Teachers recommend solving Kerala Syllabus 9th Standard Physics Question Paper Set 1 to improve time management during exams.
Kerala Syllabus Std 9 Physics Model Question Paper Set 1 English Medium
Time: 1½ Hours
Max Score: 40 Marks
Instructions:
- First 15 minutes is given as cool off time. This time is to b. spent for reading and understanding the questions.
- Answer the questions according to the directions.
- Score and time to be considered while answering.
I. Answer any three questions from 1 to 4. (1 score each) (3 × 1 = 3)
Question 1.
A body of mass 4 kg is moving with a velocity of 4m/s. What is its momentum?
(16 kgm/s, 12 kgm/s, 24 kgm/s, 48 kgm/s)
Answer:
16 kgm/s
Question 2.
Value of ‘g’ at the centre of the Earth is _____
Answer:
Zero
Question 3.
Identify the relation between the first pair and complete the second
Velocity : m/s
Acceleration : _____
Answer:
m/s2
Question 4.
Speed of light in a medium having high optical density will be _____.
Answer:
lower
II. Answer any Seven questions from 5 to 13. (2 score each) (7 × 2 = 14)
Question 5.
An object weighs 60 kg. wt. on the Earth. What is its weight on the Moon?
(gmoon = 1.62 m/s2)
Answer:
Weight on moon = \(\frac { 60 }{ 9.8 }\) × 162 = 9.918 kg.wt.
Question 6.
Write down equations of motion.
Answer:
v = u + at
s = ut + 1/2 at2
v2 = u2 + 2as
Question 7.
Match the following.
| Twinkling of stars | Total internal reflection |
| periscope | Reflection and total internal reflection |
| Mirage | Refraction |
| Straw in water appears bent | Atmospheric refraction |
Answer:
| Twinkling of stars | Atmospheric refraction |
| periscope | Total internal reflection |
| Mirage | Reflection and total internal reflection |
| Straw in water appears bent | Refraction |
Question 8.
A car and a lorry are travelling at the same velocity.
Which has greater momentum? Why?
Answer:
Lorry has greater momentum. This is because. as mass increases. momentum Increases.
Question 9.
II a ball of mass 20 kg and another ball of mass 10 kg fall simultaneously from the top of a tower, which one will reach the ground first? Why?
Answer:
Both the balls will reach the ground at the same time. It is because the acceleration due to gravity does not depend on the mass of the body.
Question 10.
Write down the action and reaction while we are walking on a floor?
Answer:
Action – When we are walking on a floor, we apply a force on the floor.
Reaction – The floor applies a force n the opposite direction.
Question 11.
The massof the Earth is 6 × 1024kg, and that of the Moon is 7.4 × 1022kg. The distance between Earth and. Moon is 3.84 × 105 km. Calculate the force of attraction. (G = 6.7 × 10-11 Nm2 kg-2)
Answer:
M = 6 × 1024kg
m = 7.4 × 1022kg
d = 3.84 × 105km
F = \(\frac{G M m}{d^2}\)
= \(\frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 7.4 \times 10^{22}}{\left(3.84 \times 10^5\right)^2}\)
= 20.17 × 105N
Question 12.
Find out the appropriate law for the given situations.
a) When the tip of an inflated bauoon is released, the balloon moves in a direction opposite to the direction of motion of air.
b) When a bus moves forward suddenly from rest, the standing passengers tend to fall back.
c) During a pole wait jump, the impact is reduced by falling on the foam bed.
d) Gas at high pressure pushed out from the chambers of rocket causes rocket propulsion.
Answer:
a) Newton’s third law of motion
b) Newton’s first law of motion
c) Newton’s second law of motion
d) Newton’s third law of motion
Question 13.
The refractive index of glass is 1.5. If the speed of light in vacuum is 3 × 108 m/s, find speed of light in glass.
Answer:
Refractive Index of glass n = 1 .5
Speed of light in vacuum = 3 × 108 m/s
Speed of light in glass,
v = \(\frac { c }{ n }\) = \(\frac{3 \times 10^8}{1.5}\) = 2 × 108 m/s
III. Answer any five questions from 14 to 19. (3 score each) (5 × 3 = 15)
Question 14.
A car starting from rest attains a velocity of 30 m/s within 3 s.
a) Calculate the acceleration of the car.
b) Find displacement of the car within 3 s.
Answer:
u = 0
v = 30 m/s
t = 3s
a) a = \(\frac{v-u}{t}\) = \(\frac{30-0}{3}\) = 10m/S2
b) S = Ut + \(\frac{1}{2}\)at2 = 0 × 3 + \(\frac{1}{2}\) × 10 × 32 = 45m
Question 15.
Assume that a body of mass 1 kg is kept at the poles, equator and centre of the Earth.
a) In which place does this body experience maximum weight?
b) In which place does this body experience minimum weight?
c) Justify your answer.
Answer:
a) At the poles.
b) Weight at equator will be less than that at poles.
Weight at the centre of Earth will be zero:
c) Weight = mxg, but g = GM/R2
Here, R is the radius of the earth. As R decreases, weight increases and as R Increases, weight decreases. Radius of earth (R) is not the same everywhere because earth is not a perfect sphere.
Question 16.
A car moving with a speed of 54 km/h comes to rest after 3s on applying brake. If the mass of the car, including the passengers, is 1000kg. What will be the force applied when the brake is applied?
Answer:
u = 54 Km/h = \(\frac{54 \times 5}{18}\) = 15m/s
t = 3s
m = 1000 kg
v = 0
F = ma = \(m\left(\frac{v-u}{t}\right)\)
= 1ooo × \(\left(\frac{0-15}{3}\right)\) = –\(\frac { 15000 }{ 3 }\) = -5000N
Question 17.
Why do stars appear to twinkle?
Answer:
The light coming from the stars reaches our eyes by traveling through the atmosphere. The optical density of the medium through which the light travels changes as the physical conditions (pressure, temperature etc.) of the layers of the atmosphere change continuously. Hence, the light undergoes an irregular refraction. Therefore, when the light rays from the stars reach the eyes after refracted several times, the star cannot be seen continuously at the same position. This is the reason for the twinkling of stars.
Question 18.
State and prove the impulse-momentum principle.
Answer:
The product of the force and the time is the impulse of the force. The impulse of force is equal to the change in momentum. This is the impulse-momentum principle. According to Newtons second law of motion.
F = \(\frac{m v-m u}{t}\)
F × t = m (v – u)
F × t = mv – mu
i.e. impulse = change in momentum
This is known as impulse – momentum principle.
Question 19.
An object starting from rest travels with a uniform acceleration of 6 m/s2. Calculate the velocity and distance travelled after 1 minute?
Answer:
u = 0
a = 6m/s2
t = 1 minute = 60s
y = u + at
= 0 + 6 × 60
= 360 m/s
s = ut + 1/2 at2
= 0 × 60 + 1/2 × 6 × 602
= 10800 m
IV. Answer any two questions from 20 to 22. (4 score each) (2 × 4 = 8)
Question 20.
Give reason.
a) Karate experts move their hands with great speed to break strong bricks.
b) While catching a cricket ball, the player moves his hands backwards along with the ball.
c) A running athlete cannot stop himself abruptly at the finishing line of a race.
d) It Is easy to stop a rolling empty drum but difficult to stop a rolling drum filled with tar.
Answer:
a) The force acting on a body will be inversely proportional to the time taken. That is, as time decreases, force increases.
b) The force acting on a body will be inversely proportional to the time taken. That is, as time increases force decreases.
c) Due to inertia of motion
d) As mass increases, inertia increases.
Question 21.
a) Define centripetal acceleration and centripetal force. What is their direction?
b) What happens to the hammer, which is under circular motion, if it loses the centripetal force?
c) A hammer of mass 2 kg is subjected to circular motion along a circular path of radius 2 m with a velocity of 2 m/s. Calculate its centripetal force.
Answer:
a) The acceleration experienced by an object in a circular motion, along the radius, towards the centre of the circle, is known as centripetal acceleration. The force that creates a centripetal acceleration is called centripetal force. Centripetal acceleration and centripetal force are directed towards the centre.
b) It will be thrown off along the tangent at that point.
c) m = 2kg.
v = 2m/s
r = 2m
Fc = \(\frac{m v^2}{r}\) = \(\frac{2 \times 2^2}{2}\) = 4N
Question 22.
a) Define critical angle. What change in the path of light will be observed if the angle of incidence is increased from critical angle?
b) Will total internal reflection occur for a ray of light entering from water to air at an angle of 49.8°? Why?
Answer:
a) Critical angle -when a ray of light enters from an optically denser medium to a rarer medium the angle of incidence at which the angle of refraction becomes 90° is the critical angle.
If the angle of incidence is increased from critical angle, then the ray will undergo total internal reflection, ie., the ray will be reflected back to the same medium.
b) Total internal reflection will occur for a ray of light entering from water to air at an angle of 49.8° because the angle of incidence is greater than the critical angle of water.