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Fundamental and Derived Units

A motor lifts...

Question

A motor lifts $100 k g$ of water in $2 m i n$ from a well of $60 m$ depth then the electric power of the motor is $(T a k e n g = 10 m / s^{2})$

1000 W

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750 W

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1200 W

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500 W

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Solution

The correct option is D $500 W$
A motor lifts $= 100 k g$ of water
Time $= 2 m i n = 2 \times 60 = 120 s$
Depth $= 60 m$ depth then,
electric power of the motor $= ?$
Taking $= 10 m / s^{2}$
$P = P o w e r = \frac{m g h}{t} = \frac{100 \times 10 \times 60}{120} = 500 W$

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