Sand falls vertically on a conveyor belt at a rate of 0.1 kg / s . In order to keep the belt moving at a uniform speed of 2 ~m / s , the force required to be applied on the belt is :

Correct answer: (B) 0.2 N. According to question Speed, v=2 ~m / s dm/dt=0.1 ~kg / s Force =d/dt_1(mv) =v (d ~m)/dt =2 × W .1=0.2 ~N

Q122 / 270 · NDA 2023 (Phase 1) · General Ability

Sand falls vertically on a conveyor belt at a rate of 0.1 $\mathrm{kg} / \mathrm{s}$. In order to keep the belt moving at a uniform speed of $2 \mathrm{~m} / \mathrm{s}$, the force required to be applied on the belt is :

A
0 N
Correct answer
B
0.2 N
C
1.0 N
D
2.0 N

Solution

According to question Speed, $v=2 \mathrm{~m} / \mathrm{s}$ $\frac{\mathrm{dm}}{\mathrm{dt}}=0.1 \mathrm{~kg} / \mathrm{s}$ Force $=\frac{\mathrm{d}}{\mathrm{dt}_1}(\mathrm{mv})$ $=v \frac{d \mathrm{~m}}{\mathrm{dt}}$ $=2 \times \mathbb{W} .1=0.2 \mathrm{~N}$

Hence, the correct answer is option: 'B'.

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