Force Analysis

The total driving force $F$ supplied by the motor consists of two components:

1. Momentum Change: The force required to accelerate the sand from rest to velocity $v$. $$ F_1 = \frac{dp}{dt} = v \frac{dm}{dt} = \mu v $$
2. Gravity Component: The force required to balance the component of weight of the sand currently on the belt down the incline.

   Let $L$ be the length of the belt. The mass of sand on the belt at any instant is $M = \mu \times (\text{time to travel } L) = \mu \frac{L}{v}$.

   The component of gravity is $F_2 = M g \sin\theta$.

   Since $L \sin\theta = h$, we have:

   $$ F_2 = \left( \frac{\mu L}{v} \right) g \sin\theta = \frac{\mu g (L \sin\theta)}{v} = \frac{\mu g h}{v} $$

Total Force:

$$ F(v) = F_1 + F_2 = \mu v + \frac{\mu g h}{v} $$

Optimization

To find the speed $v$ for the least possible driving force, we differentiate $F$ with respect to $v$ and set it to zero:

$$ \frac{dF}{dv} = \mu – \frac{\mu g h}{v^2} = 0 $$ $$ \mu = \frac{\mu g h}{v^2} \implies v^2 = gh $$ $$ v = \sqrt{gh} $$

Correct Option: (b) $\sqrt{gh}$