Question

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?

• A. \( P \propto v^3 \)
• B. \( P \propto \sqrt{v} \)
• C. \( P \propto v \)
• D. \( P \propto v^5 \)

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 \( v^5 \).

Answer 1

The Correct Option is D

Step 1: The power delivered to the conveyor belt is given by the force times the velocity. \[ P = F \cdot v \]

Step 2: From the given condition \( \frac{dm}{dt} \propto \sqrt{v} \), the rate of change of mass is proportional to the square root of the velocity. This means the rate of change of momentum is proportional to the square root of the velocity. Thus, we have: \[ F = \frac{dp}{dt} = \frac{dm}{dt} \cdot v \] Since \( \frac{dm}{dt} \propto \sqrt{v} \), we get: \[ F \propto v^{3/2} \]

Step 3: Therefore, the power delivered is: \[ P \propto F \cdot v \propto v^{3/2} \cdot v = v^{5/2} \] Hence, the correct relationship is \( P \propto v^5 \), and the correct answer is option (4).