Physics Test 1

Question 1 ( 1.0 marks)

Under what condition do the distance and displacement of a moving object have the same magnitude?

Solution:

Distance and displacement of a moving object have the same magnitude when the object keeps moving in the same direction in a straight line.

Question 2 ( 1.0 marks)

What does the slope of a distance−time graph indicate?

Solution:

The slope of a distance−time graph indicates the velocity of a moving object.

Question 3 ( 1.0 marks)

Is uniform circular motion an accelerated motion?

Solution:

Yes. A body moving in a uniform circular motion has acceleration towards the centre of the circular path.

Question 4 ( 1.0 marks)

Is acceleration a scalar or a vector quantity?

Solution:

Acceleration is a vector quantity.

Question 5 ( 1.0 marks)

If you hit a wall with a force of 100 N, then how much force will the wall exert on you?

Solution:

The wall will exert a force of 100 N in the opposite direction.

Question 6 ( 1.0 marks)

Give an example of an event that involves Newton’s third law of motion as well as the principle of conservation of momentum.

Solution:

The recoil of a gun when fired involves both Newton’s third law of motion and the principle of conservation of momentum.

Question 7 ( 1.0 marks)

What do you understand by acceleration due to gravity? What is its unit?

Solution:

The uniform acceleration produced in a freely falling body because of the gravitational force of the earth is known as acceleration due to gravity.

The unit of acceleration due to gravity is m/s².

Question 8 ( 1.0 marks)

Explain why the tip of a sewing needle is sharp.

Solution:

The tip of a sewing needle is made sharp so that the pressure at the tip becomes very high.

Question 9 ( 1.0 marks)

Ravinder tries pushing a block by applying a force of 100 N, but the block does not move. How much work is done by him?

Solution:

As the block does not displace, the amount of work done by Ravinder is zero.

Question 10 ( 1.0 marks)

When is the work done on an object negative?

Solution:

When the direction of displacement of an object is opposite to that of the applied force, the work done on the object is said to be negative.

Question 11 ( 1.0 marks)

Name the three bones present in the human ear.

Solution:

The three bones present in the human ear are hammer, anvil and stirrup.

Question 12 ( 1.0 marks)

Which wave characteristic determines the pitch of a sound?

Solution:

The pitch of a sound is determines by frequency.

Question 13 ( 2.0 marks)

Find the initial velocity of a car which is stopped in 5 seconds by applying brakes. The retardation because of the brakes is 2.5 m/s².

Solution:

According to the 1^st equation of motion,

v = u + at

Here,

v = 0

t = 5 s

a = −2.5 m/s²

Therefore,

0 = u − 2.5 × 5

u = 12.5 m/s

Thus, the initial velocity of the car is 12.5 m/s².

Question 14 ( 2.0 marks)

An air bubble rises from the bottom of a container with an acceleration of 0.1 m/s². If the height of the liquid column is 1 m, then find the time taken by the air bubble to reach the top of the container.

Solution:

Here,

Acceleration, a = 0.1 m/s²

Height of the liquid columns, S = 1 m

Initial velocity of the bubble, u = 0

According to the 2^nd equation of motion,

t = 4.5 seconds.

Therefore, the air bubble takes 4.5 s to reach the top of the container.

Question 15 ( 2.0 marks)

A bullet of mass 20 g is fired horizontally with a velocity of 200 m/s from a pistol of mass 2.5 kg. What is the recoil velocity of the pistol?

Solution:

Here,

Mass of the bullet, m₁ = 20 g = 0.020 kg

Mass of the gun, m₂ = 2.5 kg

Initial velocity of the bullet, u₁ = 0

Final velocity of the bullet, v₁ = 200 m/s

Initial velocity of the gun u₂ = 0

Final velocity of the bullet = v₂ (say)

According to the principle of conservation of momentum,

Initial momentum = Final momentum

∴m₁ × u₁ + m₂ × u₂= m₁ × v₁ + m₂ × v₂

Or, 0 = 0.020 × 200 + 2.5 × v₂

Or, 2.5 × v₂= −(0.020 × 200)

Or,

Or, v₂ = −16 m/s

Therefore, the recoil velocity of the pistol is 16 m/s.

Question 16 ( 2.0 marks)

How is ultrasound used for cleaning?

Solution:

The object to be cleaned is placed in a cleansing solution and ultrasound waves are passed through the solution. Due to its high frequency, ultrasound stirs the cleansing solution. As a result of the stirring, the dust particles or grease present on the surface of the object vibrate with a very high frequency, and become loose. These loose particles are easily taken off from the object by the cleansing solution. When the vibration stops, the particles fall into the solution and the solution is then rinsed to remove the dust particles or grease.

Question 17 ( 2.0 marks)

What is the potential energy of a body of mass 4 kg, if it is kept at a height of 5 metres above the ground level? (Take g = 9.8 m/s²)

Solution:

The potential energy of a body is calculated by using the formula

E_p = m × g × h

Here,

Mass, m = 4 kg

g = 9.8 m/s²

Height, h = 5 m

∴ E_p = 4 × 9.8 × 5

= 20 × 9.8

= 196 J

Question 18 ( 2.0 marks)

Give two examples of the application of Archimedes’ principle.

Solution:

Two examples of the application of Archimedes’ principle:

(i) The hydrometers used for determining the density of liquids are based on Archimedes’ principle.

(ii) The lactometers used for determining the purity of milk are based on Archimedes’ principle.

Question 19 ( 2.0 marks)

A coolie pushes a box of mass 50 kg with a force of magnitude 100 N. What will be the acceleration of the box?

Solution:

Here,

Mass of the box = 50 kg

Force = 100 N

Force = Mass × Acceleration

∴ Acceleration

= 2 m/s²

Hence, the acceleration of the box will be 2 m/s².

Question 20 ( 2.0 marks)

A car of mass 1000 kg, moving at a velocity of 10 m/s, is stopped in 10 seconds by the application of brakes. What is the retardation of the car?

Solution:

Here,

Mass of the car = 1000 kg

Initial velocity, u = 10 m/s

Time taken to stop, t = 10 s

Final velocity, v = 0

v = u + at

∴ 0 = 10 + a × 10

10a + 10 = 0

Or, a = − 1 m/s²

Therefore, the retardation of the car is 1 m/s².

Question 21 ( 3.0 marks)

An unloaded truck of mass 10000 kg can move at a maximum acceleration of 0.2 m/s². What will be its maximum acceleration if it carries a load of 20000 kg?

Solution:

Here,

Mass of the unloaded truck = 10000 kg

Acceleration of the unloaded truck = 0.2 m/s²

Therefore, the maximum force that can be applied by the engine of the truck can be determined as:

Force = 10000 kg × 0.2 m/s²= 2000 N

Total mass of the loaded truck = 10000 kg + 20000 kg = 30000 kg

The maximum force applied by the engine of the truck remains the same in this case, i.e., 2000 N.

Hence, the maximum acceleration of the loaded truck can be determined as:

Therefore, the maximum acceleration of the loaded truck will be 0.067 m/s².

Question 22 ( 3.0 marks)

Explain the factors on which the speed of sound depends.

Solution:

The factors on which the speed of sound depends are as follows:

(i) Temperature − The speed of sound depends upon the temperature of the medium through which it travels. Its speed increases with the increase in the temperature of the medium.

(ii) Humidity − The speed of sound is directly proportional to the humidity of the medium through which it travels.

(iii) Nature of the medium − The speed of sound depends upon the nature of the medium through which it travels. The speed of sound in a gas is less than its speed in a liquid. Again, the speed of sound in a liquid is less than its speed in a solid.

Question 23 ( 3.0 marks)

Rajat is riding a motorcycle of mass 150 kg. What is the change in kinetic energy when the velocity of the motorcycle increases from 10 m/s to 20 m/s? Rajat’s mass is 80 kg.

Solution:

Combined mass of Rajat and his motorcycle, m = (150 + 80) = 230 kg

Initial velocity, v₁ = 10 m/s

Therefore, the initial kinetic energy is:

Final velocity of the motorcycle, v₂ = 20 m/s

Therefore, the final kinetic energy is:

Hence, change in kinetic energy

= 34500 J

Question 24 ( 3.0 marks)

A train, moving at a speed of 180 km/h, comes to a stop with a constant acceleration in 15 minutes, after covering a distance of 25 km. What is the acceleration of the train?

Solution:

Here,

Initial velocity of the train, u = 180 km/h

Distance travelled by the train before it stops, S = 25 km

Time taken to stop, t = 15 minutes = 0.25 h