JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Hard
- Numerical
M and R be the mass and radius of a disc. A small disc of radius \(R / 3\) is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is \(\frac{4}{x} M R^2\). The value of \(x\) is ________.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 9
Step-by-step Solution
Detailed explanation
Without cavity \(\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}\)
\(\begin{aligned}
\text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\
& =\left(\frac{\mathrm{M}}{9}\right)
\end{aligned}\)
M.I. of removed disc \(I_2=\frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2}+\frac{M}{9} \times\left(\frac{2 R}{3}\right)^2\)
\(\begin{aligned}
& =\frac{\mathrm{MR}^2}{18} \\
& \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \\
& (\mathrm{n}=9)
\end{aligned}\)
\(\begin{aligned}
\text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\
& =\left(\frac{\mathrm{M}}{9}\right)
\end{aligned}\)
M.I. of removed disc \(I_2=\frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2}+\frac{M}{9} \times\left(\frac{2 R}{3}\right)^2\)
\(\begin{aligned}
& =\frac{\mathrm{MR}^2}{18} \\
& \mathrm{I}=\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{MR}^2}{2}-\frac{\mathrm{MR}^2}{18}=\frac{4 \mathrm{MR}^2}{9} \\
& (\mathrm{n}=9)
\end{aligned}\)
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