From a disc of mass ' m 'and radius ' r ' a circular portion of radius $\frac{r}{3}$ is removed from the edges. Find the M.I of the remaining system about a normal through centre of original disc

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{9}{4} {mr}^{2}$

$\frac{4 {mr}^{2}}{9}$

$\frac{3}{2} {mr}^{2}$

$\frac{2}{3} {mr}^{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let I₂ be the M.I of remaining portion, M.I of the removed portion about a normal axis through centre of the original disc is
$\begin{array}{l} I_{1} = \frac{\frac{m}{9} {(\frac{r}{3})}^{2}}{2} + \frac{m}{9} {(\frac{2 r}{3})}^{2} \\ I_{1} = \frac{{mr}^{2}}{18} \\ I_{2} = I - I_{1} = \frac{{mr}^{2}}{2} - \frac{{mr}^{2}}{18} \\ I_{2} = \frac{4 {mr}^{2}}{9} \end{array}$