JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(285\)
- B \(185\)
- C \(65\)
- D \(140\)
Answer & Solution
Correct Answer
(D) \(140\)
Step-by-step Solution
Detailed explanation
\({I_1} = \frac{{\left( {\frac{{7M}}{8}} \right){{\left( {ZR} \right)}^2}}}{2} = \frac{{7M \times 4{R^2}}}{{2 \times 8}} = \frac{{7M{R^2}}}{4}\)…
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