A solid sphere of mass \(M\) and radius \(R\) is divided into two unequal parts. The first part has a mass of \(\frac {7M}{8}\) and is converted into a uniform disc of radius \(2R.\) The second part is converted into a uniform solid sphere. Let \(I_1\) be the moment of inertia of the disc about its axis and \(I_2\) be the moment of inertia of the new sphere about its axis. The ratio of \(I_1/I_2\) is given by

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Correct Answer

(D) \(140\)

Detailed explanation

\({I_1} = \frac{{\left( {\frac{{7M}}{8}} \right){{\left( {ZR} \right)}^2}}}{2} = \frac{{7M \times 4{R^2}}}{{2 \times 8}} = \frac{{7M{R^2}}}{4}\)

\({I_2} = \frac{2}{5}\frac{M}{8}{\left( {\frac{R}{2}} \right)^2} = \frac{{2M}}{{5 \times 8}}\frac{{{R^2}}}{4} = \frac{{M{R^2}}}{{80}}\)

\(\frac{{{I_1}}}{{{I_2}}} = \frac{{7M{R^2} \times 80}}{{4M{R^2}}} = 140\)

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