From a solid sphere of mass M and radius R, a… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Mechanics

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From a solid sphere of mass $M$ and radius $R$ , a cube of the maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

Options

A
$\frac{4M{R}^{2}}{3\sqrt{3}\pi }$
B
$\frac{M{R}^{2}}{32\sqrt{2}\pi }$
C
$\frac{M{R}^{2}}{16\sqrt{2}\pi }$
D
$\frac{4M{R}^{2}}{9\sqrt{3}\pi }$

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Solution

Let $a$ be the length of edge for the cube, with maximum possible volume diagonal length $=2R$ $\Rightarrow \sqrt{3}a=2R\Rightarrow a=\frac{2R}{\sqrt{3}}$ . As densities of sphere and cube are equal. Let $M'$ be mass of the cube, $\frac{\text{M}}{\frac{4}{3}\pi {\text{R}}^{3}}=\frac{{\text{M}}^{'}}{{\text{a}}^{3}}$ $\Rightarrow {M}^{'}=\frac{3M{a}^{3}}{4\pi {R}^{3}}$ . Moment of inertia of cube about an axis passing through its center is, $I=\frac{{M}^{'}(2{a}^{2})}{12 }$ $=\frac{3M{a}^{3}}{4\pi {R}^{3}}\times \frac{2{a}^{2}}{12 }$ $=\frac{M{a}^{5}}{8\pi {R}^{3}}$ .

also, $a=\frac{2}{\sqrt{3}}R$ $\Rightarrow I=\frac{M\times 32 {R}^{5}}{8\pi \times 9\sqrt{3}{R}^{3 }}$ $=\frac{4M{R}^{2}}{9\sqrt{3}\pi }$ .

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Solution

also, $a=\frac{2}{\sqrt{3}}R$ $\Rightarrow I=\frac{M\times 32 {R}^{5}}{8\pi \times 9\sqrt{3}{R}^{3 }}$ $=\frac{4M{R}^{2}}{9\sqrt{3}\pi }$ .

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Options

Did you get this right?

Solution

Your note

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More JEE Main Mechanics questions

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