Q63 / 90 · JEE Main 2018 (08 Apr) · Physics

From a uniform circular disc of radius R and mass 9 M, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

Solution

By parallel axis theorem,

Mass of smaller disc

= \frac{9M}{π R^{2}} π {(\frac{R}{3})}^{2} = M

MI of smaller disc

= \frac{M {(\frac{R}{3})}^{2}}{2} + M {(\frac{2 R}{3})}^{2}

M I = \frac{M R^{2}}{18} + \frac{4 M R^{2}}{9}

M I = \frac{M R^{2}}{2}

M I = M I (B i g g e r) - M I (S m a l l e r)

= \frac{9 M R^{2}}{2} - \frac{M R^{2}}{2}

= 4 M R^{2}

Hence, the correct answer is option: 'B'.