JEE Main · 2015 · 04 Apr 2015 · Offline · Rotational Mechanics

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

(A) $\frac{MR ^{2}}{16 2 π}$
(B) $\frac{4 MR ^{2}}{9 3 π}$ ✓ Correct
(C) $\frac{4 MR ^{2}}{3 3 π}$
(D) $\frac{MR ^{2}}{32 2 π}$

Answer: (B)

Solution

The largest cube that fits inside the sphere has its body diagonal equal to the diameter,

3 a = 2 R

, so

a = \frac{2 R}{3}

. Its mass is the cube's volume times the sphere's density,

m = ρ a^{3}

with

ρ = \frac{M}{\frac{4}{3} π R ^{3}}

. For a cube, the moment of inertia about an axis through the centre perpendicular to a face is

I = \frac{m a ^{2}}{6}

. Assembling the pieces,

I = \frac{ρ a ^{5}}{6} = \frac{3 M}{4 π R ^{3}} \cdot \frac{1}{6} (\frac{2 R}{3})^{5} = \frac{4 M R ^{2}}{9 3 π}

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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JEE Main · 2015 · 04 Apr 2015 · Offline · Rotational Mechanics

Solution

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