**Question:**

**A circular hole of diameter $\mathrm{R}$ is cut from a disc of mass $M$ and radius $R$; the circumference of the cut passes through the centre of the disc. The moment of inertia of the remaining portion of the disc about an axis perpendicular to the disc and passing through its centre is**

$\left(\frac{15}{32}\right) M R^{2}$

$\left(\frac{1}{8}\right) M R^{2}$

$\left(\frac{3}{8}\right) M R^{2}$

$\left(\frac{13}{32}\right) M R^{2}$

Question of from chapter.

JEE Main Previous Year May 7, 2012

Correct Option: 4

**Solution:**

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