A sand dropper drops sand of mass $\mathrm{m}(\mathrm{t})$ on a conveyer belt at a rate proportional to the square root of speed $(\mathrm{v})$ of the belt, i.e. $\frac{\mathrm{dm}}{\mathrm{dt}} \propto \sqrt{\mathrm{v}}$. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Work, Power & Energy

A sand dropper drops sand of mass

m (t)

on a conveyer belt at a rate proportional to the square root of speed

(v)

of the belt, i.e.

\frac{dm}{dt} \propto v

. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

(A) $P^{2} \propto v^{3}$
(B) $P \propto V$
(C) $P \propto v$
(D) $P^{2} \propto v^{5}$ ✓ Correct

Answer: (D)

Solution

Let

\frac{d m}{d t} = C v

for some constant

C

. To keep the belt moving at the constant speed

v

, the motor must supply a force equal to the rate at which momentum is delivered to the newly arriving sand (which is accelerated from rest up to speed

v

F = \frac{d p}{d t} = v \frac{d m}{d t} = C v^{3/2}

Power delivered:

P = F v = C v^{3/2} \cdot v = C v^{5/2}

Squaring,

P^{2} = C^{2} v^{5} ⟹ P^{2} \propto v^{5}

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Work, Power & Energy →

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Work, Power & Energy

Solution

More JEE Main questions on Work, Power & Energy