CBSE Sample Papers for Class 11 Physics Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 5 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 pair of physical quantities does not have same dimensional formula? [1]
(A) Work and torque.
(B) Angular momentum and Planck’s constant.
(C) Tension and surface tension.
(D) Impulse and linear momentum.
Answer:
Option (C) is correct.

Explanation:
Tension = Force = [MLT^-2]
Surface Tension = \(\frac{\text { Force }}{\text { Length }}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}\)
= [ML⁰T^-2]

Question 2.
Three vectors \(\vec{\mathrm{A}}\), \(\vec{\mathrm{B}}\) and \(\vec{\mathrm{C}}\) add up to zero. Find which is false. [1]
(A) \(\vec{(A} \times \vec{B}) \times \vec{C}\) is not zero unless \(\vec{B}, \vec{C}\) are parallel.
(B) \(\vec{(A} \times \vec{B}) \cdot \vec{C}\) is not zero unless \(\vec{B}, \vec{C}\) are parallel.
(C) If \(\vec{A}, \vec{B}, \vec{C}\) define a plane. \(\vec{(A} \times \vec{B}) \times \vec{C}\) is in that plane.
(D) \(\vec{(A} \times \vec{B}) \cdot \vec{C}=|\vec{A}||\vec{B}||\vec{C}|\) and \(\vec{\mathrm{C}^2}=\vec{\mathrm{A}^2}+\vec{\mathrm{B}^2}\).
Answer:
Option (C) is correct.

Question 3.
A body with mass 5 kg is acted upon by a force, \(\vec{\mathrm{F}}=-(3 \hat{i}+4 \hat{j})\) N. If its initial velocity at t = 0 is \((\vec{u})=(6 \hat{i}-12\hat{j})\) ms^-1, the time at which it will just have velocity along the y-axis is [1]
(A) Never
(B) 10 s
(C) 2 s
(D) 15 s
OR
A car of mass m starts from rest and acquires a velocity along east υ = υ\(\hat{i}\) (υ > 0) in two seconds. Assuming the car
moves with uniform acceleration, the force exerted on the car is :
(A) \(\frac{m \vec{v}}{2}\) eastward and is exerted by the car engine.
(B) \(\frac{m \vec{v}}{2}\) eastward and is due to the friction on the tyres exerted by the road.
(C ) More than \(\frac{m \vec{v}}{2}\) eastward exerted due to the engine and overcomes the friction of the road.
(D) \(\frac{m \vec{v}}{2}\) Exerted by the engine.
Answer:
Option (B) is correct.

Explanation:
Given
m = 5 kg, \(\vec{\mathrm{F}}=-(3 \hat{i}+4 \hat{j})\) N, u = (6 \hat{i}-12\hat{j})[/latex] m/s
The acceleration of the body is
\(\vec{a}=\frac{\overrightarrow{\mathrm{F}}}{m}=\frac{-(3 \hat{i}+4 \hat{j}) \mathrm{N}}{5 \mathrm{~kg}}=-\left(\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right)\) m/s²
Velocity of the body along x-axis at any time t is
υ_x = u_x + a_xt = 6 – \(\frac{3}{5}\) t
As the body will have a velocity along y-axis, its velocity along x-axis will be zero.
i.e., υ_x= 0
or 6 – \(\frac{3}{5}\)t = 0 or t = \(\frac{30}{3}\) = 10 s

OR

Option (B) is correct.

Explanation:
Given, mass of the car = m

Question 4.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v as shown in Fig. [1]

If the collision is elastic, which of the following is a possible result after collision ?,

Answer:
Option (B) is correct.

Explanation:
Let m be the mass of each ball bearing.
Total kinetic energy of the system before collision,
= \(\frac{1}{2}\) mυ² + 0
= \(\frac{1}{2}\) mυ² = K = kinetic energy
In (A), K.E of the system after collision,
K₁ = \(\frac{1}{2}\) (2m) \(\left(\frac{v}{2}\right)^2\) = \(\frac{1}{4}\) mυ²
In (B), K.E of the system after collision,
K₂ = \(\frac{1}{2}\) (m) (υ)² = \(\frac{1}{2}\) mυ²
In (C), K.E of the system after collision,
K₃ = \(\frac{1}{2}\) (3m) \(\left(\frac{v}{3}\right)^2\) = \(\frac{1}{6}\) mυ²
In (D), K.E of the system after collision,
K₄ = \(\frac{1}{2}\) mυ² + \(\frac{1}{2}\) m \(\left(\frac{v}{2}\right)^2\) + \(\frac{1}{2}\) m \(\left(\frac{v}{3}\right)^2\)
= \(\frac{49}{72}\) mυ²
K.E is only conserved in (B).
∴ (B) is the only possibility.

Question 5.
Choose the wrong option : [1]
(A) Inertial mass is a measure of difficulty of accelerating body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass.
(B) That the gravitational mass and inertial mass are equal is an experimental result.
(C) That the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass.
(D) Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot.
Answer:
Option (D) is correct.

Question 6.
Particles of masses 2M, m and M are respectively at points A, B and C with AB = 1/2 (BC). m is much-much smaller than M and at time t = 0. They are all at rest. [1]
At subsequent times before any collision takes place :

(A) m will remain at rest.
(B) m will move towards M.
(C) m will moves towards 2M.
(D) m will have oscillatory motion.
Answer:
Option (C) is correct.

Explanation:

Gravitational force, F = \(\frac{\mathrm{GMm}{\mathrm{r}^{2}}\)
Let AB = r
∴ Force on B due to A,
∴ F_BA = \(\frac{\mathrm{G}(2 \mathrm{M} m)}{r^2}\) = 2F
Now, Force on B due to C, –
∴ F_BC = \(\frac{\mathrm{GM} m}{(\mathrm{BC})^2}=\frac{\mathrm{GM} m}{4 r^2}\) [∵ BC = 2 AB]
F_BC = \(\frac{\mathrm{F}}{\mathrm{4}}\) or F_BA < F_BC
Hence, m will move towards A i.e. 2M.

Question 7.
A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars. A mass m is suspended from the mid point of the wire. Strain in the wire is [1]

(A) \(\frac{x^2}{2 \mathrm{~L}^2}\)
(B) \(\frac{x}{\mathrm{L}}\)
(C) \(\frac{x^2}{\mathrm{L}}\)
(D) \(\frac{x^2}{2 \mathrm{~L}}\)
Answer:
Option (A) is correct.

Explanation:

Increase in length,
∆L = (PR + RQ) – PQ
= 2PR-PQ
∆L = 2 (L² + x²)^1/2 – 2L
= 2L \(\left(1+\frac{x^2}{\mathrm{~L}^2}\right)^{1 / 2}\) – 2L
= 2L \(\left[1+\frac{1}{2} \frac{x^2}{L^2}\right]\) – 2L (by binomial therorem)
strain = \(\frac{\Delta \mathrm{L}}{2 \mathrm{~L}}=\frac{x^2}{2 \mathrm{~L}^2}\)

Question 8.
A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in figure, indicate the one that represents the velocity (i>) of the pebble as a function of time (t). [1]

Answer:
Option (C) is correct.

Explanation:
As the pebble acquires terminal velocity after some time.

Question 9.
1 mole of an ideal gas is contained in a cubical volume. CDBAEFGH at 300 K. One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time: [1]

(A) the pressure on EFGH would be zero.
(B) the pressure on all the faces will the equal.
(C) the pressure of EFGH would be double the pressure on ABCD.
(D) The pressure on EFGH would be half that of ABCD.
OR
An aluminium sphere is dipped into water. Which of the following is true ?
(A) Buoyancy will be less in water at 0°C than that in water at 4°C.
(B) Buoyancy will be more in water at 0°C than that in water at 4°C.
(C) Buoyancy in water at 0°C will be same as that in water at 4°C.
(D) Buoyancy may be more or less in water at 4°C depending on the radius of the sphere.
Answer:
Option (D) is correct.
OR
Option (A) is correct.

Question 10.
Four pendulums A, B, C and D are suspended from the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement, [1]

(A) D will vibrate with maximum amplitude.
(B) C will vibrate with maximum amplitude.
(C) B will vibrate with maximum amplitude.
(D) All the four will oscillate with equal amplitude.
OR
Water waves produced by a motor boat sailing in water are
(A) neither longitudinal nor transverse.
(B) both longitudinal and transverse.
(C) only longitudinal.
(D) only transverse.
Answer:
Option (B) is correct.

Explanation:
As the length of A and C is same they will have same time and frequency of vibration. Due to which resonance will occur and C will vibrate with maximum
OR
Option (B) is correct.

Question 11.
Derive acceleration-time graph from the given velocity time graph. [1]

Answer:
Option (B) is correct.

Question 12.
What is gravitational unit of force ? [1]
Answer:
It is kilogram force.
1 kgf = 9.8 N

Question 13.
Why the mud from moving wheels of vehicle fly off tangentially ? [1]
OR
Why the sparks from a grinding wheel fly off tangentially ?
Answer:
Because of inertia of direction when mud (or sparks) leave the rotating wheel, they try of follow a tangential linear motion due to the absence of external force on them.

OR

It is done due to inertia of direction.

Question 14.
State the principle of conservation of total mechanical energy. [1]
Answer:
The total mechanical energy of a system remains same, if the working forces on it are of conservative nature.

Question 15.
Under what condition, the torque due to an applied force is zero ? [1]
Answer:
We know that τ = rFsin θ. If θ = 0° or 180°, or r = 0, then τ = 0, r = 0 means then applied force passes through the axis of rotation.

Question 16.
Why does a rubber ball bounce higher on hills than in plains ? [1]
Answer:
Acceleration due to gravity decreases with height so on hills g is less than plains. The ball bounces higher on hills because of this reason.

Commonly Made Error
Student could not relate the different rate of bouncing of ball on hills and in plains for acceleration due to gravity.

Answering Tip
The value of acceleration due to gravity varies at a height ‘h’, depth ‘d’, on equator and on poles.

Question 17.
Name the modules which is applicable for solids and fluids ? [1]
Answer:
Bulk modulus.

Question 18.
Is it possible that a particle moving with a constant velocity may not have a constant speed ? [1]
Answer:
No.

Question 19.
What type of motion particles of a medium execute when a wave passes though the medium ? [1]
OR
Why does evaporation cause cooling ?
Answer:
Particles of the medium execute simple harmonic motion about their mean position.

OR

In evaporation, escaping molecules have higher kinetic energy, hence the average kinetic energy of the molecules left behind decreases. As average kinetic energy is directly related with temperature, hence evaporation causes cooling.

Commonly Made Error
Student could not explain the reason for cooling caused by evaporation.

Answering Tip
Students should be familiar with the process of evaporation.

Question 20.
What is the nature of the force causing S.H.M. ? [1]
OR
Can mechanical waves travel through vacuum ?

Read the following text and answer any 4 of the following question on the basis of the same.
Molecular Perspective
In water, there are two types of molecules. Some molecules are at the surface, called exterior molecules, and some molecules are inside, called interior molecules. The interior molecules are attracted to all the molecules around them. The exterior molecules are attracted to only the other surface molecules and to those below the surface. So that the energy state of the molecules on the interior is much lower than that of the molecules on the exterior. The molecules always try to maintain a lower energy state and hence the exterior molecules experience a downward force. This force is known as cohesive force. As a result they try to maintain a minimum surface area, thus allowing more molecules to have a lower energy state. Thus surface tension is created.

The water molecules attract one another due to the water molecule’s polar property. The hydrogen ends, which are positive in comparison to the negative ends of the oxygen cause water to “stick ” together. This is why there is surface tension. Water has very high surface tension. It is 72.8 milli newton per meter at 20°C.
Answer:
A restoring force which is directly proportional to the displacement of a particle from its position and acts towards the centre (mean position) of the particle.

OR

No, mechanical waves cannot travel through vacuum.

Question 21.
The energy state of the interior molecules of a fluid is [1]
(A) Higher than the exterior molecules of the fluid
(B) Lower than the exterior molecules of the fluid
(C) Equal to the exterior molecules of the fluid
(D) Higher than or Equal to the exterior molecules of the fluid
Answer:
Option (B) is correct.

Explanation:
In water, there are two types of molecules. Some molecules are at the surface, called exterior molecules, and some molecules are inside, called interior molecules. The interior molecules are attracted to all the molecules around them. The exterior molecules are attracted to only the other surface molecules and to those below the surface. So that the energy state of the molecules on the interior is much lower than that of the molecules on the exterior.

Question 22.
The tendency of water to maintain ……………….. surface area is known as surface tension. [1]
(A) Maximum
(B) Minimum
(C) Fixed
(D) Energetic
Answer:
Option (B) is correct.

Explanation: The energy state of the molecules on the interior is much lower than that of the molecules on the exterior. The molecules always try to maintain a lower energy state. As a result, they try to maintain a minimum surface area, thus allowing more molecules to have a lower energy state. Thus, surface tension is created.

Question 23.
The water molecules attract one another due to the water molecule’s [1]
(A) Polar covalent bond
(B) Non-polar covalent bond
(C) Ionic bond
(D) Metallic bond
Answer:
Option (A) is correct.

Explanation:
The water molecules attract one another due to the water molecule’s polar property. The hydrogen ends, which are positive in comparison to the negative ends of the oxygen cause water to “stick” together.

Question 24.
Which one of the following is the correct molecular structure of water molecule? [1]

(A) Figure (a)
(B) Figure (b)
(C) Figure (c)
(D) Figure (d)
Answer:
Option (A) is correct.

Explanation:
Water is a polar molecule having positively charged sides where the two hydrogen atoms are found and a negatively charged end where the oxygen atom is located.

Question 25.
Surface tension of water is [1]
(A) 72.8 mN
(B) 72.8 N/m
(C) 72.8 mN/m
(D) 72.8 mN/m²
Answer:
Option (C) is correct.

Explanation:
Water has very high surface tension. It is 72.8 milli newton per metre at 20°C.

Section – B (10 Marks)

Question numbers 26 to 30 carry 2 marks each.

Question 26.
What can be represented by the graph given below, where d is height and v is velocity ? [2]

OR
What is the differences between the following two data ?
(i) 85 km/h east,
(ii) (8 h) (5 km/h east).
Answer:
This graph can be for a ball dropped vertically from a height d. It hits the ground with some downward velocity and bounces upto height d/2 where its upward velocity becomes zero.

OR

1. It is the product of a pure number and a vector (velocity). Hence the unit of product is the same as that of vector, i.e., the product is a velocity of 85km/h, towards east.
2. It is the product of a scalar (time) and a vector (velocity). Hence, the unit of the product will be h × (km/h)=km. Thus, the product is a displacement of magnitude 40 km, towards east.

Commonly Made Error
Students may not understand that the first case shows the multiplication of a vector quantity with a constant, whereas the second case shows the multiplication of a scalar quantity with a vector quantity.

Answering Tip
Students should carefully find out the difference between the multiplication in both the cases and then answer the same.

Question 27.
What is a spring force ? [2]
Answer:
Spring force is the restoring force arising because of compression or extension of spring.
Spring force, F ∝ (- x)
F = – kx,
where x is the change in length
F = – kx,
where k is a constant called spring constant or force constant. The negative sign indicates that the spring force is opposite to the direction of displacement of the spring from its normal position.

Commonly Made Error
Students generally forget to mark the negative sign in the relationship between the force and the change in length.

Answering Tip
Students should remember that the spring force always acts opposite to the direction of displacement of the spring and thus, include a negative sign the expression.

Question 28.
When the horse suddenly stops, the rider falls in forward direction. Why ? Explain it. [2]
Answer:
When the horse suddenly stops, the rider falls in the forward direction due to the inertia of motion.

Explanation:
The lower portion of the rider comes to rest along with the horse while the upper portion of the rider still continues to move forward. Hence, he falls forward.

Question 29.
What are the essential points of difference between sound and light waves? [2]
OR
The velocity of sound in air at NTP is 331 m/s. Find its velocity when the temperature rise to 91°C and its pressure is doubled.
Answer:

1. Sound waves are mechanical waves which are longitudinal in nature. Light waves are electromagnetic waves which are transverse in character.
2. Sound waves need a material medium for propagation, whereas light waves do not need any medium.
3. Velocity of sound waves in air at 0°C ≈ 332 m/s, whereas velocity of light waves in air/vacuum = 3 × 10⁸ m/s.

OR

Here, υ₀ = 331 ms^-1
Rise in temperature, t = 91°C
Since velocity of sound is not affected by the change in pressure, therefore we have to see the effect of temperature alone.
As υ₁ = υ₀ \(\sqrt{\frac{273+t}{273}}\) = 331 \(\sqrt{\frac{273+91}{273}}\)
= 331 \(\sqrt{1+\frac{91}{273}}\) = 331 \(\sqrt{1+\frac{1}{3}}\)
= 331 × \(\frac{2}{\sqrt{3}}\) = 382.2 ms^-1

Question 30.
Show that in S.H.M., the acceleration is directly proportional to its displacement at the given instant. [2]
Answer:
In S.H.M., the displacement of the particle at an instant is given by :
y = rsin ωt.
Velocity, υ = \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = rωcos ωt
Acceleration, a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = – ω²rsin ωt
= – ω²y.
From the above equation, we note that a ∝ y, i.e., acceleration in S.H.M. at an instant is directly proportional to the displacement of the particle from the mean position at that instant and opposite in direction.

Section – C (21 Marks)

Question numbers 31 to 37 carry 3 marks each.

Question 31.
Give an analytical method to find the vector sum of three vectors \(\vec{\mathrm{P}}, \overrightarrow{\mathrm{Q}}\) and \(\vec{\mathrm{R}}\) . [3]
Answer:

Question 32.
Define coefficient of friction and angle of friction and hence derive a relation between them. [3]
Answer:
The coefficient of friction between any two surfaces in contact is defined as the ratio of the force of limiting friction and normal reaction between them.
µ = \(\frac{\mathrm{F}}{\mathrm{R}}\)
Angle with the resultant of force of limiting friction F and normal reaction R makes with the direction of normal reaction R is angle of friction.

Relation :
In ∆AOC tan θ = \(\frac{\mathrm{AC}}{\mathrm{OA}}=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{\mathrm{F}}{\mathrm{R}}\) = µ
Hence, µ = tan θ

Commonly Made Error
While deriving the coefficient of friction and the angle of friction, students may commit error in using the correct angle and its component arms.

Answering Tip
Students must ensure themselves about the angle they are going to find and then, find the relationship using the suitable trigonometric function.

Question 33.
In a nerve impulse, about 105 neutrons are fired. If energy associated with discharge of a single neutron is 10^-10 J, estimate the energy used. [3]
Answer:
As energy associated with discharge of a single neutron is 10^-10 J, therefore total energy in a nerve impulse, where 10⁵ neutrons are fired is 10^-10 × 10⁵ = 10^-5 J.

Question 34.
Show that the area of the triangle contained between the vectors \(\vec{\mathrm{a}}\) and \(\vec{\mathrm{b}}\) is one-half of the magnitude of \(\vec{a} \times \vec{b}\) [3]
OR
A box is placed on horizontal measuring scale which reads zero when the box is empty. A stream of pebbles is then dropped into the box from a height h above its bottom at the rate of n pebbles per second. Each pebble has a mass m. If the pebbles collide with the box such that they immediately come to rest after collision. Calculate the scale reading at time t after the pebble begins to fill the box.
Answer:
Let \(\vec{a}\) be represented by \(\vec{OP}\) and \(\vec{b}\) be represented by \(\vec{OQ}\) . Let ∠POQ = θ

Perpendicular to Join PQ Draw QN ⊥ OP.
In ∆QNO, sin θ = \(\frac{\mathrm{QN}}{\mathrm{OQ}}=\frac{\mathrm{QN}}{b}\)
QN = bsin θ
Now, by definition,
\(|\vec{a} \times \vec{b}|\) = absin θ
= (OP)(QN)
= \(\frac{2(\mathrm{OP})(\mathrm{QN})}{2}\)
= 2 × area of ∆OPQ
∴ Area of ∆OPQ = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
Proved.

OR

Suppose, υ = speed of the pebble when it strikes the box.
Then, by applying υ² – u² = 2as,
we get, v² – 0 = 2gh
(∵ here a = g, s = h, u = 0)
or v = \(\sqrt{2 g h}\)
Mass of each pebble = m
Momentum of one pebble = mv = m\(\sqrt{2 g h}\)
Each pebble comes to rest after colliding the box.
∴ Change in momentum during one collision
= m\(\sqrt{2 g h}\)
∆t = Time for one collision = \(\frac{\mathrm{1}}{\mathrm{n}}\) s
Force exerted by pebbles on the box, is given by
F = \(\frac{\Delta p}{\Delta t}=\frac{m \sqrt{2 g h}}{1 / n}\)
= mn\(\sqrt{2 g h}\)
Weight of pebbles after time t is given by
W = mass of pebbles × g
= (mnt) g
∴ Scale Reading
F ‘ = F + W
= mn \(\sqrt{2 g h}\) + (mnt) g
= mn (\(\sqrt{2 g h}\) + gt)
n = no. of bullets fired per second at the tiger so as to stop it.
p_i = 0, before firing …………… (i)
p_f = n(mv) + MV ………….. (ii)
∴ From the law of conservation of momentum,
p_i = p_f
or 0 = n(mv) + MV
or n = –\(\frac{\mathrm{MV}}{m v}\)
= \(\frac{-60 \times(-10)}{0.05 \times 150}\)
= 80.

Commonly Made Error
Students may not take one of the values of velocities as negative.

Answering Tip
Students should carefully analyze the given question and do the calculations to get the desired answer.

Question 35.
Determine the volume contraction of a solid copper cube, 10 cm an edge, when subjected to a hydraulic pressure of 7 × 10⁶ Pa. (Bulk Modulus B for copper = 140 × 10⁹ Pa). [3]
OR
(i) Why does specific heat of gases increase with their atomicity ? [3]
(ii) Which is greater CP or CV ?
Answer:
Given : L = 10 cm = 0.1 m
B = bulk modulus of Cu
= 140 × 10⁹ Pa
P = 7 × 10⁶ Pa
∆V = Volume contraction of solid copper cube
∴ V = L³ = (0.1)³ = 0.001 m³.
Using formula, B = – \(\frac{\mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)}\)
we get ∆V = – \(\frac{\mathrm{PV}}{\mathrm{B}}=\frac{7 \times 10^6 \times 0.001}{140 \times 10^9} \mathrm{~m}^3\)
= \(\frac{1}{20}\) × 10^-6 m³
= – 0.05 × 10^-6 m³
= -5 × 10^-2 cm³
Here negative sign shows volume contraction.

Question 36.
Define coefficient of friction and angle of friction and hence derive a relation between them. [3]
OR
Three blocks are connected as shown below and are on a horizontal frictionless table. They are pulled to right with a force F = 50 N. If m₁ =5 kg, m₂ = 10 kg and m₃ = 15 kg, calculate tensions T₁ and T₂.
Answer:
The coefficient of friction between any two surfaces in contact is defined as the ratio of the force of limiting friction and normal reaction between them.
µ = \(\frac{\mathrm{F}}{\mathrm{R}}\)
Angle with the resultant of force of limiting friction F and normal reaction R makes with the direction of normal reaction R is angle of friction.

Relation :
In DAOC tan θ = \(\frac{\mathrm{AC}}{\mathrm{OA}}=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{\mathrm{F}}{\mathrm{R}}\) = µ
Hence, µ = tan θ

Commonly Made Error
While deriving the coefficient of friction and the angle of friction, students may commit error in using the correct angle and its component arms.

Answering Tip
Students must ensure themselves about the angle they are going to find and then, find the relationship using the suitable trigonometric function.

OR

Given : F = 50 N, m₁ = 5 kg, m₂ = 10 kg, m₃ = 15 kg,
Since, the three blocks move with an acceleration ‘a’
So, a = \(\frac{\mathrm{F}}{m_1+m_2+m_3}\)
or a = \(\frac{50}{5+10+15}=\frac{50}{30}\)
= \(\frac{5}{3}\) ms^-2
To determine T₂: Imagine the free body

Here, \(\vec{\mathrm{F}}\) and \(\vec{\mathrm{T}}_2\) act towards right and left respectively.
Since the motion is towards the right side, so according to Newton’s Second law of motion :

F – T₂ = m₃a
or 50 – T₂ = 15 × \(\frac{5}{3}\) = 25
or T₂ = 50 – 25 = 25 N.

To determine T₁: Consider the free body

Here m₁a = T₁
or T₁1 = m₁a
= 5 × \(\frac{5}{3}\) = \(\frac{25}{3}\)
= 8.33 N.

Question 37.
Discuss strings stretched between fixed points. [3]
Answer:
Let a spring be stretched between two points as shown in figure. Vibrations will set up in this string when it is disturbed, say from middle. The figure given below shows that vibration set up in this string appears to be similar to a half wave having wavelength \(\frac{\lambda}{2}\).

Length of the spring, l = \(\frac{\lambda}{2}\)
or λ = 2l
If T is tension in the string and m is mass per unit length of the string, then velocity of disturbance is given by
υ = \(\sqrt{\frac{\mathrm{T}}{m}}\)
also v = \(\frac{\text { Wavelength }}{\text { Time }}\)
= Wavelength × Frequency
= λf
or υ = 2l × f
or f = \(\frac{v}{2 l}\)
= \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{m}}\) (∵ υ = \(\sqrt{\frac{\mathrm{T}}{m}}\))

Section – D (15 Marks)

Question numbers 38 to 40 carry 5 mark each.

Question 38.
Express cross product of two vectors in cartesian co-ordinate. [5]
Answer:

It can be written in determinant form as
\(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array}\right|\)

Question 39.
Two identical steel cubes (masses 50g, side 1 cm) collide head-on face to face with a speed of 10 cm/s each. Find the maximum compression of each. Young’s modulus for steel = Y = 2 × 10¹¹ N/m². [5]
OR
What are the main features of gravitational force ?
Answer:
Mass of cube, m = 50g = 5.0 × 10^-2 kg
Speed of cube, v = 10cm/s = 1.0 × 10^-1 m/s
Young’s modulus Y = 2.0 × 10¹¹ N / m²
Side of cube (L) = 1 cm = 1.0 × 10^-2 m
Apply Hooke’s Law,
Young modulus, Y = \(\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
So, \(\frac{\mathrm{F}}{\Delta \mathrm{L}}=\frac{\mathrm{YA}}{\mathrm{L}}\) ……………….(i)
The force acting on each side
\(\frac{\mathrm{F}}{\Delta \mathrm{L}}\) = k …………….. (ii)
From eqn (i) & eqn. (ii),
k = \(\frac{\mathrm{YA}}{\mathrm{L}}=\frac{\mathrm{YL}^2}{\mathrm{~L}}\) (Here A = L²)
k = YL ………….. (iii)
Initial KE = 2 × \(\frac{1}{2}\) mυ² = 5.0 × 10^-4 J
Final PE = 2 × \(\frac{1}{2}\)k ∆l²
= k∆l² = YL∆l²
Apply Law of conservation of energy
YL∆l² = 5.0 × 10^-4
or, ∆l = \(\sqrt{\frac{5.0 \times 10^{-4}}{\mathrm{YL}}}\)
= \(\sqrt{\frac{5.0 \times 10^{-4}}{\left(2.0 \times 10^{11} \times 10^{-2}\right)}} \mathrm{m}\)
∆l = 5.0 × 10 m^-7

OR

Following are the main features of gravitational force :

1. It is always an attractive force.
2. It is independent of the medium between the particles.
3. It holds good over a wide range of distances (i.e., from interplanetary distances to interatomic distances).
4. It is an action-reaction pair, i.e., the force of attraction exerted by body A on body B is equal to the force of attraction exerted by body B on body A. However, the acceleration of the two bodies will not be equal.
5. The gravitational force between two particles is independent of presence or absence of other particles.
6. The total gravitational force on one particle due to number of particles is the resultant of forces of attraction exerted on the given particle due to individual particles, i.e.,
   F = F₁ + F₂ + F₃ + ……………
   It proves that the principle of superposition is valid.
7. It expresses, the force between point masses.
8. It is a conservative force, i.e., the work done in moving a particle once around a closed path under the action of gravitational force is zero.

Question 40.
Derive the formula for rise of liquid in a capillary tube (Ascent formula). [5]
OR
Derive an expression for work done in isothermal process.
Answer:
When one end of capillary tube of radius r is immersed into a liquid of density ρ which wets the sides of capillary tube (say water and capillary tube of glass), the shape of the liquid is in the tube becomes concave upward in figure.

Let R = Radius of curvature of liquid meniscus
P = Atmospheric pressure
S = Surface tension of the liquid,
The pressure at point A just above the liquid meniscus in the capillary tube is atmospheric pressure = P
The pressure at point B, just below the liquid meniscus (on convex side)
= P – (2S/R).
Pressure at points C and D, just above and below the plane surface of liquid in the vessel is also P (i.e., atmospheric pressure). The point E and D are in the same horizontal plane in liquid but the pressures at these points are different. Hence, they will not be in equilibrium.

In order to maintain an equilibrium the liquid level rises in the capillary tube upto a height h, so that the pressure at points D and E which are in the same level in liquid may become equal, from figure.
Now, pressure at E = pressure at B + pressure due to height h (= BE) of the liquid column
= (P – \(\frac{\mathrm{2S}}{\mathrm{R}}\)) + hρg
As there is equilibrium, therefore
Pressure at E = Pressure at D
i.e., P – \(\frac{\mathrm{2S}}{\mathrm{R}}\) + hρg = P
or hρg = \(\frac{\mathrm{2S}}{\mathrm{R}}\)
or h = \(\frac{2 S}{R \rho g}\)
Calculation of R. Let I be the centre of curvature of liquid meniscus GXY in the tube and GS be the tangent to the liquid surface at point G.
GI = R, GO = r
∠IGO = θ = angle of contact
In ∆IGO, cos θ = \(\frac{\mathrm{GO}}{\mathrm{GI}}\) = r/R
R = \(\frac{r}{\cos \theta}\)
Putting this value in (i) eqn.
h = \(\frac{2 \mathrm{~S} \cos \theta}{r \rho g}\)

OR

Suppose 1 gm mole of an ideal gas enclosed in a cylinder of conducting walls. Let P₁, V₁, T be initial pressure, volume, and temperature. Let gas expand to volume V₂ where pressure reduces to P₂ and temperature remains constant.
If A is the area of piston
F = P × A
dW = F × dx
= P × A × dx
W = \(\int_{V_1}^{V_2}\) P dV [∴ Adx = dV]
But, PV = RT
W = \(\int_{V_1}^{V_2} \frac{R T}{V} d V\)
W = RT \(\left[\log _e V\right]_{V_1}^{V_2}\)
W = RT[log_e V₂ – log_e V₁]
W = 2·303 RT log₁₀\(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\)

Commonly Made Error
Student could not find the expression for work done in an isothermal process.

Answering Tip
Step by step derivation for finding the work done in each type of processes should be understood.