Kerala Plus One Physics Question Paper September 2021 with Answers

Reviewing Kerala Syllabus Plus One Physics Previous Year Question Papers and Answers Pdf September 2021 helps in understanding answer patterns.

Kerala Plus One Physics Previous Year Question Paper September 2021

Time: 2 Hours
Total Scores: 60

Answer any 3 questions from 1 to 5. Each carries 1 score. (3 × 1 = 3)

Question 1.
Identify the fundamental force in nature from the following:
(i) Gravitational Force
(ii) Elastic Force
(iii) Restoring Force
(iv) Frictional Force
Answer:
(i) Gravitational Force

Question 2.
Name the branch of Science that deals with the study of stars.
Answer:
Astronomy

Question 3.
Momentum is a ___________ quantity. (vector/scalar)
Answer:
Vector

Question 4.
What is the analog of mass in rotational motion?
Answer:
Moment of Inertia

Question 5.
Write a perfect gas equation.
Answer:
PV = µRT

Answer any five questions from 6 to 16. Each carries 2 scores. (5 × 2 = 10)

Question 6.
Define plane angle.
Answer:
Plane angle is defined as the ratio of the length of arc to the radius
dθ = \(\frac{ds}{r}\)

or
Plane angle is the angle between two straight lines which are intersecting at a point.

Question 7.
A player throws a ball vertically upwards. What is the velocity and acceleration of the ball at the highest point of its motion?
Answer:
a = g, v = 0

Question 8.
Two vectors A and B are given below.

Redraw the figure and show the vector sum using the parallelogram method.
Answer:

Question 9.
What are the conditions for the equality of two vectors?
Answer:
Vectors should be of the same magnitude and direction of the same physical quantity.

Question 10.
Write whether the work done in the following cases is positive, negative, or zero.
(a) The work done by the frictional force during the motion of an object.
(b) The work done by the gravitational force during the motion of an object on a horizontal surface.
Answer:
(a) Negative work
(b) Zero work

Question 11.
Write the SI unit and dimensional formula of energy.
Answer:
Unit is Joule and dimension is ML²T^-2

Question 12.
State Newton’s Universal Law of Gravitation.
Answer:
Everybody in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 13.
State Hooke’s Law.
Answer:
Within the elastic limit, stress is directly proportional to strain.
For small deformations, the stress and strain are proportional to each other.
i.e., stress ∝ strain
Stress = k × strain
Where k is the proportionality constant and is known as the modulus of elasticity.

Question 14.
A typical stress-strain curve for a metal is given below. What is denoted by the points A and B?

Answer:
A → Proportional Limit
B → Elastic Limit or Yield Point

Question 15.
State first law of thermodynamics.
Answer:
According to first law of thermodynamics, hea supplied to a system is used to increase its internal energy and to do work.
If ΔQ is heat supplied to the system, ΔW is work done by the system and ΔU is change in internal energy, then ΔQ = ΔU + ΔW.

Question 16.
What is the difference between the isobaric process and the isochoric process?
Answer:
In the isobaric process, pressure remains constant, and work done is non-zero (w ≠ 0).
In the isochoric process, volume remains constant and hence work done is zero (w = 0).

Answer any four questions from 17 to 24. Each carries 3 scores. (4 × 3 = 12)

Question 17.
Fill in the blanks.

Name of Base Quantity	Name of SI Unit
Length	Metre
Mass	____________
____________	Second
Electric Current	____________
Thermodynamic Temperature	____________
_____________	Mole
_____________	Candela

Answer:

Name of Base Quantity	Name of SI Unit
Length	Metre
Mass	Kilogram
Time	Second
Electric current	Ampere
Thermodynamics	Kelvin
Amount of substance	Mole
Luminous intensity	Candela

Question 18.
An astronaut in a space shuttle moves along a straight line. The data given below give the measured velocities starting from t = 2s.

Time (s)	2	4	6	8	10	12	13	16
Velocity (m/s)	2	4	6	2	0	-2	-4	-2

Draw the velocity-time graph from 2s to 16s (use of graph paper is not required)
Answer:

Question 19.
State Newton’s second law of motion and write the equation of force.
Answer:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Mathematically this can be written as

Question 20.
State the law of static friction.
Answer:
The force of maximum static friction is directly proportional to normal reaction.

Question 21.
Define angular velocity. Write the equation showing its relation with linear velocity.
Answer:
Angular velocity is the rate of change of angular displacement
ω = \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\)
\(\vec{v}=\vec{\omega} \times \vec{r}\) or v = rω

Question 22.
What are the factors on which the moment of inertia of a rigid body depends?
Answer:
I = mr²
The moment of inertia of the body depends on

• Mass
• Orientation of axis of rotation
• Distribution of mass

Question 23.
Derive an expression for acceleration due to gravity at any point above the surface of the earth.
Answer:
The acceleration due to gravity on the surface of the earth,
g = \(\frac{\mathrm{Gm}}{\mathrm{R}^2}\) ……….(1)
At a height of h, the acceleration due to gravity can be written as,

Question 24.
Using the kinetic theory of an ideal gas, prove that the average kinetic energy of a molecule is proportional to the absolute temperature of the gas.
Answer:
Pressure of ideal gas, P = \(\frac{1}{3} n m \bar{v}^2\)
Multiplying both sides by volume ‘V’, we get


i.e, Average kinetic energy (\(\frac{E}{N}\)) is proportional to temperature (T).

Answer any five questions from 25 to 35. Each carries 4 scores. (5 × 4 = 20)

Question 25.
Derive an expression for the maximum height, H, and horizontal range, R of a projectile using the given figure.

Answer:
Vertical Height
The vertical height of the body is decided by the vertical component of velocity (u sin θ).
The vertical displacement of the projectile can be found using the formula v² = u² + 2as
When we substitute v = 0, a = -g, s = H and u = u sin θ, we get
0 = (u sin θ)² + 2 × -g × H
⇒ 2gH = u²sin²θ
⇒ H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)

Horizontal Range
If we neglect the air resistance, the horizontal velocity (u cos θ) of the projectile will be a constant.
Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight
= \(\frac{u \cos \theta \times 2 u \sin \theta}{g}\)
= \(\frac{\mathrm{u}^2 2 \sin \theta \cos \theta}{\mathrm{~g}}\)
= \(\frac{u^2 \sin 2 \theta}{g}\) ……(3) [∵ 2 sin θ cos θ = sin 2θ]

Question 26.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and/or direction of the centripetal acceleration of the stone?
Answer:
a_c = Rω²
R = 80 cm = 0.8 m
ω = 2πV & v = \(\frac{N}{t}\)
⇒ ω = 2 × 3.14 × \(\frac{14}{25}\) = 3.5168
⇒ a_c = 0.8 × (3.5168)² = 9.89 m/s²
a_c is directed towards the center.

Question 27.
What is the difference between elastic and inelastic collisions? Derive an expression for loss in kinetic energy on completely inelastic collision of two bodies in one-dimensional motion.
Answer:
(i) In elastic collision both momentum and KE are conserved. In an inelastic collision, momentum is conserved. But KE is not conserved.
(ii) Consider two bodies of mass m1 and m2. The first body hits the second one which is at rest and after the  collision both stick and move

According to the conservation of momentum

Question 28.
Prove that the mechanical energy of a ball of mass ‘m’ dropped from a height ‘H’ is conserved.
Answer:

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
The potential energy at 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 29.
Calculate the energy used by a 100-watt bulb which is on for 10 hours.
Answer:
P = \(\frac{E}{t}\)
E = P × t
P = 100 W
t = 10 hr = 10 × 60 × 60 sec
E = 100 × 10 × 60 × 60 = 36 × 10⁵ J

Question 30.
Derive an expression for the escape velocity of an object from the surface of the earth.
Answer:
The minimum speed with which a body is projected so that it never returns to the earth is called escape speed or escape velocity.
Expression for escape speed
Force on a mass m at a distance r from the center of earth = \(\frac{\mathrm{GMm}}{\mathrm{r}^2}\)
Work done forgiving small displacement dr, dw = \(\frac{\mathrm{GMm}}{\mathrm{r}^2} \mathrm{dr}\)
Work done in taking the body to infinity from the surface of the earth,

Question 31.
State Pascal’s law. Explain the working of a hydraulic lift by drawing a figure.
Answer:
Whenever external pressure is applied to any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions.

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 = \(\frac{F_1}{A_1}\)
If F₂ is the force exerted on the area A₂, pressure P₂ = \(\frac{F_2}{A_2}\)
According to Pascal’s law P₁ = P₂
ie; \(\frac{F_1}{A_1}=\frac{F_2}{A_2}\)
F₂ = F₁ × \(\frac{A_2}{A_1}\)
When \(\frac{A_2}{A_1}\) >> 1, we get F₂ >> F
Using this method we can lift heavy loads by applying a small force.

Question 32.
With the help of the given figure state and prove Bernoulli’s theorem.

Answer:
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 (pgh) 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 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.
The work done on the liquid in a time Δt at region 1 is given by
W₁ = force × distance
= P₁A₁Δx₁
= P₁ΔV₁ (∵ A₁Δx₁ = 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 = P₁ΔV₁ – P₂ΔV₂
According the equation of continuity
ΔV₁ = ΔV₂ = ΔV
∴ ΔW = P₁ΔV – P₂ΔV
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 work-energy theorem work done is equal to the change in kinetic energy plus thte change in potential energy.
ie; Δw = ΔkE + ΔPE ………(4)
Substituting eq. 1, 2, and 3 in eq. 4, we get

i.e. 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

Question 33.
(a) What is the difference between the latent heat of fusion and the latent heat of vaporization?
(b) Why burns from steam are usually more serious than those from boiling water?
Answer:
(a) Latent heat of vaporization is the amount of heat per unit mass required to change the state of liquid to gas.
Latent heat of fusion is the amount of heat per unit mass required to change state of solid to liquid.
(b) Latent heat of vaporization for water is 22.6 × 10⁵ J Kg-1 (ie; 22.6 × 10⁵ J heat is required to convert 1 kg of water into steam at 100°C).
So at 100°C, steam carries 22.6 × 10⁵ J. (more heat than water).

Question 34.
Drive an expression for the period of a simple pendulum.
Answer:

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 = -mg sin 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 α is angular acceleration.
From eq (1) and eq (2),
Iα = -mg sin θ . L
If we assume that the displacement θ is small, sin θ ~ 0
∴ Iα = -mgθ . L
Iα + mgθ . L = 0

T = \(2 \pi \sqrt{\frac{L}{g}}\)

Question 35.
A wave traveling along a string is described by y(x, t) = 0.005 sin(80.0 x – 3.0 t). (1 + 1 + 2 = 4)
In which the numerical constants are in SI units, Calculate
(a) the amplitude,
(b) the wavelength, and
(c) the period and frequency of the wave.
Answer:
(a) 0.005 m

Answer any three questions from 36 to 40. Each carries 5 scores. (3 × 5 = 15)

Question 36.
(a) State the principled Homogeneity of dimensions.
(b) A planet moves around the Sun in nearly circular orbits. Its period depends on the radius (r) of the orbit, the mass of the Sun (M), and the gravitation constant (G). Derive the expression for the period of the planet using the method of dimensions.
Answer:
(a) Dimension of LHS = dimension of RHS
or
The dimension of each term in an equation must be the same.

Question 37.
v-t curve for an object with uniform acceleration is given below.

Using this graph derive the relation, x = v₀t + \(\frac{1}{2}\)at2 and v² = \(\mathrm{v}_0^2\) + 2ax.
Answer:

Question 38.
Various forces acting on a car in a circular motion through a banked road are shown.

(a) Write the name of forces A and B.
(b) A cyclist traveling on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tires and the roads is 0.1. Calculate the maximum permissible speed to avoid slipping.
Answer:
(a) A = normal reaction
B = Centripetal force
(b) R = 3m
µ = 0.1
V = \(\sqrt{\mu \mathrm{Rg}}\)
= \(\sqrt{0.1 \times 3 \times 9.8}\)
= 1.715 m/s

Question 39.
State theorem of perpendicular axes. Using this theorem derive the moment of inertia of the given disc about one of its diameter. The moment of inertia of the disc about the Z axis is \(\frac{\mathrm{MR}^2}{2}\).
Answer:
Theorem of Perpendicular Axes
The moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis lying in the plane of the body.
(or)
The moment of inertia of a plane lamina about the z-axis is equal to the sum of the moments of inertia about the x-axis and the y-axis if the plane lamina lies in the y-plane.
Explanation

Let Ix and Iy be the moments of inertia of the lamina about ox and oy. Let Iz be the moment of inertia of the lamina about an axis perpendicular to the lamina and passing through ‘0’ (about the z-axis). Then by perpendicular axis theorem Iz = Ix + Iy

Question 40.
(a) Write the unit and dimensional formula of the gravitational constant.
(b) Drive an expression for acceleration due to gravity at a depth below the surface of the earth.
Answer:
(a) Unit of G is Nm2/kg2

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 due to gravity on the surface,
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) ………..(2)

The above equation shows that when depth increases g decreases.