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Physics · Fluid Mechanics

JEE Main 2024 — 8 April, Shift 2 — Question 34

A cube of ice floats partly in water and partly in kerosene oil. The radio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil =0.8=0.8, specific gravity of ice =0.9=0.9 )

Question figure
  1. Option A:

    8:98: 9

  2. Option B:

    5:45: 4

  3. Option C:

    9:109: 10

  4. Option D:

    1:11: 1

    Correct

Answer: D

Step-by-step solution

v1=\mathrm{v}_{1}= volume immersed in water.

v2=\mathrm{v}_{2}= volume immersed in oil. v1ρwg+v2ρog=(v1+v2)ρcgv_{1} \rho_{\mathrm{w}} \mathrm{g}+\mathrm{v}_{2} \rho_{\mathrm{o}} \mathrm{g}=\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right) \rho_{\mathrm{c}} \mathrm{g}

v1+v2ρoρw=(v1+v2)ρcρw\mathrm{v}_{1}+\frac{\mathrm{v}_{2} \rho_{\mathrm{o}}}{\rho_{\mathrm{w}}}=\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right) \frac{\rho_{\mathrm{c}}}{\rho_{\mathrm{w}}}

=v1+0.8v2=0.9v1+0.9v2=\mathrm{v}_{1}+0.8 \mathrm{v}_{2}=0.9 \mathrm{v}_{1}+0.9 \mathrm{v}_{2}

=0.1v1=0.1v2=0.1 \mathrm{v}_{1}=0.1 \mathrm{v}_{2}

v1:v2=1:1\mathrm{v}_{1}: \mathrm{v}_{2}=1: 1

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Physics
Chapter
Fluid Mechanics
Topic
Buoyancy and Archimedes' Principle