A) \[15\,M{{R}^{2}}/32\]
B) \[13\,M{{R}^{2}}/32\]
C) \[11\,M{{R}^{2}}/32\]
D) \[9\,M{{R}^{2}}/32\]
Correct Answer: B
Solution :
\[{{\text{I}}_{\text{Total disc }}}=\frac{M{{R}^{2}}}{2}\] \[{{M}_{Removed}}=\frac{M}{4}(Mass\propto area)\] \[{{I}_{\operatorname{Re}moved}}(about\,same\,Perpendicular\,axis)\] \[=\frac{M}{4}\frac{{{(R/2)}^{2}}}{2}+\frac{M}{4}{{\left( \frac{R}{2} \right)}^{2}}=\frac{3M{{R}^{2}}}{32}\] \[{{I}_{\operatorname{Re}maing\,disc}}={{I}_{Total}}-{{I}_{\operatorname{Re}moved}}\]\[=\frac{M{{R}^{2}}}{2}-\frac{3}{32}M{{R}^{2}}=\frac{13}{32}M{{R}^{2}}\]You need to login to perform this action.
You will be redirected in
3 sec