Kerala Plus One Physics Question Paper SAY 2019 with Answers

Reviewing Kerala Syllabus Plus One Physics Previous Year Question Papers and Answers Pdf SAY 2019 helps in understanding answer patterns.

Kerala Plus One Physics Previous Year Question Paper SAY 2019

Time: 2 Hours
Total Scores: 60

Answer any three questions from 1 to 4. Each carries one Score. (3 × 1 = 3)

Question 1.
The rotational analogue of mass is called _____________
Answer:
Moment of inertia

Question 2.
Which among the following possesses the highest specific heat capacity?
(i) Metals
(ii) Ice
(iii) Water
(iv) Glass
Answer:
(iii) Water

Question 3.
A light body and a heavy body have equal kinetic energies, which one has greater momentum?
Answer:
Heavy Body

Question 4.
Select the strongest force from the following list.
(Electromagnetic force, Gravitational force, Weak nuclearforce)
Answer:
Electromagnetic force

Answer any six questions from 5 to 11. Each carries two Scores. (6 × 2 = 12)

Question 5.
Obtain the relation between linear velocity and angular velocity.
Answer:

Let the ∆θ be the angle constructed by the body during the time interval ∆t.
The angular velocity can be written as
ω = \(\frac{\Delta \theta}{\Delta t}\) ……..(1)
If the distance traveled by the object during the time ∆t is ∆r (ie. PP₁ = ∆s)
then the speed v = \(\frac{\Delta r}{\Delta t}\) ……..(2)
But ∆s = R∆θ
where R = \(|\vec{r}|=|\vec{r}|\)
Substituting ∆r = R∆θ in eq.(1) we get
v = \(\frac{\Delta \mathrm{r}}{\Delta \mathrm{t}}=\frac{\mathrm{R} \Delta \theta}{\Delta \mathrm{t}}\)
But we know, when ∆t → 0, \(\frac{\Delta \theta}{\Delta t}=\frac{d \theta}{d t}\)
∴ Instantaneous velocity, v = R \(\frac{d \theta}{d t}\) = Rω

Question 6.
The position-time graph of two objects A and B are shown below.

(a) Which body has greater velocity?
(b) Find the ratio of velocities of A and B.
Answer:
(a) A

Question 7.
The moment of inertia of a thin rod of mass M and length l about an axis perpendicular to the rod at its midpoint is \(\frac{M \ell^2}{12}\). Find the moment of inertia of the rod about an axis perpendicular to it and passing through one end of the rod.
Answer:

Question 8.
Derive an expression for the variation of g with height (h) above the earth’s surface.
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 9.
Derive the relation C_P – C_V = R where C_P and C_V are molar-specific heat capacities of an ideal gas at constant pressure and volume respectively and R is the universal gas constant.
Answer:
According to 1st law of thermodynamics
∆Q = ∆U + P∆V
If ∆Q heat is absorbed at constant volume (∆V = 0)

= \(\left[\frac{\Delta \mathrm{U}}{\Delta \mathrm{~T}}\right]_{\mathrm{P}}+P\left[\frac{\Delta \mathrm{~V}}{\Delta \mathrm{~T}}\right]_{\mathrm{P}}\) ………(2)
From the ideal gas equation for one mole PV = RT
Differentiating w.r.t. temperature (at constant pressure)
\(p\left[\frac{\Delta \mathrm{~V}}{\Delta \mathrm{~T}}\right]_{\mathrm{P}}\) = R ……..(3)
Substituting in equation (2)
C_P = \(\frac{\Delta \mathrm{U}}{\Delta \mathrm{~T}}\) ………(4)
Equation (4) – Equation (1), we get
C_P – C_V = R

Question 10.
What do you mean by Mean free path? Give an equation for the Mean free path.
Answer:
The mean free path is the average distance covered by a molecule between two successive collisions.
Mean free path l = \(\frac{1}{\sqrt{2} n \pi d^2}\)

Question 11.
A particle executes SHM of amplitude A. At what distance from the mean position is its kinetic energy equal to its potential energy?
Answer:
KE = PE

y = \(\frac{A}{\sqrt{2}}\), where A is amplitude when y = \(\frac{A}{\sqrt{2}}\), the potential and kinetic energy of SHM motion becomes equal.

Answer any six questions from 12 to 18. Each carries three Scores. (6 × 3 = 18)

Question 12.
A large force acting for a short time is called an impulsive force.
(a) What is the SI unit of impulse?
(b) Two billiard balls each of mass 0.05 kg moving in opposite directions with a speed of 6 m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer:
(a) N_S
b) Impulse = change in momentum
= m₁υ₁ – m₂υ₂
But m₁ = m₂ = 0.05 kg
υ₁ = 6 m/s, u₂ = -6 m/s
i.e, impulse = 0.05 (6 – (-6))
= 0.05(12)
= 0.6 kg m/s

Question 13.
The velocity-time graph of an object is given below.

(a) The area under this graph gives _____________
(b) Derive the relation \(V_0 t+1 / 2 a t^2\) using the above graph.
Answer:
(a) Displacement
(b)

Question 14.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surrounding is 20°C.
Answer:
According to Newton’s law of cooling,
rate of cooling ∝ temperature difference

Question 15.
Derive an expression for the period of oscillation of a loaded spring.
Answer:

Consider a body of mass m attached to a massless spring of spring constant K.
The other end of the spring is connected to a rigid support as shown in the figure.
The body is placed on a frictionless horizontal surface.
If the body is displaced towards the right through a small distance ‘x’, a restoring force will be developed.

Question 16.
The stress-strain graph of two materials A and B are shown below.

(a) State the law which relates stress with strain.
(b) Which of the two materials has the greater Young’s modulus?
(c) Which of the two materials is more ductile?
Answer:
(a) According to Hooke’s law, stress is directly proportional to strain.
(b) A (The material A has more slope than B)
(c) A

Question 17.
A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad s^-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular momentum of the cylinder about its axis?
Answer:
Mass of cylinder, m = 20 kg
Angular velocity, ω = 100 rad/s
Radius r = 0.25 m
Moment of inertia I = \(\frac{1}{2} m r^2\)
= \(\frac{1}{2}\) × 20 × (0.25)²
= 0.625 kg m²
Kinetic energy KE = \(\frac{1}{2} \mathrm{I} \omega^2\)
= \(\frac{1}{2}\) × 0.625 × (100)²
= 3125 J
Angular momentum L = Iω
= 0.625 × 100
= 62.5 kg m² s^-1

Question 18.
Carnot cycle for a heat engine with an ideal gas as the working substance is shown below.

(a) Name the four processes taking place in the Carnot cycle.
(b) Can a Carnot engine work if its sink and source are interchanged? Explain.
Answer:
(a) (i) Iso thermal expansion
(ii) Adiabatic expansion
(iii) Iso thermal compression
(iv) Adiabatic compression
(b) No, Heat always flows from a higher temperature to a lower temperature.

Answer any 3 questions from 19 to 22. Each carries four Scores. (3 × 4 = 12)

Question 19.
The figure below shows the path of a projectile motion.

(a) Obtain the expressions for maximum height and time of flight.
(b) What is the angle of projection for maximum horizontal range?
Answer:
(a)

The time taken by the projectile to cover the horizontal range is called the time of flight.
The time of flight of the projectile is decided by u sin q.
The time of flight can be found using the formula s = ut + \(\frac{1}{2}\)at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = u sin θ, we get
0 = u sin θt – \(\frac{1}{2}\)gt²
\(\frac{1}{2}\)gt2 = u sin θt
t = \(\frac{2 u \sin \theta}{g}\)
Vertical height
The vertical height of the body is decided by the vertical component of velocity (u sin q).
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 = \(\frac{u^2 \sin ^2 \theta}{2 g}\)
(b) 45°

Question 20.
Several games such as billiards, marbles, or carrom involve collision.
(a) What is meant by completely inelastic collision?
(b) Show that in a perfectly elastic collision in one dimension, the relative velocity after the collision is numerically equal to the relative velocity before the collision.
Answer:
(a) Inelastic collision, only momentum is conserved.
(b) Collisions in one Dimension: If the initial velocities and final velocities of both the bodies are along a straight line, then it is called one-dimensional motion.

Consider two bodies of masses m₁ and m₂ moving with velocities u₁ and u₂ in the same direction and in the same line.
If u₁ > u₂ they will collide. After collision let v₁ and v₂ be their velocities.
By conservation of linear momentum.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ……….(1)
m₁u₁ – m₁v₁ = m₂v₂ – m₂u₂
m₁ (u₁ – v₁) = m₂ (v₂ – u₂) ………….(2)
This is an elastic collision, hence K.E. is conserved.

Question 21.
Kepler formulated three laws of planetary motion.
(a) State Kepler’s law of periods.
(b) A Saturn year is 29.5 times the Earth year. How far is Saturn from the sun if the Earth is 1.5 × 10⁹ km away from the sun?
(c) Of which conservation law is Kepler’s second law of planetary motion, a consequence?
Answer:
(a) The square of the period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse.
ie. T² ∝ a³
(b) radius of earth R_e = 1.5 × 10⁸ km
Period of earth T_e = 1 year
Period of Saturn, T_s = 29.5 years
We know T² ∝ a³
for earth 1² ∝ (1.5 × 10⁸)³ ……….(1)
for Saturn (29.5)² ∝ Rs …………(2)
\(\frac{1^2}{(29.5)^2} \propto \frac{\left(1.5 \times 10^8\right)^3}{R_s^3}\)
R_s = 1.432 × 10¹³ m
(c) Law of conservation of momentum.

Question 22.
When a metallic sphere falls through castor oil, its velocity becomes uniform, called terminal velocity.
(a) Write the expression for terminal velocity.
(b) The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20°C is 6.5 cm/s. Compute the viscosity of the oil at 20°C. The density of oil is 1.5 × 10³ kg/m³. The density of copper is 8.9 × 10³ kg/m³.
(c) Raindrops falling under gravity do not acquire very high velocity. Why?
Answer:
(a) Terminal velocity v = \(\frac{2}{9} a^2 \frac{(\rho-\sigma) g}{\eta}\)
(b) Radius, r = 2 × 10^-3 m
Velocity v = 6.5 × 10^-2 m/s
Density of oil = 1.5 × 10³ kg/m³
Density of copper ρ = 8.9 × 10³ kg/m³

(c) Raindrops acquire terminal velocity.

Answer any three questions from 23 to 26. Each carries five scores. (3 × 5 = 15)

Question 23.
The correctness of the equation can be checked using the principle of homogeneity in dimensions.
(a) State the principle of homogeneity.
(b) Using this principle, check whether the equation f = \(2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}}\) is dimensionally correct, where f – frequency, l – length, and g – acceleration due to gravity.
(c) The velocity V of a particle depends on time ‘t’ as V = At² + Bt. Find the dimensions and units of A and B.
Answer:
(a) For the correctness of an equation, the dimensions on either side must be the same.
(b) f = \(2 \pi \sqrt{\ell / g}\)
Dimension of f = \(\frac{1}{T}\) = T^-1
Dimension of \(\frac{\ell}{g}=\frac{L}{1 / T^2}\) = T²
Dimensions on both sides are different. Hence this equation is wrong,
(c) v = At² + Bt
v = At²
\(\frac{L}{T}\) = AT²
Dimension of A = \(\frac{\mathrm{L}}{\mathrm{TT}^2}\) = LT^-3
Unit = m/s³
v = Bt
\(\frac{L}{T}\) = BT
Dimension of B = \(\frac{\mathrm{L}}{\mathrm{~T}^2}\) = LT^-2
Unit = m/s²

Question 24.
The static friction comes into play at the moment the force is applied.
(a) Write the relation between static friction and normal reaction.
(b) Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train’s floor is 0.15.
(c) State the laws of limiting friction.
Answer:
(a) f_s = μ_sN
(b) ma = μ_smg
a = μ_sg
= 0.15 × 10
= 1.5 m/s²
(c) (1) The force of maximum static friction is directly proportional to the normal reaction
(2) The force of static friction is opposite to the direction in which the body tends to move.
(3) The force of static friction is parallel to the surfaces in contact.
(4) The force of maximum static friction is independent of the area of contact (as long as the normal reaction remains constant).
(5) The force of static friction depends only on the nature of surfaces in contact.

Question 25.
While conducting a resonance column experiment in the laboratory you can hear the maximum sound at a certain height.
(a) Which phenomenon is responsible for this?
(b) Is the resonance column apparatus an open pipe or a closed pipe?
(c) Find the ratio of frequencies of the first three harmonics in the Resonance column apparatus.
Answer:
(a) Resonance
(b) Closed pipe
(c) For closed pipe

Question 26.
A fluid moving in a pipe of varying cross-sectional area is shown below.

(a) What is the difference between streamlined flow and turbulent flow?
(b) State and prove Bernoulli’s principle.
Answer:
(a) Stream line flow is regular but turbulent flow is irregular.
(b) 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