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

5 k W

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6.25 k W

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50 k W

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62.5 k W

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Solution

The correct option is D $6.25 k W$
Let input power $= P$

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

$\frac{80 P}{100}$
Mass of coal $m = 10, 000 k g$
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, $\frac{80 P}{100} = \frac{10, 000 \times 9.8 \times 180}{60 \times 60}$
Simplifying we get: $P = 6.25 k W$

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