JEE Main 2012 — Physics Mechanics

M.I. of complete disc about its centre $O$O. $$I_{\mathrm{Total}}=\frac{1}{2} M R^2$$ITotal​=21​MR2

Mass of circular hole (removed) $$=\frac{M}{4} \quad\left(\text { As } M=\pi R^2 t \therefore M \propto R^2\right)$$=4M​( As M=πR2t∴M∝R2) M.I. of removed hole about its own axis $$=\frac{1}{2}\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2=\frac{1}{32} M R^2$$=21​(4M​)(2R​)2=321​MR2 M.I. of removed hole about $O^{\prime}$O′ $$\begin{aligned} I_{\text {removed hole }} & =I_{\mathrm{cm}}+m x^2 \\ & =\frac{M R^2}{32}+\frac{M}{4}\left(\frac{R}{2}\right)^2 \\ & =\frac{M R^2}{32}+\frac{M R^2}{16}=\frac{3 M R^2}{32} \end{aligned}$$Iremoved hole ​​=Icm​+mx2=32MR2​+4M​(2R​)2=32MR2​+16MR2​=323MR2​​ M.I. of complete disc can also be written as $I_{\text {Total }}=I_{\text {removed hole }}+I_{\text {remaining disc }}$ITotal ​=Iremoved hole ​+Iremaining disc ​ $I_{\text {Total }}=\frac{3 M R^2}{32}+I_{\text {remaining disc }}$ITotal ​=323MR2​+Iremaining disc ​ From eq. (i) and (ii), $$\begin{aligned} & \frac{1}{2} M R^2=\frac{3 M R^2}{32}+I_{\text {remaining disc }} \\ & \Rightarrow I_{\text {remaining disc }} \\ & \quad=\frac{M R^2}{2}-\frac{3 M R^2}{32}=\left(\frac{13}{32}\right) M R^2 \end{aligned}$$​21​MR2=323MR2​+Iremaining disc ​⇒Iremaining disc ​=2MR2​−323MR2​=(3213​)MR2​