A sand dropper drops sand of mass $m(t)$ on a conveyer belt at a rate proportional to the square root of…

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. $\frac{\mathrm{dm}}{\mathrm{dt}} \propto \sqrt{v}$. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?
  1. $\mathrm{P} \propto \sqrt{v}$
  2. $\mathrm{P} \propto v$
  3. $\mathrm{P}^2 \propto v^5$
  4. $\mathrm{P}^2 \propto v^3$

Solution

Power $=\vec{F} \cdot \vec{V}$
and $F=\frac{d p}{d t}=$ Rate of change of linear momentum
$F=V \cdot \frac{d m}{d t}=K_1 V^{\frac{3}{2}}, K$ is constant
Power $(P)=\left(K V^{\frac{3}{2}}\right) \cdot(V)$
$=K V^{\frac{5}{2}}$
So, $P^2 \propto V^5$

Asked in: JEE Main 2025 (29 Jan Shift 2)

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