Sand is being dropped from a stationary dropper at a rate of 0.5 kg s⁻¹ on a conveyor belt moving with a velocity of 5 m s⁻¹. The power needed to keep the belt moving with the same velocity will be: (a) 1.25 W (b) 2.5 W (c) 6.25 W (d) 12.5 W

Answer: (d). Sand lands with zero horizontal speed, so the belt must supply force F = vdm/dt. Power P = Fv = dm/dtv² = 0.5 × 5² P = 12.5 W So the answer is (d) 12.5 W.

Sand is being dropped from a stationary dropper at a rate of 0.5 kg… | JEE Main 2022 (27 Jul, Shift 1)

Q24JEE Main 2022 · 27 JUL · SHIFT 1Medium

Power

Sand is being dropped from a stationary dropper at a rate of $0.5\text{ kg s}^{-1}$ on a conveyor belt moving with a velocity of $5\text{ m s}^{-1}$ . The power needed to keep the belt moving with the same velocity will be:

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Answer: (d)

Sand lands with zero horizontal speed, so the belt must supply force $F = v\frac{dm}{dt}$ .
Power $P = Fv = \frac{dm}{dt}v^2 = 0.5 \times 5^2$
$P = 12.5\text{ W}$
So the answer is (d) 12.5 W.

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