From a solid sphere of mass M and radius R,a | vedclass

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(A) For a cube of side $a$ to be cut from a sphere of radius $R$,the diagonal of the cube must be equal to the diameter of the sphere: $\sqrt{3}a = 2R

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a cube of side $a$ to be cut from a sphere of radius $R$,the diagonal of the cube must be equal to the diameter of the sphere: $\sqrt{3}a = 2R$. Thus,$a = \frac{2R}{\sqrt{3}}$. The density of the sphere is $\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{3M}{4\pi R^3}$. The mass of the cube is $m = \rho \cdot a^3 = \left(\frac{3M}{4\pi R^3}\right) \left(\frac{2R}{\sqrt{3}}\right)^3 = \left(\frac{3M}{4\pi R^3}\right) \left(\frac{8R^3}{3\sqrt{3}}\right) = \frac{2M}{\sqrt{3}\pi}$. The moment of inertia of a cube of mass $m$ and side $a$ about an axis passing through its centre and perpendicular to one of its faces is $I = \frac{ma^2}{6}$. Substituting the values of $m$ and $a$: $I = \frac{1}{6} \left(\frac{2M}{\sqrt{3}\pi}\right) \left(\frac{2R}{\sqrt{3}}\right)^2 = \frac{1}{6} \left(\frac{2M}{\sqrt{3}\pi}\right) \left(\frac{4R^2}{3}\right) = \frac{8MR^2}{18\sqrt{3}\pi} = \frac{4MR^2}{9\sqrt{3}\pi}$.

From a solid sphere of mass $M$ and radius $R$,a cube of maximum possible volume is cut. The moment of inertia of the cube about an axis passing through its centre and perpendicular to one of its faces is

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