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Question

A crane can lift up $$10000\ kg$$ of coal in $$1\ hour$$ from a mine of $$180\ m$$ depth. If the efficiency of the crane is $$80 \%$$, its input power must be:  [Take : $$g = 10  \ ms^{-2}$$]


A
5 kW
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B
6.25 kW
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C
50 kW
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D
62.5 kW
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Solution

The correct option is D $$6.25\ kW$$
Let input power $$= P$$

Since $$80 $$ % of this power is being used, so effective power is 

$$\dfrac{80P}{100}$$
Mass of coal $$m = 10,000\ kg$$
Time $$t = 1 \times 60 \times 60\ s$$
Height $$h = 180\ m$$
Gravitational acceleration $$g = 10\ m/s^2$$ (given)
Now, Power $$=$$ work / time
So, $$\dfrac{80P}{100} = \dfrac{10,000 \times 9.8 \times 180}{60 \times 60}$$
Simplifying we get: $$P = 6.25\ kW$$

Physics

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