Railway Science (Physics-Work, Energy and Power) (Part-IV)

Total Questions: 50

21. The work done across two points having a potential difference of 20 V is 60 J. Then the electric charge flowing between the two points is: [RRB Group D 22/10/2018 (Afternoon)]

Correct Answer: (b) 3 C
Solution:

Given Voltage = 20 V and Work done = 60 J.
Then electric charge = π‘Šπ‘œπ‘Ÿπ‘˜ π·π‘œπ‘›π‘’/π‘‰π‘œπ‘™π‘‘π‘Žπ‘”π‘’
= 60/20 =Β  3C.

22. Find the energy possessed by an object of mass 11 kg when it is at a height of 7 m above the ground. Given g = 9.8 ms⁻² [RRB Group D 22/10/2018 (Afternoon)]

Correct Answer: (b) 754.6 J
Solution:

Given that : Mass of object (m) = 11 kg,
Height = 7 m, Acceleration due to gravity (g)= 9.8 ms⁻² .
∡ Potential energy = m Γ— g Γ— h
= 11 Γ— 9.8 Γ— 7= 754.6 J.

23. How much work is done to transfer a charge of 4 C between two points having a potential difference of 10 V? [RRB Group D 22/10/2018 (Evening)]

Correct Answer: (c) 40 J
Solution:

Given that : Moving charge (Q) = 4 C,
Two points having a potential difference (V) =10 V, Work = W
W = V Γ— Q
W = 10V Γ— 4C = 40 J.

24. If the amount of work done to move a charge of 3 Coulomb between two points is 72 J, then what will be the potential difference between these points? [RRB Group D 22/10/2018 (Evening)]

Correct Answer: (a) 24 V
Solution:

Given : Charge (Q) = 3 C, Work (W) = 72 J.
W = Q Γ— V
V = π‘€π‘œπ‘Ÿπ‘˜/π‘β„Žπ‘Žπ‘Ÿπ‘”e = 72/3 β‡’ V = 24 J.

25. An object of mass 12 kg is kept at a certain height from the ground. If the potential energy of the object is 480 J, then find the height of the object from the ground? Given g = 10 ms⁻² [RRB Group D 22/10/2018 (Evening)]

Correct Answer: (c) 4 m
Solution:

Given that : Mass (m) = 12 kg, Potential energy (P.E.) = 480 J, Acceleration due to gravity (g) = 10 m/sΒ²
Rearranging the formula PE = mgh, to solve for h,
we get : h = P.E. / (m Γ— g)
h = 480 /(12 Γ— 10) β‡’ h = 4 m.

26. A boy weighing 50 kg climbs 40 stairs in 10 s. If the height of each ladder is 15 cm. So find its power. Let g = 10 ms⁻² . [RRB Group D 23/10/2018 (Morning)]

Correct Answer: (c) 300 W
Solution:

Given : Mass, m = 50 kg and Acceleration due to gravity, g = 10 m/sΒ²
Force (Weight) = m Γ— g = 50 kg Γ— 10 m/sΒ² = 500 N.
Total distance climbed = Height of one stair Γ— Number of stairs = 0.15 m Γ— 40 = 6 m
Now, Power = π‘Šπ‘œπ‘Ÿπ‘˜/π‘‘π‘–π‘še
Given : t = 10 s
Power =Β  300𝐽/10𝑠= 300 W.

27. A boy of mass 50 kg climbs 45 stairs in 10 seconds. If the height of each ladder is 16 cm, find its power. (Assume g =10 ms⁻²) [RRB Group D 23/10/2018 (Afternoon)]

Correct Answer: (c) 360 W
Solution:

28. A boy of mass 50 kg climbs 40 Steps in 9 seconds. If the height of each ladder is 15 cm, find its capacity. Given g = 10 ms⁻² . [RRB Group D 23/10/2018 (Evening)]

Correct Answer: (a) 333.33 W
Solution:

Given : Mass (m) = 50 kg, Time (t) = 9 s
Distance = 40 Γ— 15 = 600 cm = 6 m
Potential Energy = mgh = 50 Γ— 10 Γ— 6 = 3000 J
Power of the Body = Rate of doing work =
πΈπ‘›π‘’π‘Ÿπ‘”π‘¦/π‘‡π‘–π‘šπ‘’ = 3000/9 = 333.33 W.

29. A boy of mass 50 kg climbs 43 stairs in 10 s. If the height of each ladder is 15 cm, find its strength. Given g = 10 ms⁻² [RRB Group D 24/10/2018 (Evening)]

Correct Answer: (c) 322.5 W
Solution:

Given : Mass (m) = 50 kg, Height of each stair = 15 cm = 0.15 m, Number of stairs = 43, Total height = 0.15 m Γ— 43 = 6.45 m.
Time taken (t) = 10 s and g = 10 ms⁻² .
Now, Force = mass Γ— acceleration
= 50 kg Γ— 10 ms⁻² = 500 N
Work = Force Γ— distance
= 500 N Γ— 6.45 m = 3225 J
Power = π‘Šπ‘œπ‘Ÿπ‘˜/π‘‘π‘–π‘še = 3225𝐽/10𝑠 = 322.5 W.

30. An object in vertical position hits the floor with velocity u and bounces back at the same velocity. Find the change in motion. [RRB Group D 24/10/2018 (Evening)]

Correct Answer: (c) 0
Solution:

Change in motion = Final velocity - Initial velocity
∴ Change in motion = v - u
Here, v = u
∴ Change in motion = 0.