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

We are given that sand is dropped onto a conveyor belt at a rate proportional to the square root of the belt's speed:

$$\frac{dm}{dt} \propto \sqrt{v}$$, or, $$\frac{dm}{dt} = \lambda \sqrt{v}$$

where $\lambda$ is a proportionality constant.

Our goal is to determine the relationship between the power $P$ required to keep the belt moving at constant speed and $v$.

Step 1: Force Required to Maintain Constant Speed

When sand falls onto the conveyor belt, it initially has zero horizontal velocity. To move with the belt, it must be accelerated to speed $v$, which requires a force.

The rate of change of momentum gives the force required:

$$F = \frac{d}{dt} (m v)$$

Since $v$ is constant:

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

Substituting $\frac{dm}{dt} = \lambda \sqrt{v}$:

$$F = v (\lambda \sqrt{v}) = \lambda v^{3/2}$$

Step 2: Power Required

Power is given by:

$$P = F v$$

Substituting $F = \lambda v^{3/2}$:

$$P = (\lambda v^{3/2}) v$$

 $$P = \lambda v^{5/2}$$ 

Final Answer:

$P^2\propto v^5$

Thus, the correct relationship is:

$P^2\propto v^5$