CBSE Sample Papers for Class 11 Physics Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 8 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 mass and volume of a body are 4.237 g and 2.5 cm, respectively. The density of the material of the body in correct significant figures is : [1]
(A) 1.6048 g cm^-3
(B) 1.69 g cm^-3
(C) 1.7 g cm^-3
(D) 1.695 g cm^-3
Answer:
Option (C) is correct.
Explanation:
∴ ρ = \(\begin{gathered}
\text { mass } \\
\hline \text { volume }
\end{gathered}\) = \(\frac{4.237 \mathrm{~g}}{2.5 \mathrm{~cm}^3}\) = 1.6948 gcm^-3
So the density in correct significant figure will be 1.7 g cm^-3

Question 2.
Among the four graphs, there is only one graph for which average velocity over the time interval (0, t) can vanish for a suitably chosen t. Which one is it? [1]

A lift is coming from 8^th floor and is just about to reach 4^th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct ?
(A) x < 0, υ < 0, a > 0
(B) x > 0, υ < 0, a < 0 (C) x > 0, υ < 0, a > 0
(D) x > 0, v > 0, a < 0 Answer: Option (B) is correct Explanation: In (B) for the value of displacement, two timings are there. Therefore for one time, the average velocity is positive and for other time is equal but negative. Due to this average velocity can vanish. Commonly Made Error Most of the students are not able to understand this type problem. Answering Tip Practice is required for different graphs between different questions. OR Option (A) is correct Explanation: Lift is coming downwards. Thus, a is acting downwards, so a > 0 and the value of x becomes less, hence negative, i.e., x < 0, velocity is downwards (i.e., negative) so υ < 0.

Question 3.
A cricket ball of mass 150 g has an initial velocity (\(\vec{u}\)) = (\(3 \hat{i}\) + \(4 \hat{j}\)) ms^-1 after being hit. The change in momentum (final momentum – initial momentum) is (in kg ms^-1):  [1]
(A) Zero
(B) -(0.45\(\hat{i}\) + 0.6\(\hat{j}\))
(C) -(0.9\(\hat{i}\) + 1.2\(\hat{j}\))
(D) -5(\(\hat{i}\) + \(\hat{j}\))
Answer:
Option (C) is correct.
Explanation: Here, m = 150 g = 0.15 kg
\(\vec{u}\) = \((3 \hat{i}+4 \hat{j})\)m/s
\(\vec{v}\) = –\((3 \hat{i}+4 \hat{j})\) m/s
Initial momentum, p_i = mu
\(\vec{p}_i\) = (0.15 kg)\((3 \hat{i}+4 \hat{j})\)m/s
= \((0.45 \hat{i}+0.6 \hat{j})\) kg-m/s
Final momentum,

Question 4.
A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is :
(A) constant and equal to mg in magnitude.
(B) constant and greater than mg in magnitude.
(C) variable but always greater than mg.
(D) at first greater than mg, and later becomes equal to mg. [1]
Answer:
Option (D) is correct
Explanation: In the whole process, the man exerts a variable force (F) on the ground to set his body in motion. This force is in addition to the force required to support his weight (mg). Once the man is in standing position, F becomes, zero.

Question 5.
A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind show in figure. The moment of inertia about the z-axis is then

(A) increased
(B) decreased
(C) the same
(D) changed in unpredictable manner
Answer:
Option (B) is correct

Question 6.
Different points in earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the c.m. (centre of mass) causing translation and a net torque at the c.m. causing rotation around an axis through the c.m. For the earth-sun system (approximating the earth as a uniform density sphere)  [1]
(A) the torque is zero.
(B) the torque causes the earth to spin.
(C) the rigid body result is not applicable since the earth is not even approximately a rigid body.
(D) the torque causes the earth to move around the sun.
OR
Satellites orbiting the earth have finite life and sometimes debris of satellites fall to the earth. This is because,
(A) the solar cells and batteries in satellites run out.
(B) the laws of gravitation predict a trajectory spiralling inwards.
(C) of viscous forces causing the speed of satellite and hence height to gradually decrease.
(D) of collisions with other satellites.
Answer:
Option (A) is correct
OR
Option (C) is correct

Question 7.
The temperature of a wire is doubled. The Young’s modulus of elasticity
(A) will also double.
(B) will become four times.
(C) will remain same.
(D) will decrease.  [1]
Answer:
Option (D) is correct.

Question 8.
Consider P-V diagram for an ideal gas shown in figure

Out of the following diagrams which represents the T-P diagram ? [1]

OR
An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram. The amount of the work done by the gas is

(A) 6P₀V₀
(B) -2P₀V₀
(C) +2P₀V₀
(D) +4P₀V₀
Answer:
Option (C) is correct
OR
Option (B) is correct
Explanation: Work done = ΔP × ΔV
= (2P₀ – P₀)(3V₀ – V₀) = 2P₀V₀
∴ Cyclic process is anticlockwise,
∴ Work done by the gas is negative.

Question 9.
A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction. If the temperature is increased

(A) both P and V of the gas will change.
(B) only P will increase according to Charles’s law.
(C) V will change but not P.
(D) P will change but not V.  [1]
Answer:
Option (C) is correct.
Explanation: P = \(\frac{F}{A}\) = \(\frac{m g}{A}\) = constant
∴ V ∝ T (at constant pressure)

Question 10.
The relation between acceleration and displacement of four particles are given below :
(A) a_x = + 2x.
(B) a_x = + 2x²
(C) a_x = – 2x²
(D) a_x = – 2x.
Which one of the particles is executing simple harmonic motion ?  [1]
Answer:
Option (D) is correct.
Explanation: Because for SHM,
a_x ∝ (-x)

Question 11.
What is the sign of work done by a frictional force ? [1]
Answer:
The frictional force act opposite to direction of motion, hence the work done by a frictional force is a negative

Question 12.
Find the angle between \(\overrightarrow{\mathrm{A}}\) = \(\hat{i}\) + \(\hat{j}\) – 2\(\hat{k}\) and \(\overrightarrow{\mathrm{B}}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\). [1]
Answer:
We have

Question 13.
Can a rocket operate in free space ?
OR
Why is it easier to roll than to pull a barrel along a road ? [1]
Answer:
yes.
OR
It is due to the fact that rolling friction is less than sliding friction.

Question 14.
Convert 1 kWh in Joule.
OR
A spring is stretched. Is the work done by the stretching force positive or negative ? [1]
Answer:
1 kWh = 1000 × 3600 Ws = 3.6 × 10⁶ J.
OR
Positive, because the force and the displacement are in the same direction.

Question 15.
Name at least two quantities each whose dimensions are:
(a) [ML^-1T²],
(b) [ML²T^-1].  [1]
Answer:
(a) Pressure and stress.
(b) Planck’s constant and angular momentum.

Question 16.
Where does a body weight more at the pole or at the equator ?  [1]
Answer:
A body weighs more at the pole.

Question 17.
Name the parameter which is measure of the degree of elasticity of a body. [1]
OR
What is Van der Waals force ?
Answer:
Coefficient of restitution.
OR
It is force of attraction between molecules. The origin of this force is electrical.

Question 18.
What is Mayer’s relation ? [1]
Answer:
Mayer’s relation is C_P – C_V = R.

Question 19.
Why a point on a rotating wheel cannot be considered as executing S.H.M. ? [1]
Answer:
The motion of rotating points on wheel is not oscillatory but is only periodic.

Question 20.
What is the distance between a compression and its nearest rarefaction in a longitudinal wave ? [1]
Answer:
Distance between a compression and adjoining rerefaction is λ/2.

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

Large distances such as the distance of a planet or a star from the earth cannot be measured directly with a metre scale. An important method in such cases is the parallax method.

When you hold a pencil in front of you against some specific point on the background (a wall) and look at the pencil first through your left eye A (closing the right eye) and then look at the pencil through your right eye B (closing the left eye), you would notice that the position of the pencil seems to change with respect to the point on the wall. This is called parallax. The distance between the two points of observation is called the basis. In this example, the basis is the distance between the eyes.

To measure the distance D of a far away planet S by the parallax method, we observe it from two different positions (observatories) A and B on the Earth, separated by distance AB = b at the same time as shown in Fig.
We measure the angle between the two directions along which the planet is viewed at these two points. The ∠ASB in figure represented by symbol θ is called the parallax angle or parallactic angle.

As the planet is very far away, \(\frac{b}{D}\) < < 1 and therefore, θ is very small. Then we approximately take AB as an arc of length b of a circle with centre at S and the distance D as the radius AS = BS so that AB = b = DG where θ is in radians, then
D = \(\frac{b}{\mathrm{q}}\)
Answer:

Question 21.
In parallax method the distance between two observation points is called [1]
(A) Base
(B) Basis
(C) Distance
(D) Basic
Answer:
Option (B) is correct.
Explanation: The distance between the two observatories is the “basis”.

Question 22.
A star at a distance 8 × 10⁸ km is being observed from Vainu Bappu Observatory, Kavalur and ARIES Observatory, Nainital is being observed. The distance between the observatories is 2500 km. What is the “basis” of this measurement?  [1]
(A) 2500 km
(B) 1250 km
(C) 5000 km
(D) Data is insufficient to calculate
Answer:
Option (A) is correct.

Question 23.
A star at a distance 2.5 × 10⁸ km is being observed from Vainu Bappu Observatory, Kavalur and ARIES Observatory, Nainital. The distance between the observatories is 2500 km. What is the value of parallax angle? [1]
(A) 625 × 10⁸ radian
(B) 10^-5 radian
(C) 10⁵ radian
(D) 10^-5 degree
Answer:
Option (B) is correct.
Explanation: b = Dθ
b = 2500 km, D = 2.5 × 10⁸ km
So, θ = \(\frac{b}{D}\) = \(\frac{2500}{2.5 \times 10^8}\) = 10^-5 radian

Question 24.
Under which condition D = \(\frac{b}{q}\) relation is valid?  [1]
(A) \(\frac{b}{D}\) << 1 (B) \(\frac{b}{D}\) >> 1
(C) \(\frac{D}{b}\) << 1
(D) \(\frac{D}{b}\) = 1
Answer:
Option (A) is correct

Question 25.
When you hold an ice-cream cone in front of you against some specific point on the background (a wall) and look at it first through your left eye (closing the right eye), it seems to be at position B. Now if you look at it through your right eye (closing the left eye), you will notice that now it is at position A.
What is the “basis” of this observation?

(A) AB
(B) LR
(C) \(\frac{(\mathrm{LR}+\mathrm{AB})}{2}\)
(D) LR + AB  [1]
Answer:
Option (B) is correct.

Section – B

Question numbers 26 to 30 carry 2 marks each.

Question 26.
Why automobiles tyres have generally irregular projections over their surface ? [2]
OR
Explain how lubricants reduce friction.
Answer:
The automobile tyres have generally irregular projections over their surface so as to :

1. Increase friction,
2. Increase the grip with the ground and thus avoiding their skidding.

OR

The lubricants spread as a thin layer between the two surfaces. Lubricants like oil, grease fill up the irregularities of the surfaces, making them smoother.

Question 27.
In given figure a block is placed at height of 5m, find its speed at the bottom. [2]

A silica glass rod has a diameter of 1 cm and is 10 cm long. Estimate the largest mass that can be hung from it without breaking it. (breaking stress of glass = 50 × 10⁶ Nm^-2)
Answer:
Here, υ = \(\sqrt{2 g h}\)
where, g = 10 m/s² and h = 5m
∴ v = \(\sqrt{2 \times 10 \times 5}\) ms
= 10 ms^-1

OR

Breaking strength of glass is 50 × 10⁶ Nm^-2.
Using, Stress = \(\frac{F}{A}\),
we get, F = stress × A

Question 28.
Why is mercury preferred as a thermo-metric fluid ? [2]
Answer:

1. Mercury is a shining bright fluid so it is easily visible in the glass tube.
2. It is good thermal conductor.

Question 29.
When a pendulum clock gains time, what adjustments should be made ? [2]
Answer:
When a pendulum clock gains time, it means it has gone fast, i.e., it makes more vibrations per day than required.

This shows that the time period of oscillations has decreased. Therefore, to correct it, the length of pendulum should be properly increased.

Commonly Made Error
Mostly students are confused to solve this problem.

Answering Tip
Students should learn about concept of simple pendulum.

Question 30.
At what distance from the mean position is the K.E. in simple harmonic oscillator equal to EE. ? [2]
Answer:
When the displacement of a particle executing S.H.M. is y, then its
K.E. = \(\frac{1}{2} m \omega^2\left(A^2-y^2\right)\)
and P.E. = \(\frac{1}{2} m \omega^2 y^2\)
If K.E. = P.E.
then, \(\frac{1}{2} m \omega^2\left(A^2-y^2\right)\)
= \(\frac{1}{2} m \omega^2 y^2\)
or 2y² = A²
∴ y = \(\frac{A}{\sqrt{2}}\)

Section – C

Question numbers 31 to 37 carry 3 marks each.

Question 31.
On a 60 km straight road, a bus travels the first 30 km with a uniform speed of 30 kmh^-1. How fast must the bust travel the next 30 km so as to have average speed of 40 kmh^-1 for the entire trip?  [3]
Answer:

Question 32.
Define centripetal acceleration. Find the expression for it. Give one example of centripetal force. [3]
Answer:
Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
Expression : Consider a particle of mass m, moving with a constant speed υ and uniform angular velocity ω, on a circular path of radius r with centre at O.


when Δt → 0 then \(\frac{|\Delta \vec{v}|}{\Delta t}\) represents the magnitude of centripetal acceleration at B, which is given by

Example of centripetal acceleration is a stone moved around tied to the string.

Question 33.
Prove that instantaneous power is given by the dot product of force and velocity, i.e., P = \(\overrightarrow{\mathrm{F}} \cdot \vec{v}\)  [3]
Answer:
Suppose, ΔW be the amount of work done in a small time interval Δt, when P_av be the average power, then
P_aυ = \(\frac{\Delta W}{\Delta t}\) ….. (i)
When P be the instantaneous power, then

where, \(\vec{F}\) = constant force producing a displacement
∴ From equations (ii) and (iii),
P = \(\frac{d}{d t}(\overrightarrow{\mathrm{F}} \cdot \vec{s})\)
= \(\overrightarrow{\mathrm{F}} \cdot \frac{\overrightarrow{d s}}{d t}\) = \(\overrightarrow{\mathrm{F}} \cdot \vec{v}\)

Question 34.
What are the conditions under which a rocket fired from the earth, launches an artificial satellite of earth ?
OR
Why does moon has no atmosphere ?  [3]
Answer:
Following are the basic conditions :

1. The rocket must take the satellite to a suitable height above surface of earth.
2. From the desired height, the satellite must be projected with a suitable velocity, called the orbital velocity.
3. In the orbital path of satellite, the air resistance should be negligible so that its velocity does not decrease and it does not burn due to the heat produced.

OR

Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on surface of moon is small. Therefore, the value of escape velocity on the surface of the moon is small (only 2-5 km.s^-1). The molecules of the atmospheric gases on the surface of the moon have thermal velocities greater than the escape velocity. That is why all the molecules of gases have escaped and there is no atmosphere on moon.

Question 35.
An air bubble of radius r rises steadily through a liquid of density ρ at the rate of υ. Neglecting density of air, find the coefficient of viscosity of liquid.
OR
What is column pressure ? Derive a relation for the same.  [3]
Answer:
Buoyant force = weight of liquid displaced

OR

Pressure exerted by a liquid due to its height is called column pressure.
Consider two points X and Y to be lying on the top and bottom circular faces of an imaginary cylinder of liquid. Let area of the circular faces be a each and height of the cylinder be h. If pressure exerted at point X is P_x and at Y is P_y, then

F_x = P_xa acting downward. Weight of this cylinder, W = mg = Vρg = ahρg is also acting downward. So total downward force = F_x + W = P_xa + ahρg
The lower face of the cylinder experiences upward force given by
F_y = P_ya.
In equilibrium, F_y = F_x + W
or P_ya = P_xa + ahρg
or (P_y – P_x) = hρg
or P = hρg

Question 36.
Calculate the average kinetic energy for one molecule of gas at constant volume. [3]
Answer:
Pressure exerted by one mole of gas

∴ Total random K.E. for one mole = \(\frac{3}{2}\) RT
and average K.E. per molecule = \(\frac{3}{2}\)K_BT

Question 37.
Velocity and displacement of a body executing S.H.M. are out of phase by π/2. Why ?  [3]
Answer:
Let displacement on y-axis is given by
y = asin ωt,
where a is the amplitude, ω is the angular velocity and t is the instantaneous time.
Then velocity is given by
υ = \(\frac{d y}{d x}\) = aωcos ωt = aωsin \(\left(\frac{\pi}{2}+\omega t\right)\)

The above equation clearly shows that displacement and velocity of a body executing S.H.M. are out of phase by \(\frac{\pi}{2}\)

Section – D

Question numbers 38 to 40 carry 5 marks each.

Question 38.
Define null vector. What are its properties ? What is its physical significance ? [5]
OR
What do you understand by rectangular resolution of a vector ? Resolve it into its two rectangular components.
Answer:
It is defined as a vector having zero magnitude and acting in the arbitrary direction. It is denoted by \(\overrightarrow{0}\).

Properties of null vector:

(i) The addition or subtraction of zero vector from a given vector is again the same vector.
i.e., \(\overrightarrow{A}\) – \(\overrightarrow{0}\) = \(\overrightarrow{A}\)

(ii) The multiplication of null vector by a non-zero real number is again the zero vector.
i.e., n.\(\overrightarrow{0}\) = \(\overrightarrow{0}\)

(iii) If n₁ \(\overrightarrow{\mathrm{A}}\) = n₂ \(\overrightarrow{\mathrm{B}}\), where n₁ and n₂ are non-zero real numbers. Then the relation will hold good.
if \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{\mathrm{B}}\) = \(\overrightarrow{\mathrm{0}}\)
i.e., both \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are null vectors.

Physical significance of null vector: It is useful in describing the physical situation involving vector quantities.
e.g., \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{\mathrm{0}}\)
\(\overrightarrow{\mathrm{0}}\) × \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{\mathrm{0}}\)

OR

It is defined as the factors of splitting a given vector in two or three component vectors at right angles to each other. The components vectors are called rectangular component of the given vector.

Let \(\vec{R}\) be the given vector acting in X-Y plane at an angle θ with X-axis. Draw CA and CB on X and Y axes respectively. If \(\vec{P}\) and \(\vec{Q}\) be the rectangular components of \(\vec{R}\) along X and Y axes respectively, then

Now in right angled ΔOAC,


Also according to Δ law of vector addition,

Thus if \(\overrightarrow{\mathrm{A}_x}\) and \(\overrightarrow{\mathrm{A}_y}\) be the two rectangular components of \(\overrightarrow{\mathrm{A}}\) along X and Y axes respectively,
then \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{\mathrm{A}}_x \hat{i}\) + \(\vec{A}_y \hat{j}\)

Question 39.
A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig). The coefficient of friction between block and the in line is 0.1. If the block is pushed up by 10 m along the incline, calculate

(A) work done against gravity
(B) work done against force of friction
(C) increase in potential energy
(D) increase in kinetic energy
(E) work done by applied force.
OR
A curved surface is shown in Fig. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C.

With the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.
(A) For which ball is total mechanical energy conserved ?
(B) Which ball (s) can reach D ?
(C) For balls which do not reach D, which of the balls can reach back A ?  [5]
Answer:
m = 1 kg, θ = 30°, cos 30° = 0.866, sin 30° = 0.5
F = 10 N, μ = 0.1
Distance d = 10 m
(A) W_g = mg sin θ = 1 × 10 × 0.5 × 10 = 50J
(B) W₁ = μmg cos θ = 0.1 × 1 × 10 × 0.866 × 10
= 8.66J
(C) ΔU = mgh = 1 × 10 × 5 = 50J
(D) a = {F – (mgsin 30° + μmgcos 30°)}
or, a = 10 – (1 × 10 × 0.5 + 0.1 × 1 × 10 × 0.87)
= 4.13 m/s²
Apply 3rd Kinematic equation of motion,
υ² – u² = 2ad
Change in KE,
ΔK = \(\frac{1}{2} m v^2\) – \(\frac{1}{2} m v^2\) = mad = 41.3J

(E) Work done = Force × displacement
= 10 × 10 J
= 100 J

OR

(A) Force of friction is zero and negligible for ball 1 & 3 respectively, so energy is conserved for balls 1 and 3.
(B) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.
(C) Ball 3 have negligible friction & crosses C so ball can not reach at A.
Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.

Question 40.
State and explain Gay Lussac’s law. [5]
Answer:
It states that if the volume remains constant, the pressure of a given mass of a gas increases or decreases by its pressure at 0°C for each 1°C rise or fall in temperature.
If P₀ and P_t are the pressure of a given mass of gas at 0°C and t°C respectively, then according to Gay Lussac’s law
P_t = P₀\(\left(1+\frac{t}{273.15}\right)\)
or P_t = P₀\(\left(\frac{273.15+t}{273.15}\right)\)
or P_t = P₀\(\frac{\mathrm{T}}{\mathrm{T}_0}\)
where T₀ (K) = 273.15
and T(K) = 273.15 + t
\(\frac{P_t}{P_0}\) = \(\frac{\mathrm{T}}{\mathrm{T}_0}\)
\(\frac{\mathrm{P}}{\mathrm{T}}\) = constant
or P ∝ T