Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 4 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, Re = 6.4 × 106 m
      Universal Gravitational constant. G = 6.67 × 10-11 Nm2kg-2
      mass of electron, me = 9.1 × 10-31 kg,
      mass of neutron, mn = 1.675 × 10-27 kg
      mass of proton, mp = 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.
A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is: [1]
(A) Frictional force along westward
(B) Muscle force along southward
(C) Frictional force along south-west
(D) Muscle force along south-west
Answer:
Option (C) is correct.

Explanation:
Direction of change in momentum will represent direction of force.

Question 2.
It is found that \(|\vec{\mathrm{A}}+\vec{\mathrm{B}}|\) = \(|\vec{\mathrm{A}}|\). This necessarily implies, [1]
(A) \(\vec{\mathrm{B}}\) = \(\vec{\mathrm{0}}\)
(B) \(\vec{\mathrm{A}}\), \(\vec{\mathrm{B}}\) are antiparallel.
(C) \(\vec{\mathrm{A}}\), \(\vec{\mathrm{B}}\) are perpendicular.
(D) \(\vec{\mathrm{A}}\) . \(\vec{\mathrm{B}}\) ≤ 0.
OR
The mean length of an object is 5 cm. Which of the following measurements is most accurate ?
(A) 4.9 cm
(B) 4.805 cm
(C) 5.25 cm
(D) 5.4 cm
Answer:
Option (B) is correct.

Explanation:
If \(|\vec{\mathrm{A}}+\vec{\mathrm{B}}|\) = \(|\vec{\mathrm{A}}|\), then either \(\vec{\mathrm{B}}\) = 0 or
\(\vec{\mathrm{B}}\) = -2\(\vec{\mathrm{A}}\). Both are satisfied when \(\vec{\mathrm{A}}\) and \(\vec{\mathrm{B}}\) are anti-parallel.
OR
Option (B) is correct.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 3.
A body is moving uni-directionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in correctly shows the displacement-time curve for its motion ? [1]
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 1
Answer:
Option (B) is correct.

Explanation:
for constant power, displacement,
s ∝ t3/2

Question 4.
In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1 m/s, at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m/s2, the kinetic energy of the shotput when it just reaches the ground will be [1]
(A) 2.5 J
(B) 5.0 J
(C) 52.5 J
(D) 155.0 J
Answer:
Option (D) is correct.

Explanation:
Initial K.E. = \(\frac{1}{2}\) mv2
= \(\frac{1}{2}\) (10kg ) × (1m/s)2
= 5 J
Initial PE at height 1.5 m = mgh
= (10 kg)(10 m/s2)(1.5 m)
= 150 J
Total initial energy = 155 J

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 5.
A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports. It can be done in one of the following three ways : [1]
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 2
The tension in the strings will be
(A) The same in all cases.
(B) least in (i)
(C) least in (ii)
(D) least in (iii)
Answer:
Option (C) is correct.

Explanation:
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 8
Let m be the mass and θ be the angle which the tension T in the string makes with the horizontal.
2Tsin θ = mg
T = \(\frac{m g}{2 \sin \theta}\) or T ∝ \(\frac{1}{\sin \theta}\)
T is least if sin θ has maximum value i.e.,
sin θ = 1 = sin 90°

Question 6.
Which of the following diagrams does not represent a streamline flow? [1]
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 3
Answer:
Option (D) is correct.

Explanation:
Lines of flow do not intersect each other.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 7.
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.
Answer:
Option (A) is correct.

Question 8.
Figure (below), shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is : [1]
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 4
OR
A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s2. The period of oscillation is : [1]
(A) π s
(B) \(\frac{\pi}{2}\) s
(C) 2π s
(D) \(\frac{\pi}{t}\) s
Answer:
Option (A) is correct.

Explanation:
Projection of OP on x-axis at time t is
x(t) = B cos\(\left(\frac{\pi}{2}-\frac{2 \pi}{\mathrm{T}} t\right)\) = B sin \(\left(\frac{2 \pi}{\mathrm{T}} t\right)\)
x(t) = B sin \(\left(\frac{2 \pi}{30} t\right)\) {∵ T = 30 s}

OR

Option (A) is correct.

Explanation:
υmax = ωA ……………… (i)
amax = ω2A ……………….. (ii)
Dividing (ii) by (i),
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 9

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 9.
Sound wave of wavelength λ travelling in a medium with a speed of υ m/s enters into another medium where its speed is 2υ m/s. Wavelength of sound wave in the second medium is : [1]
(A) λ
(B) \(\frac{\lambda}{2}\)
(C) 2λ
(D)4λ
Answer:
Option (C) is correct.

Explanation:
Wavelength in 1st medium,
λ = \(\frac{v}{v}\)
Wavelength in IInd medium,
λ’ = \(\frac{2 v}{\mathrm{v}}\)
{∴ frequency remains unchanged}
(ii) + (i), we get
\(\frac{\lambda^{\prime}}{\lambda}\) = 2
or λ’ = 2λ

Question 10.
Which of the following statements are true for wave motion ? [1]
(A) Mechanical transverse waves can propagate through all media.
(B) Longitudinal waves can propagate through solids only.
(C) Mechanical transverse waves can propagate through solids only.
(D) Longitudinal waves can propagate through vacuum.
OR
Speed of sound wave in air
(A) is independent of temperature.
(B) increases with pressure.
(C) increases with increase in humidity.
(D) decreases with increase in humidity.
Answer:
Option (C) is correct.
OR
Option (C) is correct.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 11.
What do you mean by surface tension? [1]
Answer:
The property of a liquid due to which its free surface tries to have minimum surface area is called surface tension. A small liquid drop has spherical shape due to surface tension.

Question 12.
The displacement is given by x = 2 + 4t + 5t2. Find the value of instantaneous acceleration.
(Here x is in m and t is s) [1]
OR
Will the displacement of a particle change on changing the position of origin of the coordinate system?
Answer:
Velocity, \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 4 + 10t
Acceleration, \(\frac{\mathrm{d}^{2} \mathrm{~x}}{\mathrm{dt}^{2}}\) = 10 m/s2

OR

No, it will not change.

Question 13.
If water in a flask and caster oil in other flask are violently shaken and kept on a table, then which one will come to rest earlier ? [1]
Answer:
Caster oil will came to rest earlier as its viscosity is more as compared to water.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 14.
When a pendulum clock gains time, what adjustments should be made ? [1]
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.

Question 15.
A ballet dancer stretches her hands out for slowing down. It is based on which principle of conservation ? [1]
Answer:
This is based on the principle of conservation of angular momentum.

Question 16.
Mention one difference between g and G. [1]
OR
A body has a mass of 10 kg; Find the distance of zero intensity of gravitational field.
Answer:
The value of ‘G’ remains the same throughout the universe while the value of ‘g varies from place to place.

Commonly Made Error
Confusion between the terms ‘g’ and ‘G’.

Answering Tip
Student should have basic understanding about the terms, their corresponding numerical values and variation of those values from place to place.
OR
I ∝ \(\frac{\mathrm{1}}{\mathrm{r}}\)
∴ For I = 0,
or r = ∞
Thus, intensity of gravitational field is zero at infinity.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 17.
What is the value of bulk modulus for an incompressible liquid ? [1]
Answer:
K = \(\frac{\text { stress }}{\text { strain }}=\frac{\text { stress }}{0}\) = ∞ (infinity)

Question 18.
Can mechanical energy be converted continuously and completely into heat ? Is the reverse also possible ? [1]
Answer:
The mechanical energy can be completely and continuously converted into heat. But the reverse is not possible.

Question 19.
Plot a curve to show Charles’s law. [1]
Answer:
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 10

Question 20.
When an air bubble rises in water what happens to its potential energy ? [1]
OR
Can potential energy of an object be negative ?
Read the following text and answer any 4 of the following questions on the basis of the same:
Newton’s Cradle : The device consists of a row of five metal balls positioned to just barely touch one another suspended from a frame by thin wires.
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 5
On a small cradles, the balls are hung from the crossbars by light wire, with the balls at the point of an inverted triangle. This ensures that the balls can only swing in one plane, parallel to the crossbars. If the ball could move on any other plane, it would impart less energy to the other balls in the impact or miss them altogether, and the device wouldn’t work as well, if at all.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

All the balls are, ideally, exactly of the same size, weight, mass and density. As long as the balls are all the same size and density, they can be as big or as small as you like. The balls must be perfectly aligned at the centre to make the cradle to make the cradle work the best.

When a ball on one end of the cradle is pulled away from the others and then released, it strikes the next ball in the cradle, which remains motionless. But the last ball on the opposite end of the row is thrown into the air, then swings back to strike the other balls, starting the chain reaction again in reverse.

This device illustrates the three main principles of Physics – conservation of energy, conservation of momentum and friction.

Everything that moves has momentum equal to its mass multiplied by its velocity. Like energy, momentum is also conserved. Momentum is a vector quantity, when 1st ball hits 2nd ball, it’s travelling in a specific direction, let’s say east to west. This means that its momentum is also moving east to west. Any change in direction of the motion brings a change in the momentum, which cannot happen without the influence of an outside force.

That is why 1st ball doesn’t simply bounce off 2nd ball, the momentum carries the energy through all the balls in a westward direction.

It is to remember that the law of conservation only works in a closed system, which is free from any external force. The Newton’s cradle is not a closed system. When 5th ball swings out away from the rest of the balls, it is affected by the force of gravity, which brings the ball down.

But, the horizontal line of balls at rest, functions as a closed system, free from any influence of any force other than gravity. It’s here, during the small time between the first ball’s impact and the 5th ball swinging out, that momentum is conserved.
Answer:
Potential energy of air bubble decreases and kinetic energy increases, because work is done by upthrust on the bubble.
OR
Yes, it can be negative when forces involved are attractive.

Question 21.
Newton’s cradle illustrates the three main principles of Physics: [1]
(A) Law of conservation of energy,
(B) Law of conservation of momentum and friction
(C) Law of conservation of mass
(D) Both (A) and (B)
Answer:
Option (D) is correct.

Explanation:
This device illustrates the three main principles of Physics-conservation of energy, conservation of momentum and friction.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 22.
In Newton’s cradle, the balls are hung from the crossbars by light wire, with the balls at the point of an inverted triangle. This ensures [1]
(A) that the balls can only swing in one plane, parallel to the crossbars
(B) that the balls can only swing in one plane, perpendicular to the crossbars
(C) that the balls do not fall down
(D) that the balls execute simple harmonic motion
Answer:
Option (A) is correct.

Explanation:
The balls are hung from the crossbars by light wire, with the balls at the point of an inverted triangle. This ensures that the balls can only swing in one plane, parallel to the crossbars.

Question 23.
When the 1st ball at one end of the cradle is pulled away and then released, then [1]
(A) the 1st ball strikes the 2nd ball in the cradle, which goes into motion. The 2nd ball strikes the 3rd ball which goes into motion and so on
(B) the 1st ball strikes the 2nd ball in the cradle, which remains motionless. But the last ball on the opposite end of the row is thrown into the air, then swings back to strike the other balls, starting the chain reaction again in reverse
(C) the 1stball strikes the 2nd ball in the cradle and no other movement is observed
(D) the 1st ball executes a simple harmonic motion. Other balls remain at rest
Answer:
Option (B) is correct.

Explanation:
When a ball on one end of the cradle is pulled away from the others and then released, it strikes the next ball in the cradle, which remains motionless. But the last ball on the opposite end of the row is thrown into the air, then swings back to strike the other balls, starting the chain reaction again in reverse.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 24.
What is a closed system? [1]
(A) System which is at rest
(B) System which has zero energy
(C) System which is free from any external force
(D) System which is free from any friction
Answer:
Option (C) is correct.

Explanation:
The law of conservation only works in a system, which is free from any external force. Such a system is called closed system. The Newton’s cradle is not a closed system throughout its operation.

Question 25.
When the momentum is conserved in Newton’s cradle? [1]
(A) Throughout the operation
(B) During the small time between the first ball’s impact and the 5th ball swinging out, that momentum is conserved.
(C) At the time when the 1st ball strikes the 2nd ball
(D) At the time when the 5th ball swing’s out
Answer:
Option (B) is correct.

Explanation:
Law of conservation only works in a closed system. But the Newton’s cradle is not a closed system throughout its operation. When 5th ball swings out away from the rest of the balls, it is affected by the force of gravity, which brings the ball down.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

But, the horizontal line of balls at rest, functions as a closed system, free from any influence of any force other than gravity. It’s here, during the small time between the first ball’s impact and the 5th ball swinging out, that momentum is conserved.

Section – B (10 Marks)

Question numbers 26 to 30 carry 2 marks each.

Question 26.
Which one of the following is greater ? [2]
(A) The angular velocity of the hour hand of a watch?
(B) The angular velocity of the earth around its own axis ? Why ?
OR
Suggest a suitable physical situation for the following graph :
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 6
Answer:
The angular velocity of hour hand of a watch is greater than the angular velocity of earth around its own axis.

Explanation:
We know that angular velocity (ω) of an object having time period (T) is given by 2n
ω = \(\frac{2 \pi}{T}\) ……………… (i)
(a) T for hour hand of a watch is 12 h
∴ ωh = \(\frac{2 \pi}{12}=\frac{\pi}{6}\) rad h-1 ……………. (ii)
(b) T for earth is 24 h
∴ ωe = \(\frac{24 \pi}{24}=\frac{\pi}{12}\) rad h-1 ……………. (iii)
Equations (ii), (iii) gives,
\(\frac{\omega_h}{\omega_e}=\frac{\pi / 6}{\pi / 12}\) = 2
or ωh = 2ωe
or ωh > ωe

Commonly Made Error
Students generally commit error in calculating the angular velocities for the watch and the earth.

Answering Tip
The correct formula and values for the watch and the earth should be used to calculate the angular velocities. Then, students should carefully calculate the ratio of angular velocities to know which one is greater.
OR
Suitable physical situation – A ball is thrown up with some initial velocity. It is rebounding from the floor with reduced speed after each hit. 2

Commonly Made Error
Students may write about the speed here instead of velocity.

Answering Tip
Students should observe the graph carefully. Since, the y-axis of the graph have some negative values, it cannot be speed.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 27.
The stress-strain graphs for materials A and B are shown in figure. The graphs are drawn on the same scale. [2]
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 7
(A) Which of the materials has the greater Young’s modulus ?
(B) Which of the two is stronger material ?
Answer:
(a) From graph it is clear that for a given strain, stress for A is more than that of B. Hence Young’s modulus (= stress/strain) is greater for A than that of B.
(b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question 28.
State second law of thermodynamics. [2]
OR
Derive Boyle’s law on the basis of kinetic theory of gases.
Answer:
Kelvin’s statement : No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. 1 Clausius statement : It is not possible to transfer heat from a body at lower temperature to another at higher temperature without the help of some external energy.

OR

According to kinetic theory, pressure, P exerted by a gas is
P = \(\frac{1}{3}\)ρυ2 = \(\frac{1}{3}\)Mυ2
∴ PV = \(\frac{1}{3}\) Mυ2
But at constant temperature total K.E. of gas \(\frac{1}{3}\) (M)υ2 or υ2 is constant.
∴ At a constant temperature, PV = constant.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 29.
The length of a seconds pendulum on the surface of earth is 100 cm. What will be the length of a seconds pendulum on the surface of moon ? [2]
Answer:
For a simple pendulum, time period (T) is given by
T = 2π \(\sqrt{\frac{l}{g}}\)
At Earth, T = 2π \(\sqrt{\frac{l}{g}}\)
At Moon, T’ = 2π \(\sqrt{\frac{l^{\prime}}{g^{\prime}}}\)
Here, T = T’ and g’ = \(\frac{1}{6}\) g
So, from equation (i) & (ii),
\(\frac{l}{g}=\frac{l^{\prime}}{g^{\prime}}\)
or, l’ = \(\frac{g^{\prime}}{g}\) × l
= \(\frac{1}{6}\) g × \(\frac{\mathrm{1}}{\mathrm{g}}\) × l
= \(\frac{1}{6}\) × l
= \(\frac{1}{6}\) × 100 cm
= 16.7 cm

Commonly Made Error
Students could not evaluate the length of seconds pendulum on the surface of moon.

Answering Tip
Acceleration due to gravity on the surface of moon = (1/6) acceleration due to gravity on the surface of earth.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 30.
What is phase in wave equation : [2]
y = r sin \(\left(2 \pi \frac{t}{\mathrm{~T}}+2 \pi \frac{x}{\lambda}\right)\)
Answer:
Rewriting y = r sin 2π\(\left(\frac{t}{\mathrm{~T}}+\frac{x}{\lambda}\right)\)
Here 2π\(\frac{x}{\lambda}\) is phase of the wave.

Section – C (21 Marks)

Question numbers 31 to 37 carry 3 marks each.

Question 31.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? [3]
Answer:
From the formula the horizontal range is given by
R = \(\frac{u^2 \sin 2 \theta}{g}\) ………………… (i)
For R = Rmax, θ = 45°, i.e., sin 2θ = sin 90° = 1
Putting the given value in eq. (i),
∴ Rmax = \(\frac{u^2}{g}\)
⇒ 100 = \(\frac{u^2}{g}\) (∵ Rmax = 100 given) ………….. (ii)
Suppose H = height upto which the ball goes when the cricketer throws it with velocity u. Since the final velocity of the ball, υ = 0.
∴ Applying the relation, υ2 – u2 = 2as,
(∵ here, υ = 0, a = – g, s = H)
or H = \(\frac{u^2}{2 g}\)
H = \(\frac{1}{2}\left(\frac{u^2}{g}\right)\)
= \(\frac{1}{2}\) × (100) [by using (ii)]
H = 50 m.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 32.
A bus moving with a velocity of 60 kmh-1 has a weight of 50 tonnes. Find out the force required to stop it in 10 s. [3]
Answer:
Given,
m = mass of bus = 50 tonnes
= 50 × 1000 kg
u = initial velocity of bus
= 60 kmh-1
= 60 × \(\frac{5}{18}\) ms-1 = \(\frac{50}{3}\) ms-1
υ = final velocity of bus = 0
t = 10 s, a = ? and retarding force, F = ?
∴ From relation,
υ = u + at,
we get, a = \(\frac{v-u}{t}\)
= \(\frac{0-\frac{50}{3}}{10}\) = –\(\frac{5}{3}\) ms-2
or -a = acceleration
= \(\frac{5}{3}\) ms-2
∴ F = m(-a) = retarding force
or F = 50000 × \(\frac{5}{3}\)
= \(\frac{250000}{3}\) = 83333.3 N.

Question 33.
A solid sphere of mass m and radius r is rolling on a horizontal surface. [3]
What fraction of total energy of the sphere is :
(i) Kinetic energy of rotation ?
(ii) Kinetic energy of translation ?
Answer:
Mass of sphere = m, radius = r
Moment of Inertia = \(\frac{2}{3}\) mr2
Total energy = KR + KT
KTotal = \(\frac{1}{2}\) Iω2 + \(\frac{1}{2}\) mυ2
= \(\frac{1}{2}\) × \(\frac{2}{5}\) mr2\(\left(\frac{v^2}{r^2}\right)\) + \(\frac{1}{2}\) mυ2 (∵ υ = rω)
KTotal = \(\frac{1}{2}\left(\frac{7}{5}\right)\) mυ2
(i) Fraction of K.E. of rotation = \(\frac{\mathrm{K}_{\mathrm{R}}}{\mathrm{K}_{\text {Total }}}\)
= \(\frac{\frac{1}{2}\left(\frac{2}{5}\right) m v^2}{\frac{1}{2}\left(\frac{7}{5}\right) m v^2}\) = \(\frac{2}{7}\)
(ii) Fraction of K.E. of translation
= \(\frac{\mathrm{K}_{\mathrm{T}}}{\mathrm{K}_{\text {Total }}}\) = \(\frac{\frac{1}{2} m v^2}{\frac{1}{2}\left(\frac{7}{5} m v^2\right)}\) = \(\frac{5}{7}\)

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 34.
Derive an expression for the gravitational potential energy above the surface of earth. [3]
OR
A saturn year is 29.5 times the earth year. How far is the Saturn from the Sun if the earth is 1.5 × 108 km away from the Sun ?
Answer:
Let the body of mass m be taken at height h above the surface of earth. At any instant of time t it reaches at a distance x from the centre of earth. The work done in raising through dx is.
dW = \(\frac{\mathrm{GM} m}{x^2}\) . dx
= ∆(P.E.)
Hence the work done in taking the body from surface of the earth (x = R) to a height h (x = R + h) is
P.E. = W
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 11
P.E. above the surface of earth = mgh

Commonly Made Error
Students lack clear understanding about integration of integrand and in use of binomial theorem.

Answering Tip
Students should have clear knowledge about mathematics which can help in solving numerical.
OR
Saturn Time period of saturn = Ts
Here, Ts = 29.5 Te
radius of earth Re = 1.5 × 108 km;
Rs = ? (Radius of saturn)
Using the relation,
\(\frac{\mathrm{T}_s^2}{\mathrm{R}_s^3}=\frac{\mathrm{T}_e^2}{\mathrm{R}_e^3}\)
or Rs = Re\(\left(\frac{T_s}{T_e}\right)^{2 / 3}\)
= 1.5 × 108 \(\left[\frac{29.5 \mathrm{~T}_e}{\mathrm{~T}_e}\right]^{2 / 3}\)
= 1.43 × 109 km

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 35.
Explain why small drops of mercury are spherical and large drops become flat ? [3]
Answer:
In case of a small drop of mercury force of gravity is small and force of surface tension plays a vital role. Therefore, the free surface of drop tends to have minimum surface area. For given volume the sphere has minimum area. Hence the small drops are of spherical shape.

In the case of large mercury drop, the gravitational pull becomes more effective than the surface tension and exerts downward pull on the drop so that its centre of gravity may lie at lowest possible position. Hence, the large drop of mercury becomes elliptical or flat in shape.

Question 36.
A geyser heats water flowing at the rate of 3 litres per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heating of combustion is 4.0 × 104 J/g ? [3]
OR
Estimate the fraction of molecular volume to the actual volume occupied by Oxygen gas at STE Take the diameter of an Oxygen molecule to be 3Å.
Answer:
The mass of water flowing per unit time
= 3000 g per min.
= \(\frac{3000}{60}\) = 50 g/s
= \(\frac{50}{1000}\) kg/s
Heat required to heat this water from 27°C to 77°C = mc ∆T.
= \(\frac{50}{1000}\) × 4.2 × 103 × 50 J/s
= 1.05 × 104 J/s
Heat generated by the combustion of 1 g of fuel
= 4.0 × 104 J
∴ Fuel required per second = \(\frac{1.05 \times 10^4}{4.0 \times 10^4}\)
= 0.2625 g
Fuel required to be burnt per min.
= 0.2625 × 60
= 15.75 g

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

OR

Given: diameter,
d = 3Å, radius = r = \(\frac{\mathrm{d}}{\mathrm{2}}\) = 1.5Å
= 1.5 × 10-8 cm
Molecular volume,
V = \(\frac{4}{3}\) πr3N
(Here, N = Avogadros number)
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × (1.5 × 10-8)3 × (6.023 × 1023)
= 8.52 cc
Let V’ be actual volume occupied by 1 mole of O2 at STP = 2.2400 cc.
∴ \(\frac{\mathrm{V}}{\mathrm{V}^{\prime}}=\frac{8.52}{22400}\) = 3.8 × 10-4

Commonly Made Error
Students lack understanding of the term STP.

Answering Tip
Students should be acquainted with the fact that STP stands for Standard temperature and pressure which is 1 atm pressure at 0°C.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 37.
A simple harmonic motion is represented by  [3]
x(t) = 10 sin (20t + 0.5)
Find its amplitude, frequency and initial phase.
Answer:
Comparing the given equation with standard equation of S.H.M.
x(t) = a sin (ωt + Φ)

  1. Amplitude, a = 10 m
  2. Angular frequency,
    ω= 20 rad s-1
  3. Frequency is given by
    v = \(\frac{\omega}{2 \pi}\) = 3.18 Hz
  4. Initial phase, Φ0 = 0.5 rad.

Section – D (15 Marks)

Question numbers 38 to 40 carry 5 marks each.

Question 38.
A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy ? You can neglect viscous drag of air and assume that density of air is constant. [5]
OR
Prove that coefficient of restitution/resilidnce of perfectly elastic collision in one dimension is unity.
Answer:
As dragging viscous force of air on balloon is neglected so there is net buoyant force = Vρg
Let m, V, ρHe, denote respectively the mass, volume and density of helium balloon and ρair density of air
Volume V of balloon displaces volume V volume of air.
So,
V(ρair – ρHe) g = ma,
or, V(ρair – ρHe)g = m\(\frac{\mathrm{dv}}{\mathrm{dt}}\) …………….. (i)
Integrating equation (i) with respect to t, we have
V(ρair – ρHe)gt = mυ
\(\frac{1}{2}\) mυ2 = \(\frac{1}{2} \frac{\mathrm{V}^2}{\mathrm{~m}}\) (ρair – ρHe)2g2t2
= \(\frac{1}{2m}\) V2air – ρHe)2g2t2 ……………. (ii)
If the balloon rises to height h, from h = ut + \(\frac{1}{2}\) at2
We get,
h = \(\frac{1}{2}\) at2 = \(\frac{1}{2} \frac{\mathrm{V}\left(\rho_{\text {air }}-\rho_{\text {He }}\right)}{m}\) gt2 (∵ u = 0) ………………… (iii)
From Eqs. (iii) and (ii),
\(\frac{1}{2}\) mυ2 = [V(ρair – ρHe) g] [\(\frac{1}{2m}\) V(ρair – ρHe) gt2]
= V(ρair – ρHe) gh
Rearranging the terms.
⇒ \(\frac{1}{2}\) mυ2 + VρHegh = Vρairgh
or KEballoon + PEballoon = change in PE of air.
So, as the balloon goes up, and equal volume of air comes down, increase in PE and KE of the balloon is at cost of PE of air [which comes down].

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

OR

Suppose two balls A and B of masses m1 and m2 are moving initially along the same straight line with velocities u1 and u2 respectively.
When u1 > u2, relative velocity of approach before collision,
= u1 – u2
Hence the two balls collide, Let the collision be perfectly elastic. After collision, suppose υ1 is velocity of A and υ2 is velocity of B along the same straight line. When υ2 > υ1, the bodies separate after collision.
Relative velocity of separation after collision
= υ2 – υ1
Linear momentum of the two balls before collision
= m1u1 + m2u2
Linear momentum of the two balls after collision
= m1υ1 + m2υ2
As linear momentum is conserved in an elastic collision, therefore
m1υ1 + m2υ2 = m1u1 + m2u2 ……………….(i)
or m22 – u2) = m1(u1 – υ1) …………. (ii)
Total K.E. of the two balls before collision
= \(\frac{1}{2}\) m1u12 + \(\frac{1}{2}\) m2u2>2 …………… (iii)
Total K.E. of the two balls after collision
= \(\frac{1}{2}\) m1υ12 + \(\frac{1}{2}\) m2υ2>2 ……………… (iv)
As K.E. is also conserved in an elastic collision, therefore from (iii) and (iv),
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 12
Hence, in one dimensional elastic collision relative velocity of separation after collision is equal to relative velocity of approach before collision.
From \(\frac{v_2-v_1}{u_1-u_2}\) = 1
by definition, \(\frac{v_2-v_1}{u_1-u_2}\) = e = 1

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 39.
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown with a velocity of half the first. The vertical gap between first and second balls is +15 m at f = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw. [5]
Answer:
Let the speeds of two balls (1 & 2) be υ1 and υ2, if υ1 = 2v, υ2 = v
If y1 and y2 the displacement covered by the balls 1 and 2, respectively, then
y1 = \(\frac{v_1^2}{2 g}=\frac{4 v^2}{2 g}\) and y2 = \(\frac{v_2^2}{2 g}=\frac{v^2}{2 g}\)
since, y1 – y2 = 15 m,
\(\frac{4 v^2}{2 g}-\frac{v^2}{2 g}\) = 15 m
or \(\frac{3 v^2}{2 g}\) = 15 m
υ2 = \(\sqrt{5 \mathrm{~m} \times(2 \times 10)}\)
= 10 m/s
clearly, υ1 = 20 m/s, υ2 = 10 m/s.
Time interval = 1 s.

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

Question 40.
Determine the numerical values of R and KB where in these alphabets are in their usual meanings. [5]
OR
State and explain Charles’s law.
Answer:
Numerical value of R : Consider one mole of a gas at STP, then
R = \(\frac{\mathrm{P}_0 \mathrm{~V}_0}{\mathrm{~T}_0}\)
Standard pressure
P0 = 0.76 m of Hg column
= 0.76 × 13.6 × 103 × 9.8 N/m2
Standard temperature = T0 = 273.15K
Volume of one mole of gas at
R = 22.4 × 10-3 m3
= \(\frac{0.76 \times 13.6 \times 10^3 \times 9.8 \times 22.4 \times 10^{-3}}{273.15}\)
= 8.31 J mole-1 K-1
In the C.G.S. system.
R = \(\frac{8.31}{4.2}\) cal mole-1 °C-1
= 1.98 cal mole-1 °C-1
Numerical value of KB :
We know that
KB = \(\frac{\mathrm{R}}{\mathrm{N}}\)
KB = \(\frac{8.31 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}}{6.02 \times 10^{23} \mathrm{~mole}^{-1}}\)
= 1.38 × 10-23 J/K

CBSE Sample Papers for Class 11 Physics Set 4 with Solutions

OR

It states that if the pressure remains constant, then the volume of a given mass of a gas increases or decreases by \(\frac{1}{273.15}\) times its volume at 0°C for each 1°C rise or fall in temperature.

Let V0 be the volume of the given mass of a gas at 0°C. According to Charles’s law, its volume at 1°C is
V1 = V0 + \(\frac{\mathrm{V}_0}{273.15}\)
= V0[1 + \(\frac{1}{273.15}\)]
Volume of the gas at 2°C
V2 = V0[1 + \(\frac{2}{273.15}\)]
∴ Volume of the gas at t°C
Vt = V0 [1 + \(\frac{t}{273.15}\)]
= V0\(\left(\frac{273.15+t}{273.15}\right)\)
If T0 and T are temperatures on kelvin scale corresponding to 0°C and t°C, then
T0 = 273.15 + 0 = 273.15
T = 273.15 + t
Vt = V0 \(\frac{\mathrm{T}}{\mathrm{T}_0}\)
\(\frac{\mathrm{V}_t}{\mathrm{~T}}=\frac{\mathrm{V}_0}{\mathrm{~T}_0}\)
\(\frac{\mathrm{V}}{\mathrm{T}}\) = constant
i.e., V ∝ T
CBSE Sample Papers for Class 11 Physics Set 4 with Solutions 13