Q73 / 75 · JEE Main 2026 (23 January Shift 2) · Physics
Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ 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 $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$.
Correct answer: 109
Solution
The moment of inertia of the complete disc about axis A is
$I_{disc} = \frac{1}{2}Mr^2 = \frac{128Mr^2}{256}$.
Each removed circular region has radius $r/4$ and mass $m_i = M/(r^2/16r^2) = M/16$.
For each removed disc at distance $3r/4$ from axis A, using the parallel axis theorem:
$I_{removed} = \frac{1}{2}m_i(r/4)^2 + m_i(3r/4)^2 = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{19Mr^2}{512}$.
For two removed discs:
$I_{2removed} = \frac{38Mr^2}{512} = \frac{19Mr^2}{256}$.
Therefore:
$I_{remainder} = \frac{128Mr^2}{256} - \frac{19Mr^2}{256} = \frac{109Mr^2}{256}$, so $x = 109$.
$I_{disc} = \frac{1}{2}Mr^2 = \frac{128Mr^2}{256}$.
Each removed circular region has radius $r/4$ and mass $m_i = M/(r^2/16r^2) = M/16$.
For each removed disc at distance $3r/4$ from axis A, using the parallel axis theorem:
$I_{removed} = \frac{1}{2}m_i(r/4)^2 + m_i(3r/4)^2 = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{19Mr^2}{512}$.
For two removed discs:
$I_{2removed} = \frac{38Mr^2}{512} = \frac{19Mr^2}{256}$.
Therefore:
$I_{remainder} = \frac{128Mr^2}{256} - \frac{19Mr^2}{256} = \frac{109Mr^2}{256}$, so $x = 109$.
Hence, the correct answer is 109.