JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2026
- Easy
- Numerical
Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by \(\frac{ x }{256} Mr ^2\). The value of x is ___________.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 109
Step-by-step Solution
Detailed explanation
\(M =\sigma \pi R ^2\)
\(\sigma \pi R ^2=16 m\)
\(m =\frac{\sigma \pi R ^2}{16}\)
\(I _{\text {system }}=\frac{ MR ^2}{2}-2\left(\frac{ mR ^2}{2 \times 16}+\frac{9 mR ^2}{16}\right)\)
\(=\frac{ MR ^2}{2}-2 \times \frac{19 mR ^2}{32}\)
\(=\frac{ MR ^2}{2}-\frac{19}{16} mR ^2\)
\(=\frac{ MR ^2}{2}-\frac{19}{256} MR ^2 \quad\)becoz \(m =\frac{ M }{16}\)
\(=\frac{(128-19)\left( MR ^2\right)}{256}\)
\(=\frac{109 MR ^2}{256}\)
\(\sigma \pi R ^2=16 m\)
\(m =\frac{\sigma \pi R ^2}{16}\)
\(I _{\text {system }}=\frac{ MR ^2}{2}-2\left(\frac{ mR ^2}{2 \times 16}+\frac{9 mR ^2}{16}\right)\)
\(=\frac{ MR ^2}{2}-2 \times \frac{19 mR ^2}{32}\)
\(=\frac{ MR ^2}{2}-\frac{19}{16} mR ^2\)
\(=\frac{ MR ^2}{2}-\frac{19}{256} MR ^2 \quad\)becoz \(m =\frac{ M }{16}\)
\(=\frac{(128-19)\left( MR ^2\right)}{256}\)
\(=\frac{109 MR ^2}{256}\)
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