Kerala Plus One Physics Question Paper June 2022 with Answers

Reviewing Kerala Syllabus Plus One Physics Previous Year Question Papers and Answers Pdf June 2022 helps in understanding answer patterns.

Kerala Plus One Physics Previous Year Question Paper June 2022

Time: 2 Hours
Total Scores: 60

Answer any five questions from 1 to 7. Each carries 1 score. (5 × 1 = 5)

Question 1.
The branch of physics that deals with the study of light is __________ (Mechanics, Optics, Electrodynamics)
Answer:
Optics

Question 2.
The restoring force developed in a spring extended by a length x is, F = -kx. The dimensional formula of k is?
Answer:

Question 3.
\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|\). The angle between vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is __________
Answer:
90°

Question 4.
Write the value of an object’s escape velocity from the moon’s surface.
Answer:
2.37 km/s

Question 5.
Pick the correct one:
When the temperature of a liquid increases, its surface tension __________ (increases, decreases, remains the same)
Answer:
decreases

Question 6.
A light body and a heavy body have equal momentum. Which one has greater kinetic energy?
Answer:
A lighter body has greater KE. (KE ∝ \(\frac{1}{m}\))

Question 7.
In the case of stationary waves, the displacement of a particle at positions of node is __________
Answer:
0

Answer any 5 questions from 8 to 14. Each carries 2 scores. (5 × 2 = 10)

Question 8.
The position-time graph of an object is shown below. Draw the velocity-time graph and find the total displacement.

Answer:

Displacement = Area
= 2 × 3 + (-3 × 2)
= 0

Question 9.
A javelin is projected at an angle of 30° with an initial velocity of 5 ms-1 from the ground. What are its velocity and acceleration at the highest point?
Answer:
At the highest point,
V_y = 0
V_x = u cos θ
u = 5 m/s
θ = 30°
V_x = 5 × cos 30°
= 5 × \(\frac{\sqrt{3}}{2}\)
= 4.33 m/s²
a = -g = -9.8 m/s²

Question 10.
A person firing a bullet from a gun experiences a backward jerk:
(a) Write the principle behind this. (1½)
(b) A bullet of mass 15 g fired with a velocity of 100 ms^-1 from a gun of mass 2 kg. Find the recoil speed of the gun. (1½)
Answer:
(a) Newton’s 3rd law
or
Newton’s third law
(b) Recoil velocity, V = \(\frac{mv}{M}\)
m = 15 g = 15 × 10^-3 kg
v = 100 ms^-1
M = 2 kg
V = \(\frac{15 \times 10^{-3} \times 100}{2}\) = 0.75 m/s

Question 11.
Draw the variation of kinetic and potential energy of a freely falling body, with height.
Answer:

Question 12.
The stress-Strain curve of steel wire is shown below. From the plot, identify the following:

(a) A
(b) B
(c) C
(d) OO’
Answer:
(a) A → Proportional limit
(b) B → Elastic limit
(c) C → Breaking or tractive point
(d) OO’ → Residual strain

Question 13.
Draw the block diagram of a refrigerator and write an expression for its coefficient of performance.
Answer:

Question 14.
The angular speed of a rotating body changes from ω₁ to ω₂ without applying an external torque due to a change in moment of inertia. Find the ratio of radii of gyration in the two cases.
Answer:
According to the law of conservation of angular momentum

Answer any 6 questions from 15 to 22. Each carries 3 scores. (6 × 3 = 18)

Question 15.
Check the dimensional correctness of the relation:
v = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\)
where v is orbital velocity, G is gravitational constant and R is the radius of Earth.
Answer:

= (M⁰ L² T^-2)^1/2
= M L T^-1
Dimension of LHS = Dimension of RHS
∴ This equation is dimensionally correct.

Question 16.
Obtain an expression for the centripetal acceleration of a body in a uniform circular motion.
Answer:
Acceleration in uniform circular motion

The velocity vectors \(\vec{v}\) and \(\overrightarrow{\mathrm{v}^1}\) are always perpendicular to \(\vec{r}\) and \(\overrightarrow{\mathrm{r}^1}\). Therefore the triangle CPP1 (formed by the position Vectors \(\vec{r}\) and \(\overrightarrow{\mathrm{r}^1}\)) and the triangle GHI (formed by velocity vectors \(\vec{v}\) and \(\overrightarrow{\mathrm{v}^1}\)) are similar triangles. Therefore the ratio of base length to side length for one of the triangles is equal to the other triangle.

Question 17.
Show that the total mechanical energy of a freely falling body is constant.
Answer:

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
The potential energy of A,
PE = mgh
Kinetic energy, KE = \(\frac{1}{2} m v^2\) = 0
(since the body is at rest, v = 0)
∴ Total mechanical energy = PE + KE
= mgh + 0
= mgh

Total energy at the point B
The body travels a distance x when it reaches B.
The velocity at B can be found using the formula.
v² = u² + 2as
v² = 0 + 2gx
∴ KE at B = \(\frac{1}{2} m v^2\)
= \(\frac{1}{2}\)m2gx
= mgx
P.E. at B = mg(h – x)
Total mechanical energy = PE + KE
= mg(h – x) + mgx
= mgh
Total energy at C
Velocity at C can be found using the formula
v² = u² + 2as
v² = 0 + 2gh
∴ KE at C = \(\frac{1}{2} m v^2\)
= \(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE
= 0 + mgh
= mgh

Question 18.
The value of acceleration due to gravity decreases with depth.
(a) Derive an expression for acceleration due to gravity (g) at a depth ‘d’ below the surface of Earth. (2)
(b) What is the weight of a body at the center of Earth? (1)
Answer:
(a)

If we assume the earth as a sphere of radius R with uniform density r,
mass of earth = volume × density
M = \(\frac{4}{3} \pi \mathrm{R}^3 \rho\) ……..(1)
We know acceleration is due to gravity on the surface,

(b) At the center of the earth, g = 0
Weight = mg = 0

Question 19.
(a) State Pascal’s law. (1)
(b)

The schematic diagram of a hydraulic lift is shown in the figure. Using Pascal’s law, find the value of F₂ in terms of A₁ and A₂. (A₁ and A₂ are the areas of pistons). (2)
Answer:
(a) Pascal’s Law: Whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions.
(b) A hydraulic lift is used to lift heavy loads. Consider a liquid enclosed in a vessel with two cylinders C₁ and C₂ attached as shown in the figure. The cylinders are provided with two pistons having areas A₁ and A₂ respectively.
If F₁ is the force exerted on the area A₁, pressure P₁ = \(\frac{\mathrm{F}_1}{\mathrm{~A}_1}\)
If F₂ is the force exerted on the area A₂, pressure P₂ = \(\frac{\mathrm{F}_2}{\mathrm{~A}_2}\)
According to Pascal’s law, P₁ = P₂
i.e., \(\frac{F_1}{A_1}=\frac{F_2}{A_2}\)
F₂ = F₁ × \(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\)
When \(\frac{\mathrm{A}_2}{\mathrm{~A}_1}\) >>> 1, we get F₂ >>> F
Using this/method we can lift heavy loads by applying a small force.

Question 20.
(a) Among the various modes of heat transfer, which one is the fastest? (1)
(b) A pan filled with hot food cools from 94°C to 86°C in 2 minutes, when the room temperature is at 20°C. How long will it take to cool from 71°C to 69°C? (2)
Answer:
(a) Radiation
(b) T₁ = 94°C, T₂ = 86°C, t = 2 minutes = 120 s

Question 21.
The pressure exerted by a gas is given by the equation:
P = \(\frac{1}{3} \mathrm{~nm} \overline{\mathrm{~V}}^2\)
n is the number density of molecules, m is the mass of each molecule and \(\overline{\mathrm{V}}^2\) is the mean of squared speed. Find the average kinetic energy of a molecule.
Answer:

Question 22.
The displacement of a particle executing SHM is y = A sin ωt
A = amplitude
ω = angular frequency
(a) Draw the variation of velocity of the particle with time. (1)
(b) A 2 kg body performs SHM of amplitude 20 cm and the restoring force acting at that position is 50 N. Find the acceleration and kinetic energy of the particle when the displacement is 10 cm. (2)
Answer:

Answer any 3 questions from 23 to 27. Each carries 4 scores. (3 × 4 = 12)

Question 23.
A particle is moving along the X-axis with uniform positive acceleration.
(a) Obtain an expression for displacement by drawing v-t graph. (2)
(b) A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a tower of height 25 m from the ground. How long does it remain in the air? (g = 10 ms^-2) (2)
Answer:

t = 5 or -1
time can not be negative hence t = 5s

Question 24.
A child of mass 30 kg is standing inside a lift.
(a) What happens to the apparent weight of the child if the lift is moving up with a uniform acceleration? (1)
(b) Write an expression for the apparent weight of the child. (1)
(c) Find the apparent weight of the child, if the lift is moving down with constant acceleration of 5 ms-2. (2)
Answer:
(a) Apparent weight increases
(b) F = ma = R – mg
Apparent weight, R = ma + mg
(c) For lift moving down
f = ma = mg – R
R = mg – ma = m(g – a)
g = 10 m/s², a = 5 m/s², m = 30kg
∴ R = m(g – a)
= 30(10 – 5)
= 150 N

Question 25.
(a) ‘τ’ is the torque and ‘L’ is the angular momentum of the rotating rigid body, showing that τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}\). (2)
(b) Calculate a day when Earth shrinks to \(\frac{1}{8}\)th of its initial volume. (2)
Answer:
(a) Relation between angular momentum and torque
Angular momentum of a particle, \(\vec{l}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}\)
When differentiating on both sides, we get

Question 26.
A solid sphere of radius ‘r’ and density ‘ρ’ is falling through a viscous medium of density ‘σ ‘ and coefficient of viscosity ‘η’.
(a) What are the different forces acting on the body? (1½)
(b) Derive an expression for terminal velocity. (2½)
Answer:
(a) 1. Viscous drag, Fv = 6πηrv
2. Weight = mg
3. Bouyant force = mlg
(b) Consider a sphere of radius ‘a’ density ρ falling through a liquid of density s and viscosity η. The viscous force acting on the sphere can be written as F = 6πaηv
Where v is the velocity of the sphere. This force is acting in an upward direction. When the body comes down its velocity will increase. Hence the viscous force will also increase. When the viscous force ’ is equal to the weight of the body in the medium. The net force on the body is zero. Now the body falls without acceleration. It moves with a constant velocity called the terminal velocity. The weight of a body in a medium,

Question 27.
(a) Draw the pattern of waveforms of the first two harmonics in a closed pipe. (2)
(b) Show that in a closed pipe, the frequencies of the first two harmonics are in the ratio 1 : 3. (2)
Answer:

Answer any 3 questions from 28 to 32. Each carries 5 scores. (3 × 5 = 15)

Question 28.
Banking of roads helps to increase the limit of the maximum safe speed of a vehicle at a curve.
(a) Draw the schematic diagram of a vehicle on a banked road and mark the various forces acting on it. (2)
(b) Obtain an expression for the maximum safe speed of a vehicle at a banked road with frictional force. (3)
Answer:
(a)

(b) Consider a vehicle along a curved road with the angle of banking θ. Then the normal reaction on the ground will be inclined at an angle θ with the vertical. The vertical component can be divided into N cos θ (vertical component) and N sin θ (horizontal component). Suppose the vehicle tends to slip outward. Then the frictional force will be developed along the plane of the road as shown in the figure. The frictional force can be divided into two components. F cos θ (horizontal component) and F sin θ (vertical component).
From the figure get
N cos θ = F sin θ + mg
N cos θ – F sin θ = mg ………(1)
The component N sin q and F sin q provide centripetal force. Hence

Dividing both numerator and denominator of L.H.S by N cos θ. We get

This is the maximum speed at which a vehicle can move over a banked curved road.

Question 29.
Assuming satellite orbits around a planet to be circular,
(a) Obtain expressions for the orbital velocity and period of a satellite around Earth. (3)
(b) Write one use each for geostationary and polar satellites. (2)
Answer:
(a) Consider a satellite of mass m revolving with orbital velocity ‘v’ around the earth at a height ‘h’ from the surface of the earth.
Let M be the mass of the earth and R be the radius of the earth.

The gravitational force of attraction between earth and satellite.
F = \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}\)
Centripetal force required for the satellite
F = \(\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})}\)
For stable rotation,
Centripetal force = Gravitational force
i.e., \(\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}\)
v = \(\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}\) …………(1)
The above equation shows that orbital velocity decreases as h increases.
Period of satellite
The period of a satellite is the time taken by the satellite to revolve once around the planet in a fixed orbit.

(b) Geostationary Satellite – Communication
Polar satellite – spy work, mapping

Question 30.
The flow of an ideal fluid in a pipe of varying cross-sections is shown below:

(a) State and prove Bernoulli’s principle. (4)
(b) Write any two characteristics of a fluid that obey Bernoulli’s principle. (1)
Answer:
(a) Bernoulli’s theorem: As we move along a streamline the sum of the pressure (p), the kinetic energy per unit volume \(\frac{\rho v^2}{2}\) and the potential energy per unit volume (ρgh) remains a constant.
(OR)
The total energy of an incompressible non-viscous liquid flowing from one place to another without friction is a constant.
Mathematically Bernoulli’s theorem can be written as \(p+\frac{1}{2} \rho v^2+\rho g h\) = constant.
Proof

Consider an incompressible liquid flowing through a tube of non-uniform cross-section from region 1 to region 2.
Let P₁ be the pressure, A₁ the area of the cross-section, and V₁ the speed of flow at the region 1.
The corresponding values in Region 2 are P₂, A₂, and V₂ respectively.
Region 1 is at a height of h₁ and region 2 is at a height of h₂.
Theworkdoneontheliquidinatime Δt at the region 1 is given by
W₁ = force × distance
= P₁ A₁ Δx₁
= P₁ ΔV₁ (∵ A₁ Δx₁ = ΔV)
Where Δx₁ is the displacement produced at region 1, during the time interval Δt.
Similarly, the work done in a time Δt at the region 2 is given by,
W₂ = -P₂ A₂ Δx₂
W₂ = -P₂ ΔV₂
[Here -ve sign appears as the direction of \(\overrightarrow{\mathrm{p}}\) and Δx are in opposite directions.]
Net workdone DW = P₁ ΔV₁ – P₂ ΔV₂
According the equation of continuity
ΔV₁ = ΔV₂ = ΔV
ΔW = P₁ ΔV – P₂ ΔV
∴ ΔW = (P₁ – P₂) ΔV ………..(1)
This work changes the kinetic energy, pressure energy, and potential energy of the fluid.
If Δm is the mass of liquid passing through the pipe, in a time Δt, the change in Kinetic energy is given by

Change in gravitational potential energy is given by
Δp.E = Δmgh₂ – Δmgh₁
Δp.E = Δmg(h₂ – h₁) ………..(3)
According to the work-energy theorem work done is equal to the change in kinetic energy plus the change in potential energy.
ie; ΔW = ΔKE + ΔPE ………..(4)
Substituting eq. 1, 2, and 3 in eq. 4, we get


ie. Total energy at the region (1) = The total energy at the region (2)
∴ \(P+\frac{1}{2} \rho V^2+\rho g h\) = a constant
(b) 1. When pressure decreases, speed increases.
2. Flow is streamlined.

Question 31.
(a) Write the equation of an isothermal process. (1)
(b) Obtain an expression for work done in an isothermal process. (3)
(c) A Carnot engine operates between the temperatures 398 K and 293 K. Find its efficiency. (1)
Answer:
(a) PV = constant
(b) For an isothermal process, the equation of state is PV = constant = μRT
P = \(\frac{\mu \mathrm{RT}}{\mathrm{~V}}\)
Let a system undergo an isothermal process at temperature T, from the state (P₁, V₁) to (P₂, V₂).
Let ‘ΔV’ be a small charge in volume due to pressure P. Then work done (for ΔV)
ΔW = PΔV

Question 32.
(a) Show that the oscillations produced in a simple pendulum are simple harmonic. (2)
(b) Writeanexpressionfortimeperiodofoscillation. (1)
(c) What is a seconds pendulum? Find the length of seconds pendulum. (2)
Answer:
(a)

Consider a mass m suspended from one end of a string of length L fixed at the other end as shown in the figure.
Suppose P is the instantaneous position of the pendulum. At this instant, its string makes an angle θ with the vertical.
The forces acting on the bob are (1) the weight of bob Fg (mg) acting vertically downward. (2) Tension T in the string.
The gravitational force Fg can be divided into a radial component Fg cos θ and a tangential component Fg sin θ.
The radial component is canceled by the tension T. But the tangential component Fg sin θ produces a restoring torque.
Restoring torque τ = -Fg sin θ. L
τ = -mgsin θ. L ………(1)
The  -ve sign shows that the torque and angular displacement θ are oppositely directed.
For the rotational motion of Bob,
τ = Iα ………(2)
Where I is a moment of inertia about the point of suspension and a is angular acceleration.
From eq (1) and eq (2),
Iα = -mgsin θ. L
If we assume that the displacement θ is small, sin θ ~ 0.
∴ Iα = – mgθL
Iα + mgθL = 0
\(\mathrm{I} \frac{d^2 \theta}{d t^2}+\mathrm{mgL} \theta=0\)
\(\frac{d^2 \theta}{d t^2}+\frac{\mathrm{mgL}}{\mathrm{I}} \theta=0\) ……..(3)
This is the equation of SHM. Hence oscillation of simple pendulum is SHM.
(b) T = \(2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}\)
(c) If the pendulum has a period of 2 sec, it is called seconds pendulum. The length of the second pendulum is one meter.