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A circular hole of diameter 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   JEE Main Online Paper (Held On 07 May 2012)

A) \[\left( \frac{15}{32} \right)M{{R}^{2}}\]                              

B)                        \[\left( \frac{1}{8} \right)M{{R}^{2}}\]

C)                        \[\left( \frac{3}{8} \right)M{{R}^{2}}\]                   

D)                        \[\left( \frac{13}{32} \right)M{{R}^{2}}\]

Correct Answer: D

Solution :

                M.I. of complete disc about its centreO. \[{{I}_{Total}}=\frac{1}{2}M{{R}^{2}}\]                                                   ?(i) Mass of circular hole (removed) \[=\frac{M}{4}\left( As\,M=\pi {{R}^{2}}t\therefore M\propto {{R}^{2}} \right)\] M.I. of removed hole about its own axis \[=\frac{1}{2}\left( \frac{M}{4} \right){{\left( \frac{R}{2} \right)}^{2}}=\frac{1}{32}M{{R}^{2}}\] M.I. of removed hole about O' \[{{I}_{removed\,hole}}={{I}_{cm}}+m{{x}^{2}}\] \[=\frac{M{{R}^{2}}}{32}+\frac{M}{4}{{\left( \frac{R}{2} \right)}^{2}}\] \[=\frac{M{{R}^{2}}}{32}+\frac{M{{R}^{2}}}{16}=\frac{3M{{R}^{2}}}{32}\] M.I. of complete disc can also be written as \[{{I}_{Total}}={{I}_{removed\,hole}}+{{I}_{remaining\,disc}}\] \[{{I}_{Total}}=\frac{3M{{R}^{2}}}{32}+{{I}_{remaining\,disc}}\]                 ?(ii) From eq.(i) and (ii), \[\frac{1}{2}M{{R}^{2}}=\frac{3M{{R}^{2}}}{32}+{{I}_{remaining\,disc}}\]\[\Rightarrow \]\[{{I}_{remaining\,disc}}\] \[=\frac{M{{R}^{2}}}{2}-\frac{3M{{R}^{2}}}{32}=\left( \frac{13}{32} \right)M{{R}^{2}}\]