Kerala Plus One Physics Board Model Paper 2022 with Answers

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

Kerala Plus One Physics Board Model Paper 2022 with Answers

Time: 2 Hours
Total Scores: 60

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

Question 1.
Which one of the following fundamental forces in nature binds protons and neutrons?
(a) Gravitation force
(b) Electromagnetic force
(c) Strong nuclear force
Answer:
(c) Strong nuclear force

Question 2.
1 Angstrom [1Å] = ______________
(a) 10^-15 m
(b) 10^-10 m
(c) 10^-12 m
Answer:
(b) 10^-10 m

Question 3.
The magnitude of a null vector is ______________
(a) 1
(b) Zero
(c) Unpredictable
Answer:
(b) Zero

Question 4.
When a bus suddenly moves forward, a passenger is jerked backward. Name the law used to explain the above situation.
(a) Law of area
(b) Second law of motion
(c) Law of inertia
Answer:
(c) Law of inertia

Question 5.
The ratio of tensile stress to the longitudinal strain is ______________
(a) Young’s modulus
(b) Elasticity
(c) Elastomer
Answer:
(a) Young’s modulus

Question 6.
Working of a hydraulic lift is based on ______________ law.
(a) Pascal’s law
(b) Newton’s law
(c) Kepler’s law
Answer:
(a) Pascal’s law

Question 7.
The change of solid state to vapour state without passing through the liquid state is called ______________
(a) Melting
(b) Sublimation
(c) Regelation
Answer:
(b) Sublimation

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

Question 8.
Show the impulse is equal to the change in momentum. (2)
Answer:
Relation between impulse and momentum
We know from Newton’s second law
F = \(\frac{\Delta P}{\Delta t}\)
ΔP = FΔt

Question 9.

In the given stress-strain graph identify the stresses S₁ and S₂ corresponding to points B and D respectively. (2)
Answer:
B – Yield Strength
D – Ultimate Strength

Question 10.
A body falls through a fluid. (1 + 1)
(i) Name the forces acting on the falling body.
(ii) Name the velocity of the body when the net force acting on it is zero.
Answer:
(i) Weight of the body, Resultant upthrust, Viscous drag.
(ii) Terminal velocity

Question 11.
The triple point of carbon dioxide is -56.6°C. Express this temperature on a Fahrenheit scale. (2)
Answer:

Question 12.
State the law of equipartition of energy. (2)
Answer:
Law of equipartition of energy: The total kinetic energy of a molecule is equally divided among the different degrees of freedom.

Question 13.
The equation for the velocity of the simple harmonic motion is V(t) = -ω A sin(ωt + φ). Find the expression for the acceleration of simple harmonic motion. (2)
Answer:
We know y = a sin ωt
Velocity v = \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = aω cos ωt
Acceleration a = \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dt}^2}\) = -aω² sin ωt
a = -aω² sin ωt
a = -ω²y

Question 14.
What is meant by standing waves? (2)
Answer:

When two waves of the same amplitude and frequency traveling in opposite directions superimpose the resulting wave pattern does not move to either side. This pattern is called a standing wave.

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

Question 15.
The correctness of equations can be checked by the principle of homogeneity of dimensions. (1 + 2)
(i) State the principle of homogeneity of dimensions.
(ii) Using this principle, check whether the following equation is dimensionally correct.
\(\frac{1}{2}\)mv² = mgh
Answer:
(i) The dimensions of all terms on the other side of an equation are the same.
(ii) [\(\frac{1}{2}\)mv²] = M × (LT^-1)² = ML²T^-2
[mgh] = M × LT^-2 × L = ML²T^-2
∴ Dimension of LHS = dimension of RHS
Hence the equation is dimensionally correct.

Question 16.
(i) Draw the velocity time graph of a uniformly accelerated object.
(ii) Using the graph derive an equation for displacement in terms of initial velocity V(0) and acceleration (a). (1 + 2)
Answer:

Question 17.
Find the magnitude of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) in terms of their magnitudes and angle θ between them.
Answer:

Consider two vectors \(\vec{A}(=\overrightarrow{O P})\) and \(\vec{B}(=\overrightarrow{O Q})\) making an angle θ.
Using the parallelogram method of vectors, the resultant vector \(\vec{R}\) can be written as,
\(\vec{R}=\vec{A}+\vec{B}\)
SN is normal to OP and PM is normal to OS.
From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP + PN)² + SN² ……….(1)
From the triangle SPN, we get
PN = B cos θ and SN = B sin θ
Substituting these values in eq.(1), we get
OS² = (OP + B cos θ)² + (B sin θ)²
But OS = R and OP = A
R² = (A + B cos θ)² + B² sin²θ
⇒ R² = A² + 2AB cos θ + B² cos²θ + B² sin²θ
⇒ R² = A² + 2AB cos θ + B²
⇒ R = \(\sqrt{A^2+2 A B \cos \theta+B^2}\)

Question 18.
A mass rests on a horizontal plane. The plane is gradually inclined horizontally until the mass just begins to slide. Find an equation for the coefficient of static friction (μ_s) between the block and the surface.
Answer:
Consider a body placed on an inclined plane. Gradually increase the angle of inclination till the body placed on its surface just begins to slide down.
If α is the inclination at which the body just begins to slide down, then α is called the angle of repose.

The limiting friction F acts in an upward direction along the inclined plane.
When the body just begins to move, we can write
F = mg sin α …………..(1)
From the figure normal reaction,
N = mg cos α ………….(2)
dividing eq (1) by eq (2)
\(\frac{F}{N}=\frac{m g \sin \alpha}{m g \cos \alpha}\)
μ = tan α ………..(3)

Question 19.
Find the angle between force \(\overrightarrow{\mathrm{F}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})\) unit and displacement \(\overrightarrow{\mathrm{d}}=(5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\) unit.
Answer:

Question 20.
Angular momentum of a rotating body is \(\overrightarrow{\mathrm{l}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}\). Using this relation arrive at the equation, \(\frac{\mathrm{dl}}{\mathrm{dt}}=\vec{\tau}\).
Answer:
Angular momentum of a particle, \(\vec{l}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}\)
When differentiating on both sides, we get

Question 21.
(i) The value of acceleration due to gravity is maximum at the ______________
(a) Poles
(b) Centre of the earth
(c) Equator
(ii) Find the height at which g is reduced to \(\frac{g}{2}\). (Radius of earth – R_E) (1 + 2)
Answer:
(i) Poles

Question 22.
(i) Differentiate streamline flow and turbulent flow.
(ii) What is meant by critical speed in fluid dynamics? (2 + 1)
Answer:
(i) In streamlined flow, the velocity of fluid particles passing through any given point is the same.
In turbulent flow, the velocity of fluid particles passing through any given point is changing.
(ii) Critical speed is the speed of fluid particles below which flow is streamlined and above which flow is turbulent.

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

Question 23.
An object released near the surface of the earth is accelerated downward under the influence of gravity.
(i) Write down the equations of motion in this case.
(ii) Also plot a graph connecting acceleration and time in this case. (3 + 1)
Answer:
(i) u = 0, a = g, s = h
v = u + at
⇒ v = 0 + gt
⇒ v = gt
S = ut + \(\frac{1}{2}\)at2
⇒ S = \(\frac{1}{2}\)gt2
v² = u² + 2as
⇒ v² = 2gh
⇒ v = \(\sqrt{2 g h}\)
(ii)

Question 24.
(i) What is meant by work done by a force?
(ii) Write down any two conditions in which work done is zero.
(iii) Write any one example of negative work. (1 + 2 + 1)
Answer:
(i) The work done by the force is defined as the product of a component of the force in the direction of the displacement and the magnitude of this displacement.
(ii) 1. Work done by centripetal force.
2. A person carrying a load on his head walks along a level road.
(iii) Work done by frictional force.

Question 25.
A heat engine is a device by which a system is made to undergo a cyclic process that results in the conversion of heat to work. Explain briefly the operations of a Carnot’s heat engine and draw the Carnot’s cycle. (4)
Answer:
The Carnot cycle consists of two isothermal processes and two adiabatic processes.

The cylinder is placed on the source. The gas expands and the temperature increases. This is the first stroke of the heat engine. The expansion is isothermal.
During this expansion, the working substance originally at the state A (P₁, V₁, T₁) has a new variable (P₂, V₂, T₁). The variation A to B is shown by AB in the v-p graph.
During the second stroke, the cylinder is placed on the stand and the gas is allowed to expand further and reach the state C. The new coordinates are (P₃, V₃, T₂). This expansion is adiabatic.
The third stroke is carried out when the cylinder is placed on the zink. The cylinder undergoes isothermal compression and coordinate becomes (P₄, V₄, T₂).
In the fourth and final stroke, the cylinder is placed on the non-conducting stand. The gas is compressed back to state A. This is adiabatic compression.

Question 26.
(i) What do you mean by simple harmonic motion?
(ii) Prove that the projection of uniform circular motion on any diameter of the circle is simple harmonic motion. (1 + 3)
Answer:
(i) Simple harmonic motion is the simplest form of oscillatory motion.
The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the particle from the origin varies with time as x(t) = A cos (ωt + φ).
where x(t) = displacement x as a function of time t
A = amplitude
ω = angular frequency
(ωt + φ) = phase (time-dependent)
φ = phase constant or initial phase
(ii) Consider a particle moving along the circumference of a circle of radius ‘a’ and center O, with uniform angular velocity ω.
AB and CD are two mutually perpendicular diameters along X and Y axis.
At time t = 0. let the particle be at P. so that ∠P0OB = φ.
After time ‘t’ second, let the particle reach P so that ∠POP0 = ωt.
N is the foot of the perpendicular drawn from P on the diameter CD.
Similarly, M is the foot of the perpendicular drawn from P to the diameter AB.
When the particle moves along the circumference of the circle, the foot of the perpendicular executes to and fro motion along the diameter CD or AB with O as the mean position.
From the right angle triangle OMP, we get
cos (ωt + φ) = \(\frac{\mathrm{OM}}{\mathrm{OP}}\)
∴ OM = OP cos (ωt + φ)
⇒ X = a cos (ωt + φ) ……….(1)
Similarly, we get
sin (ωt + φ) = \(\frac{y}{a}\)
⇒ Y = a sin (ωt + φ) ……….(2)
Equation (1) and (2) are similar to equations of S.H.M.
Equation (1) and (2) shows that the projection of uniform circular motion on any diameter is S.H.M.
At t = 0, if the particle is at B, then φ = 0.
Then equations (1) and (2) reduce to
x = a cos ωt ……….(3)
y = a sin ωt ………..(4)

Question 27.
A resonance column is an example of a closed pipe.
(i) Sketch the pattern of waveforms of the first two harmonics formed in a closed pipe.
(ii) Show that in a closed pipe ,the frequencies of the first two harmonics are in the ratio 1 : 3. (2 + 2)
Answer:
(i)

(ii) In closed tube: In a closed tube, one end is closed and the other end is open. An air column in a glass tube partially filled with water is an example of a closed system.
The air column in the tube can be set into vibrations with the help of air excited by a tuning fork. The longitudinal waves thus generated are reflected at the closed end and a node is formed there [reflected and incident waves are out of phase and at the closed end, they superimpose to give minimum displacement]. At the open end, the displacement is maximum and an antinode is formed.
If L is the length of the air column, the antinode occurs at x = L.
We know the condition for antinode
x = \((\mathrm{n}+1 / 2)^{\lambda / 2}\)
∴ L = \((\mathrm{n}+1 / 2)^{\lambda / 2}\) for n = 0, 1, 2,……etc.
The wavelength, λ = \(\frac{2 L}{(n+1 / 2)}\) ……..(1)
for n = 0, 1, 2, 3…. etc.
The frequency υ = \((\mathrm{n}+1 / 2) \frac{\mathrm{V}}{2 \mathrm{~L}}\) ……..(2)
for n = 0, 1, 2,…. etc.
From this equation, it is clear that the air column can vibrate with different modes of frequencies (normal modes or harmonics)
Fundamental Mode(or) First Harmonic
We get fundamental mode when n = 0
Substitute this in eq(2), we get

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

Question 28.
A ball is projected at an angle θ with the horizontal
(i) What is the path followed by this ball?
(a) Circle
(b) Ellipse
(c) Parabola
(ii) Derive an equation for the path followed by the ball.
(iii) A cricket ball is thrown at a speed of 28 ms-1 at an angle θ = 30° with the horizontal. Calculate the maximum height. (sin 30 = 0.5) (1 + 2 + 2)
Answer:
(i) c, Parabola
(ii) S = ut + \(\frac{1}{2}\)at²
y = u sin θt – \(\frac{1}{2}\)gt²
But we know horizontal displacement,
x = u cos θ × t
⇒ t = \(\frac{x}{u \cos \theta}\) ………(2)
Substitute eq.(2) eq.(1), we get

In this equation g, θ and u are constants.
Hence eq.(4) can be written in the form y = ax + bx²
where a and b are constants. This is the equation of a parabola.
i.e. the path of the projectile is a parabola.
(iii) u = 28 m/s
θ = 30°
H = \(\frac{u^2 \sin ^2 \theta}{2 g} H=\frac{(28)^2 \sin ^2 30}{2 \times 9.8}\) = 10 m

Question 29.
A car is moving on a circular level road. (1 + 4)
(i) What are the three forces acting on the car?
(ii) Derive an expression for the maximum safe speed of the car.
Answer:
(i)

(ii) Consider a vehicle moving over a level curved road. The two forces acting on it are
1. Weight (mg) vertically down
2. The reaction (N)
The normal reaction can’t produce sufficient centripetal force required for circular motion. The centripetal force for circular motion is provided by friction. This friction opposes the motion of the car moving away from the circular road. Hence condition for circular motion can be written as Centripetal force ≤ force of friction

Question 30.
State and prove Bernoulli’s principle. (5)
Answer:
(i) 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

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, A_2, 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, 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 ∆W = 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 = \(\frac{1}{2} \Delta m V_2^2-\frac{1}{2} \Delta m V_1^2\)
∆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.
i.e; ∆W = ∆kE + ∆pE ……..(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 31.
(i) Define the orbital velocity of a satellite.
(ii) Obtain an equation for orbital velocity.
(iii) Write the relationship connecting orbital velocity and escape speed. (1 + 3 + 1)
Answer:
(i) Orbital velocity of a satellite: Orbital velocity of a satellite is the velocity required for a satellite to revolve around the earth in a fixed orbit.
(ii) Expression for orbital velocity: 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 = \(\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}}\) ………(1)
(iii) Ve = \(\sqrt{2} V_0\)

Question 32.
(i) What are the analogs of mass and force in rotational motion?
(ii) Derive an expression for the kinetic energy of a rotating body. (2 + 3)
Answer:
(i) Moment of inertia and torque
(ii) Consider a body rotating about an axis passing through some point O with uniform angular velocity ‘ω’.
The body can be considered to be made up of several particles of masses m₁, m₂, m₃,…… etc at distances r₁, r₂, r₃,……..etc.
All the particles will have the same angular velocity ω. But their linear velocities will be different say v₁, v₂, v₃,…….etc.