Q58 / 75 · JEE Main 2025 (29 Jan Shift 2) · Physics

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?

  • A
    $\mathrm{P} \propto \sqrt{v}$
  • B
    $\mathrm{P} \propto v$
  • Correct answer
    C
    $\mathrm{P}^2 \propto v^5$
  • D
    $\mathrm{P}^2 \propto v^3$

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, $P^2 \propto V^5$

Hence, the correct answer is option: 'C'.

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