Physics · Mechanical Properties of Matter

JEE Main 2024 — 6 April, Shift 1 — Question 49

A small ball of mass mm and density ρ\rho is dropped in a viscous liquid of density ρ0\rho_{0}. After sometime, the ball falls with constant velocity. The viscous force on the ball is :

  1. Option A:

    mg⁡(ρ0ρ−1)\operatorname{mg}\left(\frac{\rho_{0}}{\rho}-1\right)

  2. Option B:

    mg⁡(1+ρρ0)\operatorname{mg}\left(1+\frac{\rho}{\rho_{0}}\right)

  3. Option C:

    mg⁡(1−ρρ0)\operatorname{mg}\left(1-\rho \rho_{0}\right)

  4. Option D:

    mg⁡(1−ρ0ρ)\operatorname{mg}\left(1-\frac{\rho_{0}}{\rho}\right)

    Correct

Answer: D

Step-by-step solution

mg−FB−Fv=ma\mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma}

a=0\mathrm{a}=0 for constant velocity

mg−FB=Fvm g-F_{B}=F_{v} Fv=mg−vρ0 g=mg−mρρ0 g=mg(1−ρ0ρ)\mathrm{F}_{\mathrm{v}}=\mathrm{mg}-\mathrm{v} \rho_{0} \mathrm{~g}=\mathrm{mg}-\frac{\mathrm{m}}{\rho} \rho_{0} \mathrm{~g}=\mathrm{mg}\left(1-\frac{\rho_{0}}{\rho}\right)

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Viscosity