Kerala Plus One Physics Board Model Paper 2021 with Answers

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

Kerala Plus One Physics Board Model Paper 2021 with Answers

Time: 2 Hours
Total Scores: 60

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

Question 1.
Name the branch of physics which deals with light energy.
(a) Optics
(b) Electrodynamics
(c) Mechanics
(d) Thermodynamics
Answer:
(a) Optics

Question 2.
Out of the fundamental forces in nature, which is the weakest force?
Answer:
Gravitational force

Question 3.
The maximum value of static friction is called _____________
Answer:
Limiting Friction

Question 4.
The moment of linear momentum is called _____________
Answer:
Angular momentum

Question 5.
According to the Kinetic theory of gas, what is the pressure of an ideal gas molecule?
Answer:
P = \(\frac{1}{3} n m \bar{v}^2\)

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

Question 6.
Write the SI units of the following fundamental quantities:
(a) electric current
(b) plane angle
Answer:
(a) Ampere
(b) Radian

Question 7.
A body is traveling along a circular path of radius 10 m as shown below. If it travels from A to B, find the distance and displacement of the body.

Answer:
R = 10 m
Distance = πr
= 3.14 × 10
= 31.4 m
Displacement = 2r = 20 m

Question 8.
Write whether the work done in the following cases is positive, negative, or zero.
(a) Work done by frictional force.
(b) Work done by centripetal force on a body moving in a circular path.
(c) Work done by gravitational force in a freely falling body.
(d) Work done by a person carrying a load on his head and walking along a horizontal-level road.
Answer:
(a) Negative
(b) Zero
(c) Positive
(d) Zero

Question 9.
The magnitude of the Kinetic energy of a body is ‘K’. What is its kinetic energy if its velocity is doubled?
Answer:

Question 10.
State Newton’s 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.

F = \(\frac{G m_1 m_2}{r^2}\)

Question 11.

(a) Name the type of strain produced in the above figure.
(b) Write the equation for this strain.
Answer:
(a) Shear strain
(b) Shear strain, θ = \(\frac{\Delta x}{L}\)

Question 12.
What is the modulus of elasticity? What is its unit?
Answer:
The ratio of stress to strain is called the modulus of elasticity.
Modulus of elasticity = \(\frac{Stress}{Strain}\)
The unit of modulus of elasticity is N/m² or Pascal (Pa)

Question 13.
Calculate the efficiency of a heat engine working between 273 K and 373 K.
Answer:
T₁ = 273 K
T₂ = 373 K
η = 1 – \(\frac{T_2}{T_1}\)
⇒ η = 1 – \(\frac{273}{373}\)
⇒ η = 1 – 0.732
⇒ η = 0.268
⇒ η = 26.8%

Question 14.
Define the isochoric process. What is the work done during this process?
Answer:
A thermodynamic process at constant volume is called an isochoric process.
w = p∆v = 0

Question 15.
State the law of conservation of angular momentum.
Answer:
When there is no external torque acting on the body, its angular momentum is constant.
For a system of particles, \(\frac{\mathrm{d} \stackrel{\mathrm{~L}}{ }}{\mathrm{dt}}=\tau_{\mathrm{ext}}\)
if τ_ext = 0, we get \(\frac{\mathrm{d} \overline{\mathrm{~L}}}{\mathrm{dt}}\) = 0
(or) \(\overrightarrow{\mathrm{L}}\) = constant

Question 16.
Find the torque of a force \(7 \hat{i}+3 \hat{j}-5 \hat{k}\) about the origin. The force acts on a particle whose position vector is \(\hat{i}-\hat{j}+\hat{k}\).
Answer:

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

Question 17.
The distance traveled by a freely falling body is given by the equation, y = \(v_0 t-\frac{1}{2} g t^2\). Using principles of dimensions, check the correctness of the equation.
Answer:

The equation is dimensionally correct.

Question 18.
(a) Define average velocity and average speed.
(b) Draw the position-time graph of the body having zero acceleration.
Answer:
(a) The ratio of total displacement to time interval is called average velocity.
The ratio of total distance traveled to time interval is called average speed.

Question 19.
Say whether the following statements are true/false:
(a) Kinetic friction depends on the velocity of the body.
(b) Static friction is independent of the area of the contract.
(c) Kinetic friction is directly proportional to normal reaction.
Answer:
(a) False
(b) True
(c) True

Question 20.
State Newton’s second law of motion. Using it derive an equation for force.
Answer:
Newton’s Second Law of Motion: 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 21.
(a) What is the rotational analog of force in linear motion?
(b) It is difficult to close or open a door by applying force at the hinges. Why?
Answer:
(a) Torque
(b) Torque depends on the magnitude of the force and the perpendicular distance between the fixed point and line of action of force.
\(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)
when we apply force at hinges, r = 0. Hence torque will be zero. This can not produce rotation.

Question 22.
The moment of inertia of a disc of mass ‘M’ and radius R about an axis passing through its center and perpendicular to its plane is \(\frac{M R^2}{2}\).
(a) What is the radius of gyration in this case?
(b) Using the theorem of Perpendicular axes, derive an equation for the moment of inertia of the disc about a diameter.
Answer:

(b) According to the perpendicular axis theorem,
I_z = I_x + I_y
But in this case I_x = I_y
∴ I_z = 2I_x ……….(1)
But we know I_z = \(\frac{\mathrm{MR}^2}{2}\) ……..(2)
substitute (2) in (1), we get
\(\frac{\mathrm{MR}^2}{2}\) = 2I_x
I_x = \(\frac{\mathrm{MR}^2}{4}\)
The moment of inertia about any diameter is \(\frac{\mathrm{MR}^2}{4}\)

Question 23.
(a) Define the orbital velocity of a satellite.
(b) Derive an expression for orbital velocity.
Answer:
(a) Orbital velocity is the speed of a satellite in an orbit.
(b) 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
⇒ \(\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}\)
⇒ v = \(\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}\)

Question 24.
(a) Write any two postulates of the kinetic theory of gases.
(b) Write an equation for the average kinetic energy of a molecule in terms of absolute temperature.
Answer:
(a)The kinetic theory of gases has been developed by Clausius, Maxwell, Boltzmann, and others. The theory is based on the following postulates.

• The gas is a collection of a large number of molecules. The molecules are perfectly elastic hard spheres.
• The size of a molecule is negligible compared with the distance between the molecules.
• The molecules are always in random motion
• During their motion, the molecules collide with each other and with the walls of the containing vessel.
• The collisions are elastic and hence the total K.E energy and the total momentum of the colliding molecules before and after collisions are the same.
• The kinetic energy of a molecule is proportional to the absolute temperature of the gas.
• There is no force of attraction or repulsion between molecules.

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

Question 25.
(a) A and B are two vectors acting at an angle e as shown in the figure. Redraw the figure and mark the resultant vector R.

(b) Write the magnitude of the resultant vector R.
(c) Two forces A and B of magnitudes 6N and 8N act perpendicular to each other. Find the magnitude of the resultant force.
Answer:

Question 26.
A projectile is projected with a velocity V0 making an angle θ with horizontal. Derive equations for maximum height and time of flight for a projectile.
Answer:
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 = u² sin²θ
⇒ 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
R = \(\frac{u \cos \theta \times 2 u \sin \theta}{g}\)
R = \(\frac{\mathrm{u}^2 2 \sin \theta \cos \theta}{\mathrm{~g}}\)
R = \(\frac{u^2 \sin 2 \theta}{g}\) …….(3) [∵ 2 sin θ cos θ = sin 2θ]
The eq.(3) shows that R is maximum when sin 2θ is maximum, ie. When q₀ = 45°.
The maximum horizontal range,
Rmax = \(\frac{u^2}{g}\)

Question 27.
(a) Define impulsive force. Give one example of it.
(b) A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms^-1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
Answer:
(a) The force that acts on bodies for a short time is called impulsive force.
eg. In firing a gun
(b) v₁ = -12 m/s, v₂ = +12 m/s, m = 0.15 kg
Impulse = Change in momentum
Impulse = mv₂ – mv₁
= m(v₂ – v₁)
= 0.15(12 – (-12))
= 3.6 Ns

Question 28.
(a) State the principle of conservation of mechanical energy.
(b) Prove the principle of conservation of mechanical energy in the case of a freely falling body.
Answer:
(a) The total mechanical energy is conserved if forces, doing work are conservative.
(b) Conservation of mechanical energy for a freely falling body

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 = \(\frac{1}{2}\)mv² = 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 = \(\frac{1}{2}\)mv²
= \(\frac{1}{2}\)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 = \(\frac{1}{2}\)mv2
= \(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE
= 0 + mgh
= mgh

Question 29.
(a) Derive an equation for the potential energy of a stretched spring.
(b) Draw a graph showing the variation of kinetic energy and potential energy with displacement in the case of a spring obeying Hooke’s law.
Answer:
(a)

Consider a massless spring fixed to a rigid support at one end and a body attached to the other end. The body moves on a frictionless surface.
If a body is displaced by a distance dx,
The work done for this displacement dw = Fdx
∴ Total work done to move the body from x = 0 to x

This work done is by storing potential energy in the spring.
Hence potential energy of a spring

Question 30.
(a) Write the equation connecting acceleration due to gravity (g) and universal gravitational constant (G).
(b) Derive an equation for acceleration due to gravity at a depth ‘d’ from the surface of the earth.
Answer:
(a) g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
(b) Variation of g with depth

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,
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) ……….(2)
Substituting eq(1) in eq(2), we get
g = \(\frac{\mathrm{G}}{\mathrm{R}^2}\left(\frac{4}{3} \pi \mathrm{R}^3 \rho\right)\)
g = \(\frac{4}{3} \pi \mathrm{GR} \rho\) ……..(3)
Therefore the acceleration due to gravity at a depth of d is given by

The above equation shows that when depth increases g decreases.

Question 31.
(a) State Pascal’s Law.
(b) A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of the cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear?
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.

Question 32.
State and prove Bernoulli’s principle.
Answer:
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 nonuniform cross-section from region 1 to region 2.
Let P₁ be the pressure, A₁ the area of 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 ∆t at region 1 is given by
W₁ = force × distance
= P₁ A₁ ∆x₁
= P₁ ∆V₁ (∵ A₁ ∆x₁ = ∆V₁)
Where ∆x₁ is the displacement produced at region 1, during the time interval ∆t.
Similarly, the work done in a time ∆t at region 2 is given by,
W₂ = -P₂ A₂ ∆x₂
W₂ = -P₂ ∆V₂
[Here -ve sign appears as the direction of \(\overrightarrow{\mathrm{p}}\) and ∆x are in opposite directions.]
Net work done DW = P₁ ∆V₁ – P₂ ∆V₂
According to the equation of continuity
∆V₁ = ∆V₂ = ∆V
∆W = P₁ ∆V – P₂ ∆V
∴ ∆W = (P₁ – P₂) ∆V ……….(1)
This work changes the kinetic energy, pressure energy, and potential energy of the fluid.
If ∆m is the mass of liquid passing through the pipe in a time ∆t.
the change in Kinetic energy is given by ∆k.E = \(\frac{1}{2} \Delta \mathrm{mV} V_2^2-\frac{1}{2} \Delta \mathrm{mV}_1^2\)
∆k.E = \(\frac{1}{2} \Delta \mathrm{~m}\left(\mathrm{~V}_2^2-\mathrm{V}_1^2\right)\) ……….(2)
Change in gravitational potential energy is given by
∆p.E = ∆mgh₂ – ∆mgh₁
∆p.E = ∆mg(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; ∆w = ∆k.E + ∆p.E …….(4)
Substituting eq. 1, 2, and 3 in eq. 4, we get

Question 33.
(a) Explain the anomalous expansion of water.
(b) Calculate the amount of heat energy required to convert 10 J (Latent heat of vaporization of water is 22.6 × 10⁵ Jkg^-1)
Answer:
(a) Anomalous behavior of water: Generally volume of liquid increases with temperature. When water is heated, its volume starts to decrease from 0°C and reaches a minimum of 4°C. Hence density of water is maximum at 4°C.

(b) m = 10 kg
L = 22.6 × 10⁵ J kg^-1
Q = mL
Q = 10 × 22.6 × 10⁵ = 22.6 × 10⁶ J

Question 34.
With the help of a diagram derive an equation for the period of a simple pendulum.
Answer:
The Simple Pendulum

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 F_g (mg) acting vertically downward. (2) Tension T in the string.
The gravitational force F_g can be divided into a radial component F_g Cos θ and a tangential component F_g Sin θ.
The radial component is canceled by the tension T. But the tangential component F_g Sin θ produces a restoring torque.
Restoring torque τ = -F_g 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 the 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 ~ θ.

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

Question 35.
A wave traveling along a string is described by y(x, t) = 0.005 sin (80.0x – 3.0t), in which the numerical constants are in SI units (0.005 m, 80.0 rad m^-1, and 3.0 rads^-1). Calculate:
(a) the amplitude
(b) wavelength
(c) the period of the wave.
Answer:
y(x, t) = 0.005 sin(80x – 3t)
General equation is y(x, t) = A sin(kx – ωt)
(a) Amplitude, A = 0.005 m

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

Question 36.
(a) State principle of homogeneity of dimensions.
(b) The kinetic energy (K) of a body depends on its mass (M) and its velocity (V). Using the above principle derive an equation for kinetic energy.
Answer:
(a) The dimension of each term on either side of an equation is the same.
(b) Velocity v
k ∝ m^x V^y
k = C m^x V^y (∵ C is a constant)
According to the principle of homogeneity
[k] = [m^x v^y]
ML²T^-2 = M^x (LT⁺¹)y = M^x L^y T^-y
Comparing we get x = 1 and y = 2
∴ k = C m¹ v²
We know C = \(\frac{1}{2}\)
∴ k = \(\frac{1}{3}\)mv2

Question 37.
(a) Derive positon-time relation \(\left(x=v_0 t+\frac{1}{2} a t^2\right)\) for a uniformly accelerated body.
(b) The velocity-time graph of a body is given below. Find the displacement of the body during the time interval of 5 seconds to 15 seconds.

Answer:
(a)

Consider a body moving along a straight line with uniform acceleration.
Let ‘u’ be the initial velocity and ‘v’ be the final velocity.
‘S’ is the displacement traveled by the body during the time interval ‘t’.
Displacement of the body during the time interval t,
S = average velocity × time
S = \(\left(\frac{v+u}{2}\right) t\) ………(1)
But v = u + at ………..(2)
Substitute eq.(2) in eq.(1), we get

(b) Displacement in time intervel 4s to 15s is
S = Area of ABCD
= AB × AD
= 20 × (15 – 5)
= 20 × 10
= 200 m

Question 38.
A car of mass ‘m’ moves with a speed ‘v’ along a banked road of radius ‘R’.
(a) Draw a diagram showing all the forces acting on the car.
(b) Using the diagram derive an equation for the maximum safe speed.
Answer:

Consider a vehicle along a curved road with the angle of banking q. Then the normal reaction on the ground will be inclined at an angle q with the vertical.
The vertical component can be divided into N Cos q (vertical component) and N sin q (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 q (horizontal component) and F sin q (vertical component).
From the figure get
N cos q = F sin q + mg
N cos q – F sin q = mg ………..(1)
The component N sin q and F sin q provide centripetal force. Hence

Dividing both the numerator and denominator of L.H.S by N cos q, We get

Question 39.
(a) State parallel axes theorem of the moment of inertia.
(b) What is the moment of inertia of a rod of mass M and length L about an axis perpendicular to it and passing through one end?
Answer:
(a) Theorem of Parallel Axis: The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its center of mass and the product of its mass and the square of the distance between the two parallel axes.

Consider a rod of mass M and length l, rotating about an axis passing through one end.
Let dx be a small element at a distance x from the axis of rotation.
mass of the element dx, dm = \(\frac{\mathrm{M}}{l} \mathrm{dx}\)
∴ M.I of length small element, dl = \(\frac{\mathrm{M}}{l} \mathrm{dxx}\)
Total moment of inertia, I = \(\int_0^l \frac{\mathrm{M}}{l} \mathrm{dx} \mathrm{x}^2\)

Question 40.
(a) Derive an equation for acceleration due to gravity at a height ‘h’ above the surface of the earth.
(b) Calculate the value of acceleration due to gravity at a height equal to half of the radius of the earth from the surface of the earth.
Answer:
(a) 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,