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

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Question

Maximum possible volume of cube will occur when

$\sqrt{3} a=2 R(a=$ side of cube)

$	herefore \quad a=\frac{2}{\sqrt{3}} R$

Now, density of s

Vedclass · Accepted Answer

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

Vedclass · Answer

Maximum possible volume of cube will occur when

$\sqrt{3} a=2 R(a=$ side of cube)

$	herefore \quad a=\frac{2}{\sqrt{3}} R$

Now, density of sphere, $	au=\frac{M}{\frac{4}{3} \pi R^{3}}$

Mass of cube $\mathrm{m}=\left(	ext { volume of cube) }(	au)=\left(a^{3}ight)(	au)ight.$

$=\left[\frac{2}{\sqrt{3} R}ight]^{3}\left[\frac{m}{\frac{4}{3} \pi R^{3}}ight]=\left(\frac{2}{\sqrt{3} \pi}ight) M$

Now, moment of inertia of the cube about the said axis is

$I=\frac{m a^{2}}{6}=\frac{\left(\frac{2}{\sqrt{3} x}ight) M\left[\frac{2}{\sqrt{3 R}}ight]^{2}}{\sigma}=\frac{4 M R^{2}}{9 \sqrt{3} \pi}$

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 centre and perpendicular to one of its faces is

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