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

Work Power and Energy PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q56 with Solution

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Correct answer: D

Given relation for sand falling on the belt

The sand is dropped vertically onto a conveyor belt moving with constant speed $v$ .

The rate at which mass falls on the belt is proportional to $\sqrt{v}$ :

\frac{dm}{dt} \propto \sqrt{v}

Let

\frac{dm}{dt} = k\sqrt{v}

for some proportionality constant $k$ .

Force needed to keep the belt moving

Each small mass element of sand lands on the belt with zero horizontal velocity and must be accelerated to horizontal speed $v$ .

So, horizontal momentum given per unit time is

F = \frac{dp}{dt} = \frac{dm}{dt}\, v

Substituting $\frac{dm}{dt} = k\sqrt{v}$ ,

F = k\sqrt{v}\cdot v = kv^{3/2}

Thus,

F \propto v^{3/2}

Power delivered to the belt

Power is

P = Fv

So,

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

Hence,

P \propto v^{5/2}

Squaring both sides,

P^2 \propto v^5

Check options

A: $P \propto \sqrt{v}$ ❌
B: $P \propto v$ ❌
C: $P^2 \propto v^3$ ❌
D: $P^2 \propto v^5$ ✅

Final answer

The correct relationship is

P^2 \propto v^5

So the correct option is D.

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View written solutionFree

Correct answer: D

Given relation for sand falling on the belt

The sand is dropped vertically onto a conveyor belt moving with constant speed $v$ .

The rate at which mass falls on the belt is proportional to $\sqrt{v}$ :

\frac{dm}{dt} \propto \sqrt{v}

Let

\frac{dm}{dt} = k\sqrt{v}

for some proportionality constant $k$ .

Force needed to keep the belt moving

Each small mass element of sand lands on the belt with zero horizontal velocity and must be accelerated to horizontal speed $v$ .

So, horizontal momentum given per unit time is

F = \frac{dp}{dt} = \frac{dm}{dt}\, v

Substituting $\frac{dm}{dt} = k\sqrt{v}$ ,

F = k\sqrt{v}\cdot v = kv^{3/2}

Thus,

F \propto v^{3/2}

Power delivered to the belt

Power is

P = Fv

So,

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

Hence,

P \propto v^{5/2}

Squaring both sides,

P^2 \propto v^5

Check options

A: $P \propto \sqrt{v}$ ❌
B: $P \propto v$ ❌
C: $P^2 \propto v^3$ ❌
D: $P^2 \propto v^5$ ✅

Final answer

The correct relationship is

P^2 \propto v^5

So the correct option is D.

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