JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Medium
- Numerical
\(\mathrm{A}, \mathrm{B}\) and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.

The moment of inertia of the given system about \(P Q\) is \(\frac{X}{15} I\), where \(I\) is the moment of inertia of the disc about its diameter. The value of x is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 199
Step-by-step Solution
Detailed explanation

All bodies have same mass and same radius.
\(\mathrm{A} \rightarrow\) Disc
\(\mathrm{B} \rightarrow\) Solid sphere
\(\mathrm{C} \rightarrow\) Spherical shell
and, \(\mathrm{I}=\frac{\mathrm{MR}^2}{4}\)
\(\mathrm{I}_{\mathrm{PQ}}=\frac{\mathrm{MR}^2}{4}+\left(\frac{2}{5} \mathrm{MR}^2+\mathrm{MR}^2\right)+\left(\frac{2}{3} \mathrm{MR}^2+\mathrm{MR}^2\right)\)
\(\mathrm{I}_{\mathrm{PQ}}=\frac{15 \mathrm{MR}^2+24 \mathrm{MR}^2+60 \mathrm{MR}^2+40 \mathrm{MR}^2+60 \mathrm{MR}^2}{60}\)
\(\mathrm{I}_{\mathrm{PQ}}=\frac{199}{60} \mathrm{MR}^2=\frac{199}{15}\left(\frac{\mathrm{MR}^2}{4}\right)\)
\(=\frac{199}{15} \mathrm{I}\)
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