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Physics · System Of Particles

JEE Main 2025 — 29 January, Evening Shift — Question 28

A sand dropper drops sand of mass m(t)m(t) on a conveyer belt at a rate proportional to the square root of speed (v)(\mathrm{v}) of the belt, i.e. dmdt∝v\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?

  1. Option A:

    P2∝v3P^{2} \propto v^{3}

  2. Option B:

    P∝V\mathrm{P} \propto \sqrt{\mathrm{V}}

  3. Option C:

    P∝vP \propto v

  4. Option D:

    P2∝v5P^{2} \propto v^{5}

    Correct

Answer: D

Step-by-step solution

Power =F⃗⋅V⃗=\vec{F} \cdot \vec{V}

F=dpdt[p=mv]\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}[\mathrm{p}=\mathrm{mv}]

F=(dmdt)v=C(v)v\mathrm{F}=\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right) \mathrm{v}=\mathrm{C}(\sqrt{\mathrm{v}}) \mathrm{v} F=Cv32\mathrm{F}=\mathrm{Cv}^{\frac{3}{2}}

Power =C(v3/2)v=Cv5/2=\mathrm{C}\left(\mathrm{v}^{3 / 2}\right) \mathrm{v}=\mathrm{Cv}^{5 / 2}

p2∝v5p^{2} \propto v^{5}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Evening Shift
Subject
Physics
Chapter
System Of Particles
Topic
Variable Mass Systems

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