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Q.

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs^-1 on a conveyor belt moving with a velocity of 5 ms^-1. The power needed to keep the belt moving with the same velocity is ________W (Round off to the nearest integer)

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answer is 12.

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Detailed Solution

$\begin{array}{l} P = F \times v \\ P = \frac{dmv}{dt} \times v = v^{2} \frac{dm}{dt} \end{array}$

$P = 25 \times 0.5 = 12.5 W$

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