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

Total Questions: 50

31. How many times is the weight of an object on Earth its weight on the Moon? [RRB Group D 08/10/2018 (Morning)]

Correct Answer: (d) 6
Solution:

The weight of an object on the moon is about one-sixth of its weight on the earth. This is because the value of acceleration due to gravity on the moon is about one-sixth of that on the earth.

32. A 30 gm bullet with velocity 150 ms⁻¹ is fired horizontally from a 3 kg pistol. What is the reaction velocity of the pistol? [RRB Group D 08/10/2018 (Morning)]

Correct Answer: (c) -1.5 ms⁻¹
Solution:

Given : Mass of the bullet (m) = 30 g = 0.03 kg, Velocity of bullet (v)= 150 ms⁻¹ , Mass of the pistol (M) = 3 kg, Velocity of pistol (V)= ?
∵ Initial momentum of the system is ‘0’ as it is at rest.
After firing,

V = - 𝑚𝑣/m = - 0.03 × 150/3 = -1.5 ms⁻¹.

33. A car falls from the edge of the road to the ground in 0.5 s. Let g = 10 ms⁻² . How high is the edge of the road above the ground? [RRB Group D 09/10/2018 (Morning)]

Correct Answer: (b) 1.25
Solution:

34. The weight of an object is 60N when measured on the surface of the Earth. What will be its weight on the surface of the moon? [RRB Group D 09/10/2018 (Evening)]

Correct Answer: (d) 10 N
Solution:

Given that, Weight of the object on the Earth surface (Wₑ) = 60 N
Now, Weight of the object on the Moon
(Wₘ) = 1/6 × Wₑ ⇒ Wₘ = 1/6 × 60 = 10 N.

35. A car falls off a ledge and drops to the ground in 0.7 s. (Let g=10 ms⁻²) What is its speed on striking the ground? [RRB Group D 11/10/2018 (Morning)]

Correct Answer: (d) 7 ms⁻¹
Solution:

Given, Time = 0.7 s and Acceleration due to gravity (g) = 10 m/s²
Using first equation of motion, v = u + at V = 0 + 10 × 0.7 = 7 m/s

36. Which of the following represents the equation for position-time relation? [RRB Group D 11/10/2018 (Morning)]

Correct Answer: (b) s = ut + 1/2 𝑎𝑡²
Solution:

Equation of  Motion - Mathematical formula that describes the position, velocity, or acceleration of a body relative to a given frame of reference. Other Equation of Motion : v = u + at (Velocity - time relation) and v²  = u²  + 2as (Velocity - Position relation).

37. A car falls from a mountain road and comes to the ground in 0.8 seconds. (Let's say g = 10 ms⁻²). What will be its speed when it hits the ground ? [RRB Group D 11/10/2018 (Afternoon)]

Correct Answer: (d) 8ms⁻¹
Solution:

Given : As the car is in free fall, the initial speed is zero. Under free fall, initial velocity (u)= 0, g = 10 ms⁻² , Time (t) = 0.8 s, Final velocity (v) = ?
v = u + gt = 0 + 10 × 0.8 = 8 ms⁻¹

38. The mass of an object is 10 kg. What would be its weight on earth ? (g = 10 ms⁻²) [RRB Group D 11/10/2018 (Afternoon)]

Correct Answer: (b) 100 N
Solution:

Given : Mass of an object (m) = 10 kg,
Acceleration due to gravity (g) = 10 ms⁻² .
We know that, Weight = Mass × Acceleration Due to Gravity.
⇒ Weight = 10  × 10 =100 N.

39. Which of the following equations represents the position-velocity relation? [RRB Group D 11/10/2018 (Evening)]

Correct Answer: (b) 2as = v² - u²
Solution:

Three equations of motion : First Equation of motion : v = u + at (represents the velocity - time relation); Second Equation of motion : s = ut + 1/2 𝑎𝑡²
(represents position-time relation); Third Equation of motion
v² - u² = 2as  (position-velocity relation).

40. A car falls from an outcrop and hits the ground in 0.9 sec. (Assume g = 10 ms⁻²). What is its speed when it hits the ground ? [RRB Group D 11/10/2018 (Evening)]

Correct Answer: (b) 9 ms⁻¹
Solution:

Given, Initial speed (u) = 0 (As car is under free fall), t = 0.9 s, Acceleration due to gravity (g) = 10 ms⁻²
v = u + gt = 0 + 10 × 0.9 = 9 ms⁻¹ .