Kerala Plus One Physics Question Paper March 2023 with Answers

Reviewing Kerala Syllabus Plus One Physics Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus One Physics Previous Year Question Paper March 2023

Time: 2 Hours
Total Scores: 60

Section – A

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

Question 1.
1Å (angstrom) = ___________
(i) 10¹⁰ m
(ii) 10^-15 m
(iii) 10¹⁵ m
(iv) 10^-10 m
Answer:
(iv) 10^-10 m

Question 2.
___________ is the measure of inertia.
Answer:
Mass

Question 3.
True or False:
Elastomers do not obey Hooke’s law.
Answer:
True

Question 4.
The point at which the whole mass of the body is supposed to be concentrated is called ___________
Answer:
Center of Mass

Question 5.
The change of state from liquid to solid is called ___________
Answer:
Fusion

Question 6.
Kinetic energy of a gas molecule is directly proportional to the ___________ of the gas.
Answer:
Absolute Temperature

Question 7.
Most fundamental property of a wave is
(a) Temperature
(b) Pressure
(c) Frequency
(d) Wavelength
Answer:
(c) Frequency

Section – B

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

Question 8.
A ball is thrown vertically upwards with an initial velocity u. Calculate the time taken to reach maximum height.
Answer:
At maximum height, v = 0 and a = -g
v = u + at
0 = u – gt
u = gt
∴ t = \(\frac{u}{g}\)

Question 9.
What is a null (zero) vector? Give any two properties.
Answer:
A vector with zero magnitude is called a null vector.
\(\overrightarrow{\mathrm{A}}+\overrightarrow{0}=\mathrm{A}\)
\(\overrightarrow{\mathrm{A}}+-\overrightarrow{\mathrm{A}}=\overrightarrow{0}\)

Question 10.
(a) Choose the correct option:
The spring force is a conservative/non-conservative force. (1)
(b) Draw the variation of potential and kinetic energies of a spring with its displacement from the equilibrium position. (1)
Answer:
(a) Conservative

Question 11.
What are the factors affecting moment of inertia of an object?
Answer:
Mass of body, position of axis, orientation of axis, distribution of mass around the axis of rotation.

Question 12.
(a) Consider a planet moving in an elliptical orbit around the sun as shown in Fig. If t₁ and t₂ are the time taken by the earth to go from P₁ to P₂ and P₃ to P₄ respectively. The two shaded parts have equal area, then
(i) t₁ < t₂
(ii) t₁ = t₂
(iii) t₁ > t₂
(iv) t₂ = -t₁

(b) State the law which explains the above case. (1)
Answer:
(a) (ii) t₁ = t₂
(b) The line that joins any planet to the sun sweeps equal areas in equal intervals of time. ie, the aerial velocity is constant.

Question 13.
(a) A drop of liquid under no external force is always spherical. Why? (1)
(b) When a painting brush dipped in water, its hairs cling together. Why? (1)
Answer:
(a) Sphere has least surface area. To reduce potential energy due to surface tension, a liquid drop has to acquire the least surface area.
(b) When the hairs cling together, the free surface area gets reduced, which reduces potential energy due to surface tension.

Question 14.
The modes of heat transfer are shown below. Identify A and B.

Answer:
A → Conduction
B → Convection

Section – C

Answer any 6 questions from 15 to 21. Each carries 3 scores. (6 × 3 = 18)

Question 15.
(a) Give an example for a body possessing zero velocity which is accelerating. (1)
(b) Show that the area under the velocity-time graph gives displacement. (2)
Answer:
(a) (i) Extreme position of simple harmonic oscillation.
(ii) When a body is projected vertically upwards, at maximum height, the body has zero velocity but non-zero acceleration.
(b) Consider a body moving with constant velocity v. Its velocity time graph is given below.

The area of the rectangle has height v and breadth t.
Therefore, Area = vt = displacement

Question 16.
While firing a bullet, the gun recoils.
(a) Which conservation law helps you to explain this phenomenon? (1)
(b) “In the firing process, the speed of the gun is very low compared to the speed of the bullet”. Substantiate the above statement mathematically. (2)
Answer:
(a) Conservation of momentum
(b) Let m and v be the mass and velocity of the bullet.
Let M and V be the mass and velocity of the gun
According to conservation of momentum,
the momentum before firing = the momentum after firing.
0 = mv’ + MV
MV = -mv’
M = \(\frac{-m v^{\prime}}{\mathrm{M}}\)
Since m<<M, V is small compared to v’.
ie, the velocity of the gun is small compared to the velocity of the bullet.

Question 17.
(a) Which type of elasticity is involved in stretching a wire? (1)
(b) Steel is more elastic than rubber. Why? (1)
(c) If the bulk modulus of water is 2 × 10⁹ Nm^-2, find its compressibility. (1)
Answer:
(a) Young’s modulus
(b) Steel regains its original condition almost completely, when the deforming force is removed. But rubber will not regain its original condition after the removal of deforming force.
(c) Compressibility, K = \(\frac{1}{B}\)
B = 2 × 10⁹ Nm^-2
K = \(\frac{1}{2 \times 10^9}\) = 5 × 10^-10 Pa^-1

Question 18.
(a) “When water flowing in a broader pipe enters into a narrow pipe, its pressure decreases.” Do you agree with this statement? Explain. (2)
(b) Define coefficient of viscosity. (1)
Answer:
(a) Yes. I do agree
According to the continuity equation,
AV = constant
Hence, as the area decreases, the velocity increases.
According to Bernoulli’s theorem,
P + \(\frac{1}{3}\)PV² + ρgh = constant
As velocity increases, kinetic energy increases, which in turn causes a decrease in pressure.
(b) The coefficient of viscosity is defined as the ratio of shear stress to shear strain rate.

Question 19.
(a) State First law of thermodynamics. (1)
(b) One mole of an ideal gas expands from volume V1 to volume V2 at a constant temperature T. Derive an expression for the work done. (2)
Answer:
(a) According to the first law of thermodynamics, heat 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
(b) Work done in isothermal process: 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 ‘DV’ be a small charge in volume due to pressure P.
Then work done (for ΔV)
ΔW = PΔV

Question 20.
(a) State the law of equipartition of energy. (1)
(b) A box contains an equal number of molecules of hydrogen and oxygen. If there is a fine hole on top of the box, which gas will leak rapidly? Why? (1)
(c) When gas is heated, its temperature increases. Explain it based on kinetic theory of gases. (1)
Answer:
(a) The total kinetic energy of a molecule is equally divided among the different degrees of freedom.

Thus rms speed \(\bar{v}\) is proportional to \(\frac{1}{\sqrt{\mathrm{~m}}}\)
The mass of hydrogen, is less compared to that of oxygen and hence veloicty of hydrogen is large compared to oxygen. Hence hydrogen leaks rapidly.
(c) According to kinetic theory fo gases, average kinetic energy of molecule is directly proportional to temperature. When heat is given to gas molecules vibrate and kinetic energy increases and hence its temperature increases.

Question 21.
(a) Derive an expression for the period of oscillation of a Simple Pendulum. (2)
(b) Can a pendulum clock show correct time inside a spaceship? Why? (1)
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. AP this instant it’s string makes an angle θ with the vertical.
The forces acting on the bob are (1) 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 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).
-ve sign shown that the torque and angular displacement θ are oppositely directed. For rotational motion of the bob.
τ = Iα ……………..(2).
Where I is a moment of inertia about the point of suspension and a is the angular acceleration.
From eq (1) and eq (2)
Iα = -mg sin θ . L
If we assume that the displacement θ is small, sin θ ~ θ.
∴ Iα = -mgθ L
Iα + mgθ L = 0

T = \(2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}\)
(b) No. In space spaceship, g = 0
T = \(2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}\)
Hence T becomes infinity, ie, the pendulum will not oscillate.

Section – D

Answer any 3 questions from 22 to 25. Each carries 4 scores. (3 × 4 = 12)

Question 22.
(a) What do you mean by principle of homogeneity of dimensions? (1)
(b) Using this principle, check whether the following equation is dimensionally correct: (2)
\(\frac{1}{2}\)mv² = mgh
(c) State the number of significant figures in the following: (1)
(i) 0.060607 m²
(ii) 6.0320 Nm^-2
Answer:
(a) If an equation contains more than one term, the dimension of each term must be the same.
[\(\frac{1}{2}\)mv²] = (MLT^-1)² = M²L²T^-2
[mgh] = M × LT^-2 × L = ML²T²
Dimension on each side of the equation is the same.
Hence the equation is correct.
(c) (i) 5
(ii) 5

Question 23.
(a) ‘Sketch the schematic diagram of a vehicle on a banked road and mark various forces acting on it. (2)
(b) Obtain an expression for maximum safe speed for a vehicle in the banked road. (Without considering friction) (2)
Answer:
(a)

(b) Consider a vehicle along a curved road with 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 thefigureare get,
N cos θ = F sin θ + mg
N cos θ – F sin θ = mg ……….(1)
The component N sin θ and F sin θ provide centripetal force.
Hence N sin θ + F cos θ = \(\frac{\mathrm{mv}^2}{\mathrm{R}}\)
eq (1) by eq (2)

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

Question 24.
(a) A car and a lorry have equal kinetic energy, which one will have greater momentum? Explain. (1)
(b) State and prove work-energy theorem. (3)
Answer:
(a) KE_lorry = KE_car
P = \(\sqrt{2 \mathrm{~m} \times \mathrm{KE}}\)
∴ P ∝ √m
The mass of the lorry is large compared to the mass of the car.
Hence the momentum of the lorry is greater than that of the car.
(b) The change in kinetic energy of a particle is equal to the work done on it by the net force.
Proof: We know v² = u² + 2as
v² – u² = 2as
Multiplying both sides with m/2; we get

Question 25.
(a) A string is stretched between two fixed supports and a note of sound is heard when it is plucked at its middle point.
(i) How are standing waves produced? (1)
(ii) Derive an expression for the frequency of sound produced by the first two modes of vibration. (2)
(b) Where will a man hear a louder sound in the case of a stationary wave (node or anti-node)? (1)
Answer:
(a) (i) When two waves of the same amplitude and frequency travelling in opposite directions superimpose, the resulting wave pattern does not move to either side. This pattern is called standing wave.
(ii) A string of length L is fixed at two ends. The position of one end is chosen as x = 0, then the position of the other end will be x = L.
At x = 0, there will be a node.
To occur node at x = L, it must satisfy L = \(n \frac{\lambda}{2}\) (since, x = \(n \frac{\lambda}{2}\))
ie; the wavelength, λ = \(\frac{2 \mathrm{~L}}{\mathrm{n}}\)
the frequency, υ = \(\frac{v}{\lambda}=\frac{\mathrm{nv}}{2 L}\) (υλ = v)
n = 1, 2, 3, ….. etc.
The frequency of vibrations of the stretched string of length L is
υ = \(\frac{n v}{2 L}\), n = 1, 2, 3…etc.
This set of frequencies at which the string can vibrate are called natural frequencies or modes of vibration or harmonics.
The above equation shows that the modes of vibration (natural frequencies) of a string are integral multiples of the lowest frequency.
n = \(\frac{\mathrm{V}}{2 \mathrm{~L}}\) (for n = 1)
Fundamental mode (or) First harmonics
For n = 1, υ₁ = \(\frac{\mathrm{V}}{2 \mathrm{~L}}\)
This is the lowest frequency with which the string vibrates. This is called fundamental mode or first harmonic of vibration.

(b) antinode

Section – E

Answer any 3 questions from 26 to 29. Each carries 5 scores. (3 × 5 = 15)

Question 26.
A boy throws a cricket ball with velocity u at an angle θ with the horizontal.
(a) What is the trajectory of the ball? (1)
(b) Deduce an expression for maximum height reached by the ball. (2)
(c) If the boy can throw the ball to a maximum height ‘h’, what is the maximum horizontal distance to which he can throw? (2)
Answer:
(a) Parabola
(b) Vertical height of body is decided by 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 = u² sin²θ
H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)

Question 27.
(a) State the law of conservation of angular momentum. (1)
(i) outstretching her arms and legs.
(ii) folding her arms and legs. (2)

(c) Derive the relation connecting torque and angular momentum. (2)
Answer:
(a) Conservation of angular momentum and moment of inertia.
When there is no external torque, the total angular momentum of a body or a system of bodies is a constant.
τ = \(\frac{\mathrm{dL}}{\mathrm{dt}}\) (when τ = 0, we get \(\frac{\mathrm{dL}}{\mathrm{dt}}\) = 0)
L = constant
But L = Iω
∴ Iω = a constant
(b) According to the law of conservation of angular momentum
L = Iω = constant
(i) When a girl stretches her arms and legs, the moment of inertia (I = mr²) increases and angular speed decreases.
(ii) On folding arms and legs, moment of inertia decreases and angular speed increases.
(c) Angular momentum of a particle, \(\vec{l}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}\) When we differentiate on both sides, we get

Question 28.
Escape velocity from the earth for a body of mass m is 11 km/s.
(a) If the body is projected at an angle of 45° with the vertical, the escape velocity will be
(i) 11/√2 km/s
(ii) 22√2 km/s
(iii) 22 km/s
(iv) 11 km/s
(b) Derive an expression for the escape velocity of an object from the surface of the earth. (3)
(c) The moon has no atmosphere. Why? (1)
Answer:
(a) 11 km/s (escape velocity is independent of θ)
(b) Force on a mass m at a distance r from the centre of Earth = \(\frac{\mathrm{GMm}}{\mathrm{r}^2}\)
Work done for giving a 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,


This escape velocity \(\sqrt{2 \mathrm{Rg}}\) is estimated to be 11.2 km/s on the earth. But escape velocity of the moon is 2.3 km/s.
(c) Escape velocity on the moon is 2.3 km/s. The r.m.s. velocity of the gas molecules is greater than this value. Gas molecules easily escape from the surface of the moon.
Hence, there is no atmosphere on the moon.

Question 29.
Due to capillary action, water will rise inside a glass tube.
(a) The angle of contact in this case is (1)
(i) Acute
(ii) Obtuse
(iii) 90°
(iv) Zero
(b) Obtain an expression for the rise of liquid in the tube. (3)
(c) Water rises to different heights in glass tubes of different radii. Give reason. (1)
Answer:
(a) (iv) zero
(b)

Consider a capillary tube of radius ‘a’ dipped in a liquid of density r and surface tension S. If the liquid has a concave meniscus, it will rise in the capillary tube.
Let h be the height of the liquid in the tube.
Let p₁ be the pressure on the concave side of the meniscus and p0, that on the other side.
The excess pressure on the concave side of the meniscus can be written as
p_i – p₀ = \(\frac{2 \mathrm{~S}}{\mathrm{R}}\) ………(1)
Where R is the radius of the concave meniscus.
The tangent to the meniscus at the point A makes an angle q with the wall of the tube.
In the right-angled triangle ACO

Considering two points M and N in the same horizontal level of a liquid at rest,
pressure at N = pressure at M
But pressure at M = pi, the pressure over the concave meniscus, and pressure at N = p₀ + hrg
∴ p_i = p₀ + hrg
or p_i – p₀ = hrg …….(3)
From equations (2) and (3), we get
hρg = \(\frac{2 S \cos \theta}{a}\)
h = \(\frac{2 S \cos \theta}{\text { apg }}\)