JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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.......
- A \(57\)
- B \(55\)
- C \(5\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(I _{1}=\frac{2}{5} M (2 R )^{2}=\frac{8}{5} MR ^{2}\) \(I _{1}=\frac{1}{2} M (2 R )^{2}=2 MR ^{2}\) \(I _{3}=\frac{ M (2 R )^{2}}{4}= MR ^{2}\) \(I _{4}=\frac{ M (2 R )^{2}}{2}=2 MR ^{2}\) \(2\left( I _{2}+ I _{3}\right)+ I _{4}= x I _{1}\)…
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