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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 29 January · Evening Shift · Physics · Work, Power and Energy

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate...

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)(v) of the belt, i.e. dmdt∝v\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?

Options

  1. AP∝v\mathrm{P} \propto \sqrt{v}
  2. BP∝v\mathrm{P} \propto v
  3. CP2∝v5\mathrm{P}^2 \propto v^5
  4. DP2∝v3\mathrm{P}^2 \propto v^3

Correct answer

C. P2∝v5\mathrm{P}^2 \propto v^5

Step-by-step solution

Power =F⃗⋅V⃗=\vec{F} \cdot \vec{V} and F=dpdt=F=\frac{d p}{d t}= Rate of change of linear momentum F=V⋅dmdt=K1V32,KF=V \cdot \frac{d m}{d t}=K_1 V^{\frac{3}{2}}, K is constant Power (P)=(KV32)⋅(V)(P)=\left(K V^{\frac{3}{2}}\right) \cdot(V) =KV52=K V^{\frac{5}{2}} So, P2∝V5P^2 \propto V^5
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