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), \[\frac{1}{2}M{{R}^{2}}=\frac{3M{{R}^{2}}}{32}+{{I}_{remaining\,disc}}\]\[\Rightarrow \]\[{{I}_{remaining\,disc}}\] \[=\frac{M{{R}^{2}}}{2}-\frac{3M{{R}^{2}}}{32}=\left( \frac{13}{32} \right)M{{R}^{2}}\]You need to login to perform this action.
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