question_answer

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

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 :

                \[{{\text{I}}_{\text{Total disc }}}=\frac{M{{R}^{2}}}{2}\] \[{{M}_{Removed}}=\frac{M}{4}(Mass\propto area)\] \[{{I}_{\operatorname{Re}moved}}(about\,same\,Perpendicular\,axis)\] \[=\frac{M}{4}\frac{{{(R/2)}^{2}}}{2}+\frac{M}{4}{{\left( \frac{R}{2} \right)}^{2}}=\frac{3M{{R}^{2}}}{32}\] \[{{I}_{\operatorname{Re}maing\,disc}}={{I}_{Total}}-{{I}_{\operatorname{Re}moved}}\]\[=\frac{M{{R}^{2}}}{2}-\frac{3}{32}M{{R}^{2}}=\frac{13}{32}M{{R}^{2}}\]