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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 will be :

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1.25 W

2.5 W

6.25 W

12.5 W

answer is D.

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Detailed Solution

$Thrust = V_{rel} \times \frac{dm}{dt} = 2.5 N$

Now, Power = F × V = 12.5 W

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