Rotational Motion

JEE Main level, single correct, medium

90 of 280

Question 90 of 280

From a solid sphere of mass MM and radius RR, 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:

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Answer: (D) 4MR293π\dfrac{4M{R}^{2}}{9\sqrt{3}\pi }

Let aa be the length of edge for the cube, with maximum possible volume diagonal length =2R=2R⇒3a=2R⇒a=2R3\Rightarrow \sqrt{3}a=2R\Rightarrow a=\dfrac{2R}{\sqrt{3}}. As densities of sphere and cube are equal. Let M′M' be mass of the cube, M43πR3=M′a3\dfrac{\text{M}}{\dfrac{4}{3}\pi {\text{R}}^{3}}=\dfrac{{\text{M}}^{'}}{{\text{a}}^{3}}⇒M′=3Ma34πR3\Rightarrow {M}^{'}=\dfrac{3M{a}^{3}}{4\pi {R}^{3}}. Moment of inertia of cube about an axis passing through its center is, I=M′(2a2)12I=\dfrac{{M}^{'}\left(2{a}^{2}\right)}{12}=3Ma34πR3×2a212=\dfrac{3M{a}^{3}}{4\pi {R}^{3}}\times \dfrac{2{a}^{2}}{12}=Ma58πR3=\dfrac{M{a}^{5}}{8\pi {R}^{3}}.

also, a=23Ra=\dfrac{2}{\sqrt{3}}R⇒I=M×32R58π×93R3\Rightarrow I=\dfrac{M\times 32{R}^{5}}{8\pi \times 9\sqrt{3}{R}^{3}}=4MR293π=\dfrac{4M{R}^{2}}{9\sqrt{3}\pi }.

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