \(\frac{1}{32}\) × 8.314 JK^-1
= 0.259 JK^-1

Question 24.
Relation between Boltzmann constant and universal gas constant is: [1]
[k_B-Boltzmann constant, R-Universal gas constant, N-Avogadro number]
(A) R = k_B/N
(B) N = k_BR
(C) k_B = \(\frac{N}{R}\)
(D) R = k_BN
Answer:
Option (D) is correct.

Question 25.
What will be the value of \(\frac{P V}{T}\) for 1.00 × 10^-3 kg of Hydrogen gas if the oxygen gas is replaced by hydrogen gas? [1]
[Give: mass of Hydrogen = 2.02 u]
(A) 2 JK^-1
(B) 16.8 JK^-1
(C) 8.3 JK^-1
(D) 4.112 JK^-1
Answer:
Option (D) is correct.

Explanation:
No. of moles of Hydrogen atom
n = \(\) = 0.495
Since, \(\frac{\mathrm{PV}}{\mathrm{T}}\) = nR
= 0.495 × 8.314 = 4.11 JK^-1

Section – B (10 Marks)

Question numbers 26 to 30 carry 2 marks each.

Question 26.
A ball is dropped and its displacement vs. time graph is as shown in figure (displacement x is from ground and all quantities are +ve upwards). [2]

(A) Plot qualitatively velocity vs. time graph.
(B) Plot qualitatively acceleration vs. time graph.
Answer:
Under gravity, the ball is released and falling. For short time intervals in which the ball collides with ground and when the impulsive force acts, a large acceleration produces, unless it is -g.

Question 27.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How
long does the body take to stop ? [2]
OR
What is pseudo force ?
Answer:
Force F = ma
a = \(\frac{\mathrm{F}}{\mathrm{m}}\) = – \(\frac{50}{20}\)
= – 2.5 ms^-2
Using, υ = u + at
When, u = 15 ms^-1, υ = 0
We get, 0 = 15 – 2.5 × t
t = \(\frac{15}{2.5}\)
or t = 6 s.

OR

The fictitious or imaginary force used to balance an actual or material force is called pseudo force, e.g., centripetal force, F = \(\frac{-m v^2}{r}\) is a pseudo force. Pseudo forces do not have material sources, they are simply produced because of acceleration or rotation of the frame itself.

Question 28.
What are fundamental note and overtones ? [2]
Answer:
When a source is sounded, it generally vibrates in more than one mode and therefore, emits tones of different frequencies. The tone of lowest frequency is called the fundamental note and the tones of higher frequencies are called overtones.

Question 29.
Derive Boyle’s law on the basis of kinetic theory of gases. [2]
Answer:
According to kinetic theory, pressure, P exerted by a gas is
P = \(\frac{1}{3}\) ρυ²
∴ PV = \(\frac{1}{3}\) Mυ²
At constant temperature, total K.E. of gas \(\frac{1}{3}\)(M)v² or v², will be constant
∴ At constant temp, PV = constant

Question 30.
The angular velocity and amplitude of simple pendulum is ro and r respectively. At 9 displacement x from the mean position, if its kinetic energy is T and potential energy is V, find the ratio of T to V. [2]
OR
Write down the expression for speed of transverse waves in solids and in a stretched string.
Answer:
Kinetic energy, T = \(\frac{1}{2}\) mυ² = \(\frac{1}{2}\) mω²(r² – x²);
Potential energy, V = \(\frac{1}{2}\) mω²x²;
So \(\frac{\mathrm{T}}{\mathrm{V}}=\frac{r^2-x^2}{x^2}\)

OR

In a solid :
Speed of transverse wave, υ = \(\sqrt{\frac{\eta}{\rho}}\)
where η is modulus of rigidity and ρ is density of material of solid.

In a streched string :
Here υ = \(\sqrt{\frac{\mathrm{T}}{m}}\)
where T is tension in the string and m is mass per unit length of the string.

Section – C (21 Marks)

Question numbers 31 to 37 carry 3 marks each.

Question 31.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. [3]
Answer:
Given, r = 1 km = 1000 m;
υ = 900 kmh^-1
= 900 × \(\frac{1000}{3600}\) ms^-1
= 250 ms^-1
The centripetal acceleration of the aircraft is a
a = \(\frac{v^2}{r}=\frac{(250)^2}{1000}\)
= \(\frac{62500}{1000}\) = 62.5 ms^-2
Acceleration due to gravity,
g = 9.8 ms^-2
∴ \(\frac{\text { Centripetal acceleration }}{\text { Acceleration due to gravity }}=\frac{a}{g}\)
= \(\frac{62.5}{9.8}\)
or \(\frac{\mathrm{a}}{\mathrm{g}}\) = 6.38

Question 32.
Deduce the expression for position vector of centre of mass for n particle system? [3]
Answer:
If a system contains n particles of masses m₁, m₂, m₃ …………………… m_n
whose position vectors are\(\vec{r}_1, \vec{r}_2, \bar{r}_1, \ldots \ldots \ldots \ldots . \vec{r}_n\)
respectively, then position vector of centre of mass
\(\vec{r}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+m_3 \vec{r}_3+–\ldots \ldots \ldots . .—-+m_n \vec{r}_n}{m_1+m_2+m_3+-\ldots \ldots \ldots . .–+m_n}\)
If we consider two equal masses i.e., m₁ = m₂, then position vector of centre of mass \(\vec{r}=\frac{\vec{r}_1+\vec{r}_2}{2}\) , where, r₁ is position vector of mass 1, r₂ is position vector of mass 2.

Question 33.
Prove that momentum is conserved. [3]
Answer:
Using Newton’s second law of motion for a system of n particles, total force
\(\overrightarrow{\mathrm{F}}=\sum_{i=1}^{i=n} \frac{d}{d t}\left(m_i \vec{v}_i\right)\)
= \(\frac{d}{d t}\left(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}+\ldots . . m_n \overrightarrow{v_n}\right)\)
Internal forces acting on the particles cancel out in pairs. Taking external force also to be zero,
i.e., \(\vec{F}=\overrightarrow{0}\)
We get,
\(\frac{d}{d t}\left(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}+\ldots \ldots m_n \overrightarrow{v_n}\right)\) = 0
or \(\left(m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}+\ldots . . m_n \overrightarrow{v_n}\right)\) = constant
The above expression shows that linear momentum is conserved and this principle is called principle of conservation of momentum.

Question 34.
Assuming the earth to be a sphere of uniform mass density, how much would a body weight half way down to the centre of earth, if it weighs 250 N on the surface ? [3]
OR
An object of mass m is raised from the surface of the Earth to a height of equal to the radius of the Earth, that is, taken from a distance R to 2R from the centre of the Earth. What is the gain in its potential energy?
Answer:
Suppose g, g_d be the acceleration due to gravity on earth’s surface and at a depth ‘d’, from surface respectively.
Also, suppose W and W_d be the weight of a body on earth’s surface and at depth ‘d’, respectively,
∴ W = mg = 250 N …………….. (i)
and W_d = mg_d ……………….. (ii)
Now, we know that
g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) ……………. (iii)
Here, d = \(\frac{\mathrm{R}}{\mathrm{2}}\), R = radius of earth ……………… (iv)
∴ From equations (iii) and (iv), we get
g_d = g(1 – \(\frac{\mathrm{R} / 2}{\mathrm{R}}\))
= g(1 – \(\frac{1}{2}\)) = g × \(\left(\frac{1}{2}\right)\)
= \(\frac{\mathrm{g}}{\mathrm{2}}\) ………………… (v)
∴ W_d = mg_d
= m\(\frac{\mathrm{g}}{\mathrm{2}}\) [by using equation (v)]
= \(\frac{1}{2}\) mg = \(\frac{1}{2}\) W
= \(\frac{1}{2}\) × 250 = 125 N.
∴ Weight of the body half way down to the centre of earth = 125 N.

OR

Given: An object is raised distance R → 2R Potential energy of body on the surface of \(\frac{\mathrm{-GMm}}{\mathrm{R}}\)

P.E. of object at height equal to radius of \(\frac{\mathrm{-GMm}}{\mathrm{2R}}\)
Gain in Potential Energy = P.E_f – P.E_i
= \(\frac{-G M m}{2 R}-\left(\frac{-G M m}{R}\right)\)
= \(\frac{\mathrm{GMm}}{\mathrm{R}}\)[-\(\frac{1}{2}\) + 1]
= \(\frac{\mathrm{GMm}}{\mathrm{2R}}\)
As, GM = gR²
Gain in P.E. = \(\frac{g R^2 m}{2 R}\) = \(\frac{1}{2}\) mgR.

Question 35.
An electric heater supplies heat to a system at a rate of 100 W. If the system perforrqs work at a rate of 75 joules per second at what rate is the internal energy increasing ? [3]
Answer:
Given : Heat supplied per second,
dQ = 100 J
Work done by the system per second,
dW = 75 J
Increase in internal energy/second dU = ?
From first law of thermodynamics,
dQ = dU + dW
or dU = dQ – dW
or dU = 100 J – 75J
or dU = 25 J s^-1
or dU = 25 W (1 watt = 1 J)

Question 36.
A particle moves with S.H.M. in a straight line. In the first second after starting from rest, it travels a distance x₁ cm and in the next second it travels a distance x₂ cm in the same direction. Prove that the amplitude of oscillation is \(\frac{2 x_1^2}{\left(3 x_1-x_2\right)}\) [3]
OR
Obtain an equation for a progressive wave.
Answer:
As the particle starts from rest, it must start from the extreme position. Hence, when t = 0, x = r, where r is the required amplitude.
Using the relation, x = rcos ωt
or r – x₁ = rcos (ω × 1)
= rcos ω ……………… (i)
and r – (x₁ + x₂) = rcos (ω × 2)
= rcos 2ω
or r – x₁ – x₂ = r(2 cos² ω – 1) …………….. (ii)
Solving (i) and (ii),
r = \(\frac{2 x_1^2}{3 x_1-x_2}\)

OR

For a simple harmonic motion, displacement
y = rsin (ωt + Φ)
where r is amplitude, ω is angular velocity and Φ is phase difference
But ω = \(\frac{2 \pi}{\mathrm{T}}\) = 2πf
and Φ = \(\frac{2 \pi}{\mathrm{T}}\) x
where λ is wavelength and x is path difference.
∴ y = rsin 2π\(\left(\frac{t}{\mathrm{~T}}+\frac{x}{\lambda}\right)\)

Question 37.
How much should the pressure on a litre of water be changed to compress it by 0.10%. [3]
Answer:
Here V = 1 litre
∆V = -0.10% of V
= –\(\frac{0.10}{100}\) × 1 = –\(\frac{1}{1000}\) litre
Suppose ∆P = change in pressure required for compression of 1 litre of water.
K = bulk modulus of water = 2.2 × 10⁹ Nm^-2
From the relation
K = –\(\frac{\mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)}\)
we get ∆P = -K . \(\frac{\Delta \mathrm{V}}{\mathrm{V}}\)
or ∆P = 2.2 × 10⁹ × \(\frac{\left(\frac{1}{1000}\right)}{1}\)
or ∆P = \(\frac{2.2 \times 10^9}{1000}\)
= 2.2 × 10⁶ Nm^-2

Commonly Made Error
Student fail to comprehend the term compress it by 0.10%

Answering Tip
Student should have practice in solving numerical problems related to percentage.

Section – D (15 Marks)

Question numbers 38 to 40 carry 5 marks each.

Question 38.
Find the way for a swimmer who wants to cross the river in the shortest time. [5]
Answer:
Let \(\vec{v}_s\) and \(\vec{v}_r\) be the velocities of swimmer and river respectively.
Let \(\vec{v}\) = resultant velocity of υ_s and υ_r

Let the swimmer begins to swim at an angle O with the line OA where OA is perpendicular to the flow of river.
1ff = time taken to cross the river, then
t = \(\frac{l}{v_s \cos \theta}\)
where l = breadth of river
For t to be minimum, cos θ should be maximum,
i.e., cos θ = 1
This is possible if
θ = 0°
Thus, we conclude that the swimmer should swim in a direction perpendicular to the direction of flow of river.
If cos θ = 1 ⇒ θ = 90°
then shortest time,
t = \(\frac{l}{v_s}\)

Question 39.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track. Will the stones reach the bottom at the same time ? Will they reach there with the same speed ? Explain. Given θ₁ = 30°, θ₂ = 60° and h =10 m, what are the speeds and times taken by two stones ? [5]
OR
(a) Derive the second equation of motion using graphical method.
(b) On a two lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A from opposite directions with speeds of 54 km/h each. At a certain instant, when both car B and C are at a distance of 1 km from A, B decides to overtake car A before C does. What minimum acceleration of B is required to avert an accident ?
Answer:
AB and AC are two smooth planes or tracks inclined at θ₁ and θ₂ respectively.
No, two stones do not reach the bottom at the same time.

Explanation : Supposes nz1g and rn2g be the weights of the two stones on these planes respectively. The rectangular components of m₁g and m₂g are shown in given figure below.

When a₁ and a₂ be the accelerations produced in the stones 1 and 2 respectively, then clearly
m₁a₁ = m₁g sin θ₁
a₁ = g sinθ₁
Similarly a₂ = g sin θ₂
Also as θ₂ > θ₁
a₂ > a₂
it means a₁ = g sin 30° = \(\frac{g}{2}\)
and a₂ = g sin 60° = \(\frac{\sqrt{3}}{2}\)g
From the equation υ = u + at,
∴ υ = at
or t = υ/a …………………… (i)
Here u = 0 as the two stones are initially at rest
or t ∝ \(\frac{1}{a}\)
∴ \(\frac{t_2}{t_1}=\frac{a_1}{a_2}\) ………………… (ii)
Now as, a₂ > a₁ or \(\frac{a_1}{a_2}\) < 1 …………………. (iii)
From the equations (ii) and (iii),
∴ \(\frac{t_2}{t_1}\) < 1
or t₂ < t₁
It means second stone will take lesser time and reach the bottom earlier than the first stone, i.e.,stone on the steep plane reaches the bottom earlier Yes, the two stones reach the bottom with the same speed as explained follow:

Explanation: Let h = height of the plane at point
A = 10m
When υ₁ and υ₂ be the speeds of the two stones with which they reach at the bottom, then according to the law of conservation of energy at top = KE. at bottom
or m₁gh = \(\frac{1}{2}\)m₁υ₁²
and m₂gh = \(\frac{1}{2}\)m₂υ₂²
or υ₁ = \(\sqrt{2 g h}\)
and υ₂ = \(\sqrt{2 g h}\)
or υ₁ = υ₂
= \(\sqrt{2 \times 9.8 \times 10}\)
= \(\sqrt{196}\)
= 14ms^-1

OR

(a) Graphical derivation of 2nd equation of motion:
Distance travelled ‘s’ = Area of trapezium ABDO
Area of rectangle ACDO + Area of DABC
s =OD × OA + \(\frac{1}{2}\) × BC × AC
= t × u + \(\frac{1}{2}\)(υ – u)t
= ut + \(\frac{1}{2}\)(υ – u)t.

According to 1st eqn. of motion
υ = u + at
∴ υ – u =at
∴ s = ut + \(\frac{1}{2}\)at²

(b) Speed of car A
υ_A = 36km/h
= 36 × \(\frac{5}{18}\) m/s
= 10 m/s

Let υ_B and υ_C be the speed of car B & C
∴ υ_B = υ_C
= 54km/h
= 54 × \(\frac{5}{18}\) m/s
= 15m/s
∴ Relative speed of car B w.r.t. A i.e., υ_BA is given by
υ_BA = υ_B – υ_A
= (15 – 10) m/s
= 5 m/s
Also, relative speed of car C w.r.t A i.e. υ_CA is given by
υ_CA = υ_C – (-υ_A)
υ_CA = υ_C + υ_A
= (15 + 10) m/s
= 25 m/s
Also AB = AC = 1 km (given) = 1000 m
Let t = time taken by car C to travel distance AC
∴ Using relation s = ut
(∴ Car C is in uniform motion)
we get, t = \(\frac{\mathrm{AC}}{v_{\mathrm{CA}}}=\frac{1000 \mathrm{~m}}{25 \mathrm{~m} / \mathrm{s}}\)
= 40 s
Suppose a = acceleration of car B for t = 40 s; it will cover 1000 m.
s = ut + \(\frac{1}{2}\) at²
AB = υ_BA t + \(\frac{1}{2}\) at²
or 1000 = 5 × 40 + \(\frac{1}{2}\) a × (40)²
1000 = 200 + 800a
a = 1 m/s²

Question 40.
(a) Liquid drops are spherical in shape. Why ? [5]
(b) Derive the expression of excess pressure inside the liquid drop.
(c) Give two similarities and dissimilarities between friction and viscosity.
OR
Derive Dalton’s law of partial pressures on the basis of kinetic theory of gases.
Answer:
(a) Liquid drops : One consequence of surface tension is that free liquid drops and bubbles are spherical if effects of gravity can be neglected. A liquid air interface has energy, so for a given value the surface with minimum energy is the one with the least area. The sphere has this property. So, if gravity and outer forces were ineffective, liquid drops would be spherical and the pressure inside a spherical drop is more than that the pressure outside.

(b) If the drop is in equilibrium, this energy cost is balanced by the energy gain due to expansion under the pressure difference (p – P0) between the inside of the drop and the outside, the work done is
W = [P₁ – P₀]4πr²∆r = S × 4π2r∆r
i.e, [P₁ – P₀] = (2S/r)
In general, for a liquid gas interface, the concave side has a higher pressure than the convex side. For example, an air bubble in a liquid, would have higher pressure inside it.

Drop, cavity and bubble of radius r.
A bubble in fig (c) differs from a drop. Applying the above argument, we have
[P₁ – P₀] = (4S/r)
This is probably why you have to blow hard, but not too hard, to from a soap bubble. A little extra air pressure is needed inside.

(c) Similarities between friction & viscosity :

1. Viscosity is due to the friction between neighbouring particles in a fluid that are moving at different velocities. Fluid friction also called as viscosity.
2. Viscosity and friction both depends on the size and shape of its particles and the attractions between the particles.

Dissimilarities :

1. Viscosity is the measure of the resistance of a fluid, which is deformed by either shear stress or tensile stress. While friction is caused by the contact of two rough surfaces.
2. The viscosity of a fluid depends on the temperature. It decreases as the temperature is increased. While friction depends on the surface area.

Commonly Made Error
Students could not enlist similarities and dissimilarities between friction and viscosity.

Answering Tip
Students should be familiarized with the terms ‘friction’ and ‘viscosity’ in detail.

OR

Consider a mixture of gases occupying a volume V Let m₁, m₂, m₃…. be the molecular masses of the gases and n₁, n₂, n₃…. be the number of their molecules, P₁, P₂, P₃…. the pressure exerted by individual gases and υ₁, υ₂, υ₃…. be the r.m.s. velocities of the molecules of various gases.
According to kinetic theory

As the temperature of all the gases in the mixture is same, so their average K.E. will be equal that is,
\(\frac{1}{2}\) m₁υ₁² = \(\frac{1}{2}\) m₂υ₂² = \(\frac{1}{2}\) m₃υ₃² = ………… = \(\frac{1}{2}\) mυ²(say)
or m₁υ₁² = m₂υ₂² = m₃υ₃² = ………… = mυ²
Now,
P = P₁ + P₂ + P₃ + …. \(\frac{1}{3}\) (n₁ + n₂ + n₃ + ….) mυ²
= \(\frac{1}{3} \frac{m n}{\mathrm{~V}}\)υ²
where, n = n₁ + n₂ + n₃ + ………………… = total no. of molecules in the mixture.
But \(\frac{1}{3} \frac{m n}{V}\)υ² = P = total pressure exerted by the mixture.
It proves Dalton’s Law of partial pressure
∴ P = P₁ + P₂ + P₃ + ……..