Physics · Rotational Dynamics

JEE Main 2026 — 23 January, Evening Shift — Question 45

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 $\frac{\mathrm{X}}{256} \mathrm{Mr}^{2}$ . The value of x is $\_\_\_\_$ .

Answer: 109

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{M}=\sigma \pi \mathrm{R}^{2}$ $\sigma \pi \mathrm{R}^{2}=16 \mathrm{~m}$ $\mathrm{m}=\frac{\sigma \pi \mathrm{R}^{2}}{16}$ $\mathrm{I}_{\text {system }}=\frac{\mathrm{MR}^{2}}{2}-2\left(\frac{\mathrm{mR}^{2}}{2 \times 16}+\frac{9 \mathrm{mR}^{2}}{16}\right)$ $=\frac{\mathrm{MR}^{2}}{2}-2 \times \frac{19 \mathrm{mR}^{2}}{32}$ $=\frac{\mathrm{MR}^{2}}{2}-\frac{19}{16} \mathrm{mR}^{2}$ $=\frac{\mathrm{MR}^{2}}{2}-\frac{19}{256} \mathrm{MR}^{2}$ becoz $\mathrm{m}=\frac{\mathrm{M}}{16}$ $=\frac{(128-19)\left(\mathrm{MR}^{2}\right)}{256}$ $=\frac{109 \mathrm{MR}^{2}}{256}$

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Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia