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

Total Questions: 50

31. A constant force acts on an object of mass 10 kg for a duration of 2 seconds. It increases the object’s velocity from 5 m/s to 10 m/s. Find the magnitude of the applied force. Now, if the force is applied for a duration of 5 seconds what would be the final velocity of the object? [RRB NTPC CBT - I (12/03/2021) Morning]

Correct Answer: (d) Applied Force = 25 N, Final Velocity = 17.5 m/s
Solution:

Given, Mass = 10 kg, t₁= 2s, Initial velocity u = 5 m/s, Final Velocity v = 10 m/s, t₂ = 5s.
So, Let the force and acceleration be ‘F’ and ‘a’ respectively.
So, a = (𝑣−𝑢)/t = (10-5-)/2 = 2.5 m/s².
So, the magnitude of applied force is Mass × Acceleration = 10 × 2.5 = 25 N And now final velocity after 5s be v,
v = u + at = 5 + 2.5 x 5 = 17.5 m/s
The final velocity after 5s is 17.5 m/s.

32. A shell of mass 0.04 kg is fired by a gun of mass 120 kg. If the muzzle speed of the shell is 90 m/s, what is the recoil speed of the gun ? [RRB Group D 17/09/2018 (Evening)]

Correct Answer: (b) 3 × 10⁻² m/s
Solution:

33. A ball is dropped from a height of 80 m. The distance travelled by it in the fourth second will be ___. (take g=10 m/s²) [RRB Group D 18/09/2018 (Morning)]

Correct Answer: (d) 35 m
Solution:

34. An object starts moving from its rest position. It attains a speed of 5 m/s in 2 seconds. What will be its acceleration? [RRB Group D 19/09/2018 (Morning)]

Correct Answer: (a) 2.5 m/s²
Solution:

35. An object travels northwards with a constant velocity. The net force acting on the object will be ___________ . [RRB Group D 19/09/2018 (Afternoon)]

Correct Answer: (c) zero
Solution:

Constant velocity means there is no acceleration. Newton’s Second law of motion, Force= Mass x Acceleration. As acceleration = zero. Net acting force is also zero.

36. A force of 350 N is applied to a mass of 500 kg. In this case, what would be the acceleration generated in the object ? [RRB Group D 19/09/2018 (Evening)]

Correct Answer: (b) 0.7 ms⁻²
Solution:

Given : F = 350 N and m = 500 kg

∵ F = ma ⇒ acceleration (a) = f/m = 350/500 = = 0.7 ms⁻².

37. What is the mass of a girl who weighs 450 N ? [RRB Group D 20/09/2018 (Afternoon)]

Correct Answer: (d) 45.9 kg
Solution:

∵ Weight = Mass × Acceleration due to Gravity, and Acceleration due to Gravity = 9.8 m/𝑠² .
According to question,
Mass = weight/accleration due to gravity
= 450/9.8 = 45.9 kg.

38. An 800-kg car is moving at a speed of 90 km/h. It takes 5 s to stop after the brakes are applied. The force applied by the brakes will be __________ . [RRB Group D 20/09/2018 (Evening)]

Correct Answer: (b) 4000 N
Solution:

Given, mass of car (m) = 800 Kg, Initial
Velocity (u) = 90 km/h = = 90 × 5/18 m/s, Time taken to stop the car (t) = 5 s Final Velocity = 0 m/s.
∵ v = u + at ⇒ 0 = 25 + 5a ⇒ a = -5 m/s²
Force = mass × acceleration = 800 × (-5)= -4000 N [negative sign due to opposite direction]
So, The Force applied by the breaker = 4000 N.

39. When a retarding force 'F' is applied in the opposite direction, the angle between the two directions will be ________. [RRB Group D 20/09/2018 (Evening)]

Correct Answer: (c) 180°
Solution:

Work Done = Fs cosθ
Given, The Retarding Force i.e. Work Done = -F × s
Here, we get cosθ = -1
So, θ = 180°

40. A truck falls from an edge to the ground in 0.8 seconds. What height will it be from the ground? (g=10 m/s²) [RRB Group D 22/09/2018 (Morning)]

Correct Answer: (d) 3.2 m
Solution: