Kerala Plus One Physics Board Model Paper 2023 with Answers

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

Kerala Plus One Physics Board Model Paper 2023 with Answers

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.
The escape speed from the surface of the earth is ________________ (1)
Answer:
11.2 km/s

Question 2.
Which physical quantity is conserved in all types of collision? (1)
Answer:
Linear momentum

Question 3.
The radius of a capillary tube is doubled. The height of capillary rise will be ________________ (1)
(i) 2h
(ii) h
(iii) \(\frac{h}{2}\)
(iv) √h
Answer:
(iii) \(\frac{h}{2}\)

Question 4.
State true of false.
The rusting of iron is an irreversible process. (1)
Answer:
True

Question 5.
Does the sound waves in air longitudinal or transverse? (1)
Answer:
Longitudinal

Question 6.
The area under velocity-time graph gives ________________ (1)
Answer:
Displacement

Question 7.
Will two vectors be parallel or perpendicular if the cross-product between them is zero? (1)
Answer:
Parallel

Section – B

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

Question 8.
Check whether the equation \(\frac{1}{2}\)mv² = mgh is dimensionally correct by the method of dimensions. (2)
Answer:
\(\frac{1}{2}\)mv² = M(LT^-1)² = ML²T^-2
[mgh] = M × LT^-2 × L = ML²T^-2
Since the dimensions on both sides of the equation are the same, the equation is dimensionally correct.

Question 9.
Write down the type of energy present in each of the following: (4 × ½ = 2)
(i) Flowing water
(ii) Spring of a clock
(iii) Rolling ball
(iv) Raised hammer
Answer:
(i) KE
(ii) PE
(iii) KE
(iv) PE

Question 10.
(a) What is an ideal gas? (1)
(b) Write down the equation for the pressure of an ideal gas and interpret the terms used in it. (1)
Answer:
(a) The gas that obeys all gas laws at all conditions of pressure and temperature is called an ideal gas.
(b) P = \(\frac{1}{3} \mathrm{~nm} \overline{\mathrm{v}}^2\)
n → number of molecules per unit volume
m → mass of molecule
\(\bar{v}\) → average velocity of molecules

Question 11.
(a) What is the angle of contact of a liquid with a solid surface? (1)
(b) Different shapes of water drops on a lotus leaf and glass plate are given below. Redraw it and mark the angle of contact in each case. (1)

Answer:
(a) Angle of contact is the angle between the solid surface and the tangent drawn to the liquid surface at the point of contact inside the liquid.

Question 12.
(a) What is a moment of force? (1)
(b) Under what conditions, the torque due to an applied force is zero? (1)
Answer:
(a) Torque or rotating effect of force.
(b) (i) When the force is applied exactly on the origin or axis of rotation.
(i) When the angle between position vector, \(\overrightarrow{\mathrm{r}}\) and force is zero or 180°.

Question 13.
Match the following. (2)

Physical Quantity	Symbol
(i) Gravitational constant	g
(ii) Gravitational potential energy	F
(iii) Gravitational force	G
(iv) Gravitational intensity	V

Answer:

Physical Quantity	Symbol
(i) Gravitational constant	G
(ii) Gravitational potential energy	V
(iii) Gravitational force	F
(iv) Gravitational intensity	g

Question 14.
(a) What is the shape of the path followed by a projectile? (1)
(b) Bharath wants to draw the variation of velocity components of a projectile motion with time. Help him with a correct diagram. (1)
Answer:
(a) Parabola
(b) The vertical component v_y = v sin θ decreases and becomes zero at maximum height and then increases. The horizontal component v_x = v cos θ remains constant.
Variation of vertical components

Variation of horizontal components

Section – C

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

Question 15.
(a) What are significant figures? (1)
(b) State the number of significant figures of the following: (2)
(i) 0.007 m²
(ii) 2.64 × 10²⁴ kg
(iii) 0.2370 g cm^-3
(iv) 6.320 J
Answer:
(a) Significant figures are digits in the measured value that are reliable plus the uncertain digit.
(b) (i) 1
(ii) 3
(iii) 4
(iv) 4

Question 16.
(a) Define impulse. (1)
(b) Why is it more dangerous to fall on ice than fresh snow? (1)
(c) The passengers fall backward when a bus at rest, starts moving suddenly. Why? (1)
Answer:
(a) Impulse is the product of force and time for which it acts
Impulse = Force × time
(b) ice is harder compared to fresh snow. The impulse caused by fresh snow is less than ice.
(c) This is due to the inertia of rest.

Question 17.
(a) State Hooke’s law. (1)
(b) Draw the stress-strain curve for a metal and mark the following: (2)
(i) Yield point
(ii) Proportional limit
(iii) Fracture point
Answer:
(a) 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 18.
(a) What are nodes and anti-nodes? (1)
(b) Draw the first two harmonics of vibrations of a stretched string fixed at both ends and mark nodes and anti-nodes. (2)
Answer:
(a) The position of maximum amplitude in a standing wave is termed as antinode and the position of minimum amplitude (zero) is termed as node.

Question 19.
(a) State law of conservation of angular momentum. (1)
(b) What ¡s the torque of the force \(\overrightarrow{\mathrm{F}}=(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\) acting at the point \(\overrightarrow{\mathrm{r}}=(3 \mathrm{i}+3 \mathrm{j}+3 \mathrm{k})\) about the origin? (2)
Answer:
(a) If the external torque acting on a rotating body is zero, its angular momentum is constant.

τ = \(27 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-21 \hat{\mathrm{k}}\)

Question 20.
Deduce the equation for uniformly accelerated motion, S = ut + \(\frac{1}{3}\)at² where s – displacement, u – initial velocity, t – time, and a – acceleration. (3)
Answer:

Consider a body moving along a straight line with uniform acceleration a.
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 = \(\left(\frac{v+u}{2}\right) t\)
But v = u + at ………(2)
Substitute eq.(2) in eq.(1), we get
S = \(\left(\frac{u+a t+u}{2}\right) \mathrm{t}\)
S = ut + \(\frac{1}{2}\)at2

Question 21.
(a) What is a conservative force? (1)
(b) Obtain the expression for energy stored in a spring. (2)
Answer:
(a) If work done by a force is independent of the path followed and depends only on the initial and final position, the force is conservative.
(b)

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

W = \(-\frac{1}{2} k x^2\)
This work is stored as potential energy in the spring.
Hence the potential energy of a spring.
P.E. = \(\frac{1}{2} k x^2\)

Section – D

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

Question 22.
(a) What is a Carnot engine? (1)
(b) Draw the P-V curve of the Carnot engine and mark the different thermodynamic processes. (2)
(c) Is it possible to design a Carnot engine with 100% efficiency? Substantiate. (1)
Answer:
(a) A reversible heat engine operating between two temperatures is called Carnot’s engine.
(b)

(c) No. Efficiency of Carnot engine is η = 1 – \(\frac{\mathrm{Q}_2}{\mathrm{Q}_1}\)
For η = 1, Q₂ must be zero, which is impossible.

Question 23.
(a) What is retardation? (1)
(b) A car moving along a straight highway with a speed of 126 km/h is brought to a stop within a distance of 200 m. What is the retardation of car? How long does it take to stop the car? (3)
Answer:
(a) Retardation is rate of decrease in velocity or decrease in velocity in are second.

Question 24.
(a) What is a second’s pendulum? (1)
(b) The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the surface of moon? (3)
Answer:
(a) Second pendulum is the pendulum whose time period is 2s.

Question 25.
(a) Differentiate between conduction and convection of heat. (1)
(b) Write any two factors on which the rate of heat flow of an iron bar depend. (1)
(c) At what temperature do the Celsium and Fahrenheit thermometers have same numerical value? (2)
Answer:
(a) In conduction, heat transfers between two adjacent parts of a body due to temperature difference.
In convection, different parts of fluid moves from one point to other.
(b) (i) Temperature difference
(ii) Area of cross section
(iii) Length of iron bar

Section – E

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

Question 26.
Acceleration due to gravity changes with depth.
(a) Derive the expression for acceleration due to gravity at a depth d, from the surface of the earth. (3)
(b) What is the value of acceleration due to gravity at a depth 250 km from the surface of the earth (R = 6400 km)? (2)
Answer:
(a)

If we assume the earth as a sphere of radius R with uniform density ρ,
mass of earth = volume × density
M = \(\frac{4}{3} \pi R^3 \rho\) ………..(1)
We know acceleration due to gravity on thfe surface,
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) ……….(2)
Substituting eq(1) in eq(2), we get

Question 27.
(a) Find the magnitude and direction of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) in terms of their magnitude and angle θ between them. (3)
(b) The horizontal range of a projectile fired at an angle of 15° is 50 m. Calculate its velocity of projection. (2)
Answer:
(a)

Consider two vectors \(\overrightarrow{\mathrm{A}}(=\overrightarrow{\mathrm{OP}})\) and \(\overrightarrow{\mathrm{B}}(=\overrightarrow{\mathrm{OQ}})\) making an angle q.
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 q and SN = B sin q
Substituting these values in eq(1), we get
OS² = (OP + B cos q)² + (B sin q)²
But OS = R and OP = A
R² = (A + B cos q)² + B² sin²q
R² = A² + 2AB cos q + B² cos²q + B² sin²q
R² = A² + 2 AB cos q + B²
R = \(\sqrt{A^2+2 A B \cos \theta+B^2}\)
The resultant vector \(\vec{R}\) make an angle a with \(\vec{A}\).
From the right angled triangle OSN,

Question 28.
(a) State and prove Bernoulli’s principle. (3)
(b) Define viscosity. (1)
(c) On which factors does the viscous force depend? (1)
Answer:
(a) As we move along a streamline the sum of the pressure (ρ), 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 \(\mathrm{p}+\frac{1}{2} \rho \mathrm{v}^2+\rho \mathrm{gh}\) = constant
(b) Viscosity is liquid friction. When liquid layer moves over another liquid layer, there is a force of friction between the liquid layers, opposing the motion of layers. The coefficient of viscosity for a fluid is defined as the ratio of shearing stress to the strain rate.
(c) Area of contact of liquid, velocity gradient (\(\frac{V}{L}\)).

Question 29.
(a) State Newton’s second law of motion. (1)
(b) Arrive at the law of conservation of momentum from second law and state the law of conservation of momentum. (3)
(c) A bullet of mass 0.010 kg is fired by a gun of mass 100 kg. If the muzzle speed of the bullet is 50 m/s, what is the recoil speed of the gun? (1)
Answer:
(a) 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
F ∝ \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}\)
(b)

Consider two bodies A and B with initial momenta PA and PB. After collision, they acquire momenta \(\mathrm{P}_{\mathrm{A}}^1\) and \(\mathrm{P}_{\mathrm{B}}^1\) respectively.
According to Newton’s second law, the change in momentum of A due to the collision with B,
\(\mathrm{F}_{\mathrm{AB}} \Delta \mathrm{t}=\mathrm{P}_{\mathrm{A}}^1-\mathrm{P}_{\mathrm{A}}\)
F_AB = \(\frac{P_A^1-P_A}{\Delta t}\) ………(1)
Similarly the change in momentum of B due to the collision with A,
\(\mathrm{F}_{\mathrm{BA}} \Delta \mathrm{t}=\mathrm{P}_{\mathrm{B}}^1-\mathrm{P}_{\mathrm{B}}\)
F_BA = \(\frac{P_B^1-P_B}{\Delta t}\) ………..(2)
[Where ∆t is time for which the two bodies are in contact]
According Newton’s third law, we can write

Total momentum before collision = Total momentum after collision
(c) m = 0.010 kg
M = 100 kg
v = 50 m/s
V = \(\frac{\mathrm{mv}}{\mathrm{M}}\)
= \(\frac{0.010 \times 50}{100}\)
= 0.005 m/s
= 5 × 10^-3 m/s