CBSE Sample Papers for Class 11 Physics Set 10 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 10 with Solutions

Time Allowed : 3 hours
Maximum Marks : 70

General Instructions:

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

Question numbers 1 to 25 carry 1 mark each.

Question 1.
The number of significant figures in 0.06900 is : [1]
(A) 5
(B) 4
(C) 2
(D) 3
Answer:
Option (B) is correct

Question 2.
The angle between \(\overrightarrow{\mathrm{A}}\) = \(\hat{i}\) + \(\hat{j}\) and \(\overrightarrow{\mathrm{A}}\) = \(\hat{i}\) – \(\hat{j}\) is :  [1]
(A) 45°
(B) 90°
(C) -45°
(D) 180°
Answer:
Option (B) is correct.
Explanation:

Question 3.
A ball is travelling with uniform translatory motion. This means that
(A) It is at rest.
(B) The path can be a straight line or circular and the ball travels with uniform speed.
(C) All parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.
(D) The centre of the ball moves with constant velocity and the ball spins about its centre uniformly. [1]
Answer:
Option (C) is correct.

Question 4.
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,
(A) the two magnetic forces are equal and opposite, so they produce no net effect.
(B) the magnetic forces do not work on each particle.
(C) the magnetic forces do equal and opposite (but non-zero) work on each particle.
(D) the magnetic forces are necessarily negligible.  [1]
OR
A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is :
(A) + 2000 J
(B) -200 J
(C) zero
(D) -20,000 J
Answer:
Option (B) is correct.

Explanation: As the magnetic field due to motion of electron and proton act in a direction perpendicular to the direction of motion, no work is done by the forces. This is why one ignores the magnetic force of one particle on another.

OR

Option (C) is correct.
Explanation: Just because road does not move at all so the work done by the cycle on the road must be zero.

Question 5.
For which of the following does the centre of mass lie outside the body?  [1]
(A) A pencil
(B) A shotput
(C) A dice
(D) A bangle
OR
Which of the following point is the likely position of the centre of mass of the system shown in figure?  [1]

(A) A
(B) B
(C) C
(D) D
Answer:
Option (D) is correct
OR
Option (C) is correct.

Question 6.
The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity
(A) will be directed towards the centre but not the same everywhere.
(B) will have the same value everywhere but not directed towards the centre.
(C) will be same everywhere in magnitude directed towards the centre.
(D) cannot be zero at any point.  [1]
Answer:
Option (D) is correct

Question 7.
Modulus of rigidity of ideal liquids is :
(A) infinity
(B) zero
(C) unity
(D) some finite small non-constant value.  [1]
Answer:
Option (B) is correct.

Question 8.
A bimetallic strip is made of aluminium and steel (α_aluminium > α_steel). On heating, the strip will:
(A) remain straight
(B) get twisted
(C) will bend with aluminium on concave side.
(D) will bend with steel on concave side.
OR
A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
(A) its speed of rotation increases.
(B) its speed of rotation decreases.
(C) its speed of rotation remains same.
(D) its speed increases because its moment of inertia increases.  [1]
Answer:
Option (D) is correct.
Explanation:
α_aluminium > α_steel
∴ Aluminium will expand more than that of steel strip. Due to it, this steel strip will bend on concave side

OR

Option (B) is correct.

Explanation Its M.I increases.
∴ Angular momentum, L = Iω,
ω is angular speed which decreases to conserve L

Question 9.
An ideal gas undergoes four different processes from same initial state (Fig.). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic ?  [1]

(A) 4
(B) 3
(C) 2
(D) 1
Answer:
Option (C) is correct.

Question 10.
The displacement of a particle is represented by the equation y = 3 cos\(\left(\frac{\pi}{4}-2 \omega t\right)\). The motion of the particle is : 1
(A) simple harmonic with period 2π/w.
(B) simple harmonic with period π/w.
(C) periodic but not simple harmoniç
(D) non-periodic.  [1]
Answer:
Option (B) is correct. :
Explanation:
y = 3 cos\(\left(\frac{\pi}{4}-2 \omega t\right)\) = 3 cos \(\left[-\left(2 \omega t-\frac{\pi}{4}\right)\right]\)
= 3 cos \(\left(2 \omega t-\frac{\pi}{4}\right)\) [∵ cos(-θ) = cos θ]
This shows simple harmonic motion with time period, T = \(\frac{2 \pi}{2 \omega}\) = \(\frac{\pi}{\omega}\)

Question 11.
Write the dimensional formula of torque.  [1]
Answer:
[ML²r^-2].

Question 12.
What is meant by a point object in physics ?  [1]
Answer:
An object is said to be a point object if it is dimensionless.

Question 13.
Rocket works on which principle of conservation ?  [1]
Answer:
Law of conservation of linear momentum.

Question 14.
What do you mean by positive work ?  [1]
OR
What do you mean by power ?
Answer:
Positive work means that force (or its component) is parallel to displacement.
OR
Power is defined as the rate at which the body can do the work.

Question 15.
A girl is able to land on her feet after a fall. Which principle of physics is being used by her ?  [1]
OR
A particle moves in a circular path with decreasing speed. What happens to its angular momentum?
Answer:
Principle of conservation of angular momentum.
OR
As \(\overrightarrow{\mathrm{L}}\) = \(\overrightarrow{\mathrm{r}}\) × m\(\overrightarrow{\mathrm{v}}\) i.e., magnitude of \(\overrightarrow{\mathrm{L}}\) decreases but direction remains constant.

Question 16.
Name the natural satellite of earth. [1]
OR
What is the formula for escape velocity in terms of g and R ?
Answer:
Moon.
OR
v_e = \(\sqrt{(2 g R)}\)

Question 17.
Why do crystals have reproducible external shapes ?  [1]
Answer:
The flat surfaces of the crystal have their relative areas in fixed ratios and the angles between the flat faces also have fixed ratios.

Question 18.
What is an isothermal process ?  [1]
Answer:
Isothermal process is the process in which temperature variation does not exist. Such processes are to be carried in

1. conducting cylinders,
2. at a slow pace.

Question 19.
Is it possible to have longitudinal waves on a string ?  [1]
Answer:
No, because string is not stretchable. It can neither be compressed nor rarefied.

Commonly Made Error
Student could not predict the possibility of existence of longitudinal waves on a string.

Answering Tip
The distinction between longitudinal and ! transverse waves should be understood clearly.

Question 20.
How will the time period of a simple pendulum change when its length is doubled ?  [1]
Answer:
The time period becomes \(\sqrt{2}\) times the original value since T ∝ \(\sqrt{l} \text {. }\)
as we have T = \(2 \pi \sqrt{\frac{l}{g}}\)

Read the following text and answer any 4 of the following questions on the basis of the satire:

2 friends started for a picnic spot, in two different cars. A drove his car at a constant velocity 60 km/h. B drove his car at a constant velocity 50 km/h. The velocity of B relative to A is v_B – v_A. Similarly, the velocity of object A relative to object B is v_A – v_B. Their friend C was supposed to wait at a point on the road for a lift. Both of them forgot to pick up C. A and B reached the picnic spot within 2 hours and 2 hours 24 minutes respectively.

Question 21.
What was the velocity of B relative to A?  [1]
(A) 10 km/h
(B) – 10 km/h
(C) 55 km/h
(D) -55km/h
Answer:
Option (B) is correct.
Explanation: The velocity of B relative to A is
V_B – B_A = 50 – 60
= 10 km/h

Question 22.
What was the velocity of A relative to B?  [1]
(A) 10 km/h
(B) -10 km/h
(C) 55 km/h
(D) -55 km/h
Answer:
Option (A) is correct.

Explanation: The velocity of A relative to B is
V_A – V_B = 60 – 50
= 10 km/h

Question 23.
What were the velocities of A and B relative to C?  [1]
(A) 50 km/h, 60 km/h
(B) 60 km/h, 50 km/h
(C) Both 10km/h
(D) -55 km/h
Answer:
Option (B) is correct

Explanation: Since C was in stationary position, his velocity was 0. Hence the velocity of A relative to C was 60 – 0 = 60 km/h and the velocity of B relative to C was 50 – 0 = 50 km/h.

Question 24.
Which one of the following shows the Velocity vs. Time Plot for A?  [1]

Answer:
Option (C) is correct

Question 25.
How the distance of the picnic spot can be found from the appropriate graph?
(A) Length of the straight line is the required distance
(B) Area under the graph is the required distance
(C) Half of the area under the line is the required distance
(D) Distance of the origin from the end point of the line is the required distance  [1]
Answer:
Option (B) is correct
Explanation: Area under the graph gives the distance. It is 60 × 2 = 120 km.

Section – B

Question numbers 26 to 30 carry 2 marks each.

Question 26.
Is kinetic energy a scalar or a vector ? Give its S.I. unit and dimensional formula. [2]
Answer:
Kinetic energy is a scalar quantity
S.I. unit of kinetic energy is joule (J)
Dimensional formula of kinetic energy is [ML²T^-2].

Question 27.
What are absolute pressure and gauge pressure ?  [2]
Answer:
Pressure at a point is given by the relation
P = P_a + hρg
where, P_a is the atmospheric pressure and hρg is the column pressure. Here, P is the absolute pressure and (P – P_a) is the gauge pressure normally measured.

Question 28.
Is it possible to increase the temperature of a gas without adding heat to it ? If yes, then explain how ?
OR
Why does the temperature of a gas decrease, when it is allowed to expand adiabatically ?  [2]
Answer:
For an adiabatic change, first law of thermodynamics may be expressed as :
dU + PdV = 0
or dU = – PdV
The temperature of a gas will increase, if dU is positive. For this, dV has to be negative. Therefore temperature of a gas can be increased without adding heat, if it is compressed adiabatically.

OR

For an adiabatic change, the first law of thermodynamics may be expressed as
dU + VdV =0
or PdV = – dU
During expansion, dV is + ve. Therefore the equation (i) will hold, if dU is – ve, i.e., temperature decreases.

Question 29.
Liquids and gases cannot propagate transverse waves. Why ? [2]
Answer:
Liquids and gases cannot sustain shearing stress. Therefore, transverse waves in the form of crests and troughs (involving change of shape) are not possible in fluids. Rather, the fluid posses volume elasticity. Therefore, compressions and rarefactions (involving changes in volume) can be propagated through fluids.

Question 30.
What is the length of a simple pendulum which ticks seconds ?  [2]
OR
The displacement of a harmonic oscillator is given by x = αsin ωt + βcos ωt. What is the amplitude of the oscillation ?
Answer:
A simple pendulum which ticks seconds is a second pendulum. Its time period T = 2 s. If / is the length of this pendulum, then
T = \(2 \pi \sqrt{\frac{l}{g}}\)
or l = \(\frac{g \mathrm{~T}^2}{4 \pi^2}\)
= \(\frac{9.8 \times 2^2}{4 \times(22 / 7)^2}\) = 0.99m

OR

Given : x = αsin ωt + βcos ωt
Let a = r cos θ and P = r sin θ
Then x = r cos θ sin ωt + r sinθ cos ωt
= r sin (ωt + θ)
The amplitude of oscillation is r
α² = r²cos² θ ….(i)
β² = r²sin² θ ….(ii)
From (i) and (ii)
r = \(\sqrt{\alpha^2+\beta^2}\)

Section – C

Question numbers 31 to 37 carry 3 marks each.

Question 31.
A machine gun is mounted on the top of a tower 100 m high. At what angle should the gun be inclined to cover a maximum range of firing on the ground below ? The muzzle speed of bullet is 150 m/s. Take g = 10 m/s². [3]
Answer:
Step I: Horizontal range
R = ucos θ × t
where θ is the angle of inclination of gun to cover maximum range and t is the time.
i.e., R = 150cos θ × t …..(i)

Step III : Putting the value of t from eqn. (ii) in eq.

(i) we get,
R = 150cos θ(15 sin θ + \(\sqrt{225 \sin ^2 \theta-20}\))
Choose the value of θ between 43° to 47° to
calculate the value of R’.
At, θ = 43°, R = 2347 m
θ = 43.5°, R = 2347.7 m
θ = 44°, R = 2348 m
θ = 45°, R = 2346 m
θ = 46°, R = 2341 m
θ = 47°, R = 2334 m
In mean value of θ = \(\frac{43^{\circ}+43.5^{\circ}}{2}\) = 43.75° = 43.8°
Thus the required Lingle should be 43.8 to cover max. range.

Question 32.
Derive a relation for the velocity at the lowest point and the highest point for looping the loop of a vertical circle.  [3]
Answer:

Consider an object of mass m completing vertical circle of radius r with velocities υ_L at the lowest point and υ_T at the highest point.
Minimum velocity at the highest point.
Here, centripetal force = weight of the object + tension in the string
i.e., υ_T = \(\sqrt{g r}\)
Minimum velocity at the lowest point.
Here, centripetal force = tension in the string – weight of the object
then, \(\frac{m v_2^2}{r}\) = T₁ – mg
Also by using the law of conservation of energy, total energy of the object at T = kinetic energy of the object at L.

Question 33.
State and derive work- energy relationship.
OR
A block initially at rest breaks into two parts of masses in the ratio 2 : 3. The velocity of smaller part is (8\(\hat{i}\) + 6\(\hat{j}\)) m/s. Find the velocity of bigger part.  [3]
Answer:
It states that change in kinetic energy of a body is equal to work done and vice versa. Let a constant force \(\overrightarrow{\mathrm{F}}\) be applied to a body moving with initial velocity \(\vec{u}\), so that its velocity becomes \(\vec{v}\) along the direction of force when s is its displacement. Using Newton’s second law of motion we get magnitude of force F = ma and from equation of motion, we get v² – u² = 2as, where a is the acceleration of the body.
Multiplying both sides by m/2, we get
\(\frac{1}{2} m v^2\) – \(\frac{1}{2} m u^2\) = mas
i.e., \(\frac{1}{2} m v^2\) – \(\frac{1}{2} m u^2\) = Fs = W
i.e., K.E_(f) – K.E._(i) = W

where K.E_(f) is final kinetic energy and K.E_(i) is initial kinetic energy.
Thus work done on a body by a net force is equal to the change in kinetic energy of the body.

OR

Let mass of the block = m
After breaking, m₁ = \(\frac{2}{5}\)m and m₂ = \(\frac{3}{5}\)m
Linear momentum = \(m_1 \overrightarrow{v_1}\) + \(m_2 \overrightarrow{v_2}\)
According to law of conservation of momentum

Question 34.
What is binding energy of a satellite ?  [3]
Answer:
The minimum energy required to free a satellite from the gravitational attraction is called binding energy. Binding energy is the negative value of total energy of satellite. Let a satellite of mass m be revolving around earth of mass M and radius R.
∴ Total energy of satellite

∴ Binding energy of satellite = -[total energy of satellite]
= \(\frac{\mathrm{GMm}}{2 \mathrm{R}}\)

Question 35.
A steel cable with a radius of 1.5 cm supports a chair lift at a sky area. If the maximum stress is not to exceed 10⁸ Nm^-2, what is the maximum load the cable can support ?  [3]
Answer:
Given: Radius of steel cable,
r = 1.5 cm = 1.5 × 10^-2m.
Maximum Stress = 10⁸ Nm^-2.
∴ Area of cross-section of cable
A = πr² = π(1.5 × 10^-2)²
Maximum load the cable can stand = Maximum force

or Maximum force = Maximum stress × Area of cross-section
or F_max = 10⁸ × π × (1.5 × 10^-2)²
or F_max = 3.142 × 2.25 × 10⁸ × 10^-4 N.
or Maximum load the cable can withstand
= 707 × 10⁴ N.

Question 36.
Define Pascal law and give its practical applications.
OR
A razor blade can be made to float on water. What forces act on this blade ? Is Archimedes’ principle applicable ?  [3]
Answer:
Pascals Law : II states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same. Pascals law also states that the increase in pressure at one point of the gravity effect is neglected.

Applications : This principle is used to manufacture hydraulic lift. It consists of two cylinders-one of larger area of cross-section A and the other smaller piston of cross-sectional area a. Force is applied to smaller piston to produce a pressure,
P = \(\frac{\mathrm{F}}{a}\)
As per Pascal’s law same pressure is transmitted to larger piston. Then W = P × A.

Clearly large area A is producing more lifting force W.
Hydraulic brakes are also based upon Pascal’s law

OR
When a razor blade is made to float on water, three forces act on the blade:

1. Weight of the blade acting vertically downwards.
2. Reaction on blade exerted by the liquid surface acting vertically upwards.
3. Force of the surface tension on circumference of the blade acting tangentially to the liquid surface.

In this case, as no portion of razor blade is immersed in water, hence Archimedes principle is not applicable.

Commonly Made Error
A few students are not aware about the concept of surface tension.

Answering Tip
Students should learn about surface tension.

Question 37.
Given below are some examples of wave motion. State in each case, if the wave motion is transverse, longitudinal or a combination both.

1. Motion of a kink in a long coil spring produced by displacing one end of the string sideways.
2. Waves produced in a cylinder containing a liquid by moving its piston back and forth.
3. Waves produced by a motor boat sailing in water.  [3]

Answer:

1. When the spring is pulled sideways, the kink moves at 90° to the length of the spring. Waves are transverse.
2. Waves in this case are longitudinal, because molecules of the liquid will move along the direction of motion of the piston.
3. The water surface is cut laterally and pushed backwards by the propeller of motor boat.
   Therefore, the waves are a mixture of longitudinal and transverse waves.

Section – D

Question numbers 38 to 40 carry 5 marks each.

Question 38.
For an angular projection given to a projectile, find :
(i) Maximum height,
(ii) Time of flight,
(iii) Horizontal range

OR
The ceiling of a long hail is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^-1 can go without hitting the ceiling of the hall? [5]
Answer:
(i) Maximum height: It is the maximum vertical height attained by the object above the point of projection during its flight denoted by h.
y = u sin θ, a_y = -g, y₀ = 0
y = h, t = T/2 usin θ/g
Using relation,

(ii) Time of flight : The total time for which the projectile is in flight, taking vertical downward motion of object from O to C.
Let time taken to complete the trajectory = T

As thee projectile is reaching the same level of projection, vertical displacement, y = 0
We have, S = ut + \(\frac{1}{2}\)at²
0 = usin θ – \(\frac{1}{2}\)gT²
T = \(\frac{2 u \sin \theta}{g}\)

(iii) Horizontal range : It is the horizontal distance travelled by projectile during its flight i.e., horizontal distance covered by object while going
from O to C.
Distance = velocity × time
= ucos θ × T
= \(\frac{u \cos \theta \times 2 u \sin \theta}{g}\)
= \(\frac{u^2(2 \sin \theta \cos \theta)}{2 g}\)

OR

Question 39.
Find the relation between torque and angular momentum.
OR
Derive an expression for torque in polar coordinates. [5]
Answer:
In order to express torque as the rate of change of some quantity, we rewrite expression for torque of rotating a particle in XY plane as
\(\tau\) = xF_y – yF_x … (i)

If p_x = mυ_x and p = mυ_y, are the x and y components of linear momentum of the body, then
According to Newton’s 2^nd law of motion

Substituting in (i), we get

Now, Differentiating (xυ_y – yυ_x)

Substituting (iii) in (ii)

OR

Suppose the line of action of force F makes an angle α with x-axis
F_x = F cos α …. (i)
F_y = F cos α …….. (ii)
If x, y are the co-ordinates of the point p where \(\overrightarrow{\mathrm{OP}}\) = \(\vec{r}\) and ∠XOP = θ.

Then x = r cos θ
y = r sin θ …. (iii)
Substituting those values in \(\tau\) = (xF_y – yF_x),
\(\tau\) = (rcos θ)Fsin α – (r sin θ)(Fcos α)
\(\tau\) = rF[sin α cos θ – cos α sin θ] …. (iv)
\(\tau\) = rFsin(α – θ)
Let φ be the angle which the line of action of force \(\overrightarrow{\mathrm{F}}\) make, with the position vector
\(\overrightarrow{\mathrm{OP}}\) = \(\vec{r}\)
As clear from figure,
θ + φ = α
or φ = α – θ
Putting in (iv)
\(\tau\) = rFsinφ ….. (vi)
Equation (vi) is the expression for torque in polar co-ordinates

Question 40.
Explain surface energy. Establish its relation with surface tension.  [5]
Answer:
Surface energy is defined as the amount of the work done against the force of surface tension, in forming the liquid surface of a given area at a constant temperature.

To obtain an expression for surface energy, take a rectangular frame ABCD having a wire PQ which can slide along the sides AB and CD. Dip the frame in soap solution and form a soap him BCQP on the rectangular frame. There will be two free surfaces of film where air and soap are in contact.

Let S = Surface tension of the soap solution.
l = Length of the wire PQ.
Since there are two free surfaces of the film and surface tension acts on both of them, hence total inward force on the wire PQ is
F = S × 2l
To increase the area of the soap film we have to pull the sliding wire PQ outwards with a force F. Let the film be stretched by displacing wire PQ through a small distance x to the position P₁Q₁.
The increase in area of film PQQ₁P₁ in both sides
= 2(1 × x)
∴ Work done in stretching film is
W = Force applied × Distance moved
= (S × 2l) × x
= S × (2lx)
= S × a
where, 2lx = a increase in area of the film in both sides
If temperature of the film remains constant in this process. this work done is stored in the film as its surface energy.