Moment of Inertia \((M.I.)\) of four bodies having same mass ' \(M\) ' and radius ' \(2 R^{\prime}\) are as follows \(I _{1}=\) \(M.I.\) of solid sphere about its diameter \(I _{2}=\) \(M.I.\) of solid cylinder about its axis \(I _{3}=\) \(M.I.\) of solid circular disc about its diameter \(I _{4}=\) \(M.I.\) of thin circular ring about its diameter If \(2\left(I_{2}+I_{3}\right)+I_{4}=x\). \(I_{1}\) then the value of \(x\) will be.......

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Correct Answer

(C) \(5\)

Detailed explanation

\(I _{1}=\frac{2}{5} M (2 R )^{2}=\frac{8}{5} MR ^{2}\)

\(I _{1}=\frac{1}{2} M (2 R )^{2}=2 MR ^{2}\)

\(I _{3}=\frac{ M (2 R )^{2}}{4}= MR ^{2}\)

\(I _{4}=\frac{ M (2 R )^{2}}{2}=2 MR ^{2}\)

\(2\left( I _{2}+ I _{3}\right)+ I _{4}= x I _{1}\)

\(8 MR { }^{2}= x \frac{8}{5} MR ^{2}\)

\(x =5\)

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