A sand dropper drops sand of mass \( m(t) \) on a conveyor belt at a rate proportional to the square root of the 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 relationships is true?
Show Hint
Power is the rate at which work is done, and it is given by the force multiplied by the velocity. For systems where mass is added at a rate proportional to the square root of velocity, the power will scale with \( v^5 \).
Step 1: Power supplied to the conveyor belt equals force multiplied by velocity.
\[
P = F \cdot v
\]
Step 2: Given \( \frac{dm}{dt} \propto \sqrt{v} \), the mass flow rate is proportional to the square root of velocity. Consequently, the rate of change of momentum is also proportional to the square root of velocity.
This leads to:
\[
F = \frac{dp}{dt} = \frac{dm}{dt} \cdot v
\]
Since \( \frac{dm}{dt} \propto \sqrt{v} \), it follows that:
\[
F \propto v^{3/2}
\]
Step 3: Consequently, the power supplied is:
\[
P \propto F \cdot v \propto v^{3/2} \cdot v = v^{5/2}
\]
Therefore, the correct relationship is \( P \propto v^5 \), and option (4) is the correct answer.
