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. $$\frac{dm}{dt} \propto \sqrt{v}$$. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

Since $$\frac{dm}{dt} \propto \sqrt{v}$$, we write $$\frac{dm}{dt} = k\sqrt{v}$$ for some constant $$k$$.

The belt runs at constant speed $$v$$. The force needed to maintain constant speed when sand is being dropped is:

$$ F = v\frac{dm}{dt} $$

This is because the sand needs to be accelerated from rest to speed $$v$$.

The power delivered is:

$$ P = Fv = v^2 \frac{dm}{dt} = v^2 \cdot k\sqrt{v} = kv^{5/2} $$

Therefore: $$P \propto v^{5/2}$$

Squaring both sides: $$P^2 \propto v^5$$

The correct answer is Option 3: $$P^2 \propto v^5$$.