From a solid sphere of mass M and radius R , a cube of the maximum possible volume is cut. Moment of inertia…

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:
  1. 4MR233π
  2. MR2322π
  3. MR2162π
  4. 4MR293π

Solution

Let a be the length of edge for the cube, with maximum possible volume diagonal length =2R⇒3a=2R⇒a=2R3. As densities of sphere and cube are equal. Let M' be mass of the cube, M43πR3=M'a3⇒M′=3Ma34πR3. Moment of inertia of cube about an axis passing through its center is,  I=M′2a212 =3Ma34πR3×2a212 =Ma58πR3.

also, a=23R⇒I =M×32 R58π×93R3 =4MR293π.

Asked in: JEE Main 2015 (04 Apr)

Practice more Rotational Motion questions on Aicharya