QJEEM15Ph1e72: JEE Main 2015 Physics - Class 11 JEEM15Ph1e72 Solution: From a solid sphere of mass and radius a cube of maximum…

1. Determine the side length of the cube:

The cube with the maximum possible volume that can be cut from the sphere will have its diagonal equal to the diameter of the sphere. Let the side length of the cube be 'a'. Using Pythagoras in 3D, we have:

$a^2 + a^2 + a^2 = (2R)^2$

$3a^2 = 4R^2$

$a = \sqrt{\frac{4R^2}{3}} = \frac{2R}{\sqrt{3}}$

2. Calculate the mass of the cube:

The volume of the cube is $V = a^3 = \left(\frac{2R}{\sqrt{3}}\right)^3 = \frac{8R^3}{3\sqrt{3}}$

The density of the sphere (and hence the cube) is $\rho = \frac{M}{\frac{4}{3}\pi R^3}$

The mass of the cube is $m = \rho V = \frac{M}{\frac{4}{3}\pi R^3} \cdot \frac{8R^3}{3\sqrt{3}} = \frac{2M}{\sqrt{3}\pi}$

3. Find the moment of inertia of the cube:

The moment of inertia of a cube about an axis passing through its center and perpendicular to one of its faces is given by:

$I = \frac{1}{6}ma^2$

Substituting the values we found:

$I = \frac{1}{6} \cdot \frac{2M}{\sqrt{3}\pi} \cdot \left(\frac{2R}{\sqrt{3}}\right)^2$

$I = \frac{4MR^2}{9\sqrt{3}\pi}$

Therefore, the correct answer is Option A: ${{4M{R^2}} \over {9\sqrt {3\pi } }}$