JEE Main 2015 · 4 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

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 :

\dfrac{MR^2}{32\sqrt2\pi}

\dfrac{MR^2}{16\sqrt2\pi}

\dfrac{4MR^2}{9\sqrt3\pi}

\dfrac{4MR^2}{3\sqrt3\pi}

Answer and worked solution

Official answer: C

The largest cube has body diagonal $2R$ , so side $a=\dfrac{2R}{\sqrt3}$ . Its mass is
$m=M\frac{a^3}{\frac43\pi R^3}=\frac{2M}{\sqrt3\pi}.$
$I=\frac{ma^2}6=\frac{2M}{\sqrt3\pi}\cdot\frac{4R^2}{3\cdot6}=\frac{4MR^2}{9\sqrt3\pi}.$

Topics:Moment of inertia

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 6