Railway Science (Physics-Force and Pressure) (Part-IV)

Total Questions: 50

11. The momentum of an object of mass 'm' moving with velocity 'V' will be: [RRB Group D 27/09/2018 (Afternoon)]

Correct Answer: (b) mv
Solution:

(Momentum = Mass × Velocity). Momentum is the quantity of motion possessed by an object. The SI unit of momentum is kilogram meter per second (kg/s). Newton’s second law of motion states that the rate of change of momentum is equal to the force acting on the particle.

12. If an object with a mass of 25 kg is moving with a uniform acceleration of 8 ms⁻² , then the force exerted by the body is _______ . [RRB Group D 28/09/2018 (Afternoon)]

Correct Answer: (a) 200 N
Solution:

Given that: Mass = 25 kg and Acceleration = 8 ms⁻²
Force (F) = mass (m) × acceleration (a) = 25 × 8 = 200 N.

13. A boy standing at the top of a 20-metre-high tower drops a stone. Assuming g= 10 m/s², the velocity with which the stone hits the ground is : [RRB Group D 28/09/2018 (Afternoon)]

Correct Answer: (b) 20 m/s
Solution:

Given: Height of the tower (h) = 20 m, g = 10 m/s², initial velocity (u) = 0, final velocity (v) = ?
Using formula of motion:
v²  - u² = 2gh
V²  = u² + 2gh
v²  = (0)² + 2 × 10 × 20
v² = 400 ⇒ v = 20 m/s.

14. If a stone is released from the top of a tower with zero velocity, it reaches the ground in 4 seconds. What is the height of the tower? (take g = 10 m/s²) [RRB Group D 28/09/2018 (Evening)]

Correct Answer: (a) 80m
Solution:

15. A constant force acts on an object of mass 5 kg for a period of 2 seconds. This increases the velocity of the object from 4 ms⁻¹ to 7 ms⁻¹ . Find the amount of force used. [RRB Group D 28/09/2018 (Evening)]

Correct Answer: (a) 7.5 N
Solution:

16. An object weighs 60 N when measured on the surface of the earth. Its weight on the surface of the Moon will be ______. [RRB Group D 28/09/2018 (Evening)]

Correct Answer: (b) 10 N
Solution:

Given that, weight of the object on the earth surface (Wₑ) = 60 N
Weight of the object on the moon (Wₘ) = 1/6 × Wₑ = 1/6 × 60 N = 10 N.

17. If a ball is thrown upward with an initial velocity of 25 m/s, then how much time will it take to reach its highest point. The value of g can be taken as 10 m/s. [RRB Group D 01/10/2018 (Morning)]

Correct Answer: (b) 2.5 seconds
Solution:

Given that : Initial velocity of the ball, u= 25 m/s, final velocity of the ball, v = 0 m/s , Acceleration due to gravity, g = -10
m/s² (upward motion).
By using the formula, v = u + at
0 = 25 + (-10) × t
10t = 25 ⇒ t = 2.5 s.

18. The speed of a body of mass 100 kg changes from 5 ms⁻¹ to 15 ms⁻¹ in 5 s and a uniform acceleration is applied on it. Calculate the force applied on the body? [RRB Group D 01/10/2018 (Afternoon)]

Correct Answer: (d) 200 N
Solution:

Given that, Mass of the object (m) = 100 kg, initial velocity (u)= 5 m/s, final velocity (v) = 15 m/s, Time (t)= 5 sec.
We know that, Force
= mass × acceleration = m × (𝑣 − 𝑢)/t
⇒ Force = 100 × (15-5)/5 = 200 N.

19. The force exerted on an object is 200 N and its mass is 100 kg. Find the acceleration of the object. [RRB Group D 03/10/2018 (Morning)]

Correct Answer: (b) 2 ms⁻²
Solution:

Given, Force = 200 N, Mass = 100 kg.
Force = Mass × Acceleration.
According to the question, Acceleration
= 𝐹𝑜𝑟𝑐e/𝑀𝑎𝑠s = 200/100 = 2 ms⁻²

20. The formula to determine the height of the rocket above the ground at any time during the rocket's flight is given by: h = 119t - 7t² where t = the time, in seconds, that has elapsed since the rocket was launched, and h = the height, in meters, of the rocket above the ground at time t. The value of h was 0 when the rocket hit the ground. How many seconds after the rocket was launched did it hit the ground? [RRB Group D 03/10/2018 (Afternoon)]

Correct Answer: (a) 17
Solution:

Given that, h = 119t - 7t²
∵ h= 0
⇒ 0 = 119t - 7t² ⇒ t = 17 sec.