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  • Solve! 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 center and perpendicular to one of its faces is :

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 center and  perpendicular to one of its faces is :

Option 1)

\frac{M\! R^{2}}{32\sqrt{2}\pi }\;

 Option 2)

\frac{M\! R^{2}}{16\sqrt{2}\pi }\;
Option 3)

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

 Option 4)

\frac{4M\! R^{2}}{3\sqrt{3}\pi }\;

Answers (1)

best_answer

As we learnt in

Moment of inertia for system of particle -

I= m_{1}r_{1}^{2}+m_{2}r_{2}^{2}+.........m_{n}r_{n}^{2}

\dpi{100} = \sum_{i=1}^{n}\: m_{i}r_{i}^{2}

 

- wherein

Applied when masses are placed discretely.

 

 

a=\frac{2}{\sqrt{}3}R

\frac{M}{M'}=\frac{\frac{4}{3}\pi R^{3}}{a^{3}}=\frac{\frac{4}{3}\pi R^{3}}{\left (\frac{2}{\sqrt{}3}R \right )^{3}}

\frac{M}{M'}=\frac{\frac{4}{3}\pi R^{3}}{\frac{8}{3\sqrt{3}}R^{3}}=\frac{4\pi}{3}\times\frac{3\sqrt{3}}{8}

\frac{M}{M'}=\frac{\sqrt{3}\pi}{2}\Rightarrow M'=\frac{2M}{\sqrt{3}\pi}

\therefore  M.O.I. of the cube about the given axis.

I=\frac{M'a^{2}}{6}=\frac{\frac{2M}{\sqrt{3}\pi}\times\left ( \frac{2}{\sqrt{3}}R \right )^{2}}{6}=\frac{4MR^{2}}{9\sqrt{3}\pi}


Option 1)

\frac{M\! R^{2}}{32\sqrt{2}\pi }\;

This is an incorrect option.

Option 2)

\frac{M\! R^{2}}{16\sqrt{2}\pi }\;

This is an incorrect option.

Option 3)

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

This is the correct option.

Option 4)

\frac{4M\! R^{2}}{3\sqrt{3}\pi }\;

This is the correct option.

Posted by

perimeter

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