Physics · Fluid Mechanics

JEE Main 2024 — 9 April, Shift 1 — Question 36

A sphere of relative density $\sigma$ and diameter $D$ has concentric cavity of diameter $d$ . The ratio of $\frac{\mathrm{D}}{\mathrm{d}}$ , if it just floats on water in a tank is:

Option A:
$\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}$
Correct
Option B:
$\left(\frac{\sigma+1}{\sigma-1}\right)^{\frac{1}{3}}$
Option C:
$\left(\frac{\sigma-1}{\sigma}\right)^{\frac{1}{3}}$
Option D:
$\left(\frac{\sigma-2}{\sigma+2}\right)^{\frac{1}{3}}$

Answer: A

Step-by-step solution

weight $(w)=\frac{4}{3} \pi\left(\frac{D^{3}-d^{3}}{8}\right) \sigma g$ Buoyant force $\left(F_{b}\right)=1 \times \frac{4}{3} \pi\left(\frac{D^{3}}{8}\right) \cdot g$ For Just Float $\Rightarrow \mathrm{w}=\mathrm{F}_{\mathrm{b}}$

$\Rightarrow\left(\mathrm{D}^{3}-\mathrm{d}^{3}\right) \sigma=\mathrm{D}^{3}$

$\Rightarrow 1-\frac{\mathrm{d}^{3}}{\mathrm{D}^{3}}=\frac{1}{\sigma}$

$\Rightarrow 1-\frac{1}{\sigma}=\left(\frac{d}{D}\right)^{3}$

$\Rightarrow\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}=\left(\frac{\mathrm{D}}{\mathrm{d}}\right)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Fluid Mechanics
Topic: Buoyancy and Archimedes' Principle