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
B
6.25 kW
C
50 kW
D
62.5 kW

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 / timeSo, $$\dfrac{80P}{100} = \dfrac{10,000 \times 9.8 \times 180}{60 \times 60}$$Simplifying we get: $$P = 6.25\ kW$$Physics

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