Question

A crane of efficiency 80% is used to lift 8000 kg of coal from a mine of depth 108 m. If the time taken by the crane to lift the coal is one hour, then the power of the crane (in kW) is (Acceleration due to gravity \( = 10 \, \text{m s}^{-2} \))

• A. 5
• B. 4
• C. 6
• D. 3

Hint:

Work done to lift a mass \(m\) to a height \(h\) is \(mgh\). Power = Work / Time. Efficiency \( \eta = \frac{\text{Useful Output Power}}{\text{Total Input Power}} \). Units: 1 kW = 1000 W. 1 hour = 3600 seconds. The "power of the crane/engine" typically refers to its input power rating.

Answer 1

The Correct Option is D

Mass of coal \( m = 8000 \) kg.
Depth (height) \( h = 108 \) m.
Time taken \( t = 1 \) hour \( = 1 \times 60 \times 60 = 3600 \) seconds.
Acceleration due to gravity \( g = 10 \, \text{m s}^{-2} \).
Work done in lifting the coal (Output Work) = Potential energy gained by coal.
\( W_{out} = mgh = 8000 \times 10 \times 108 = 80000 \times 108 = 8640000 \) Joules.
Output Power \( P_{out} = \frac{W_{out}}{t} = \frac{8640000 \text{ J}}{3600 \text{ s}} \).
\( P_{out} = \frac{86400}{36} = \frac{21600}{9} = 2400 \) Watts.
\( P_{out} = 2.
4 \) kW.
Efficiency \( \eta = \frac{\text{Output Power}}{\text{Input Power}} = \frac{P_{out}}{P_{in}} \).
Given efficiency \( \eta = 80% = 0.
8 \).
The "power of the crane" usually refers to its input power \( P_{in} \).
\[ P_{in} = \frac{P_{out}}{\eta} = \frac{2400 \text{ W}}{0.
8} = \frac{24000}{8} = 3000 \text{ Watts} \] \[ P_{in} = 3 \text{ kW} \] This matches option (4).