2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    A circular hole of radius \[\frac{R}{4}\] is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point \[\text{O}\] and perpendicular to the plane of the disc is [JEE Online 09-04-2017]

    A)  \[\frac{219{{R}^{2}}}{256}\,\]                   

    B)  \[\frac{237M{{R}^{2}}}{512}\]

    C)  \[\frac{197M{{R}^{2}}}{256}\,\]                               

    D)  \[\frac{19M{{R}^{2}}}{512}\]

    Correct Answer: B

    Solution :

                    \[{{I}_{D}}\,=\frac{m{{r}^{2}}}{2}\] \[{{I}_{removed}}\,=\frac{1}{2}\frac{m}{16}\,\frac{{{r}^{2}}}{16}+\frac{m}{16}\,\frac{9{{r}^{2}}}{16\,}(\operatorname{Im}+md)\] \[=\frac{m{{r}^{2}}+18m{{r}^{2}}}{512}\] \[=\frac{19m{{r}^{2}}}{512}\] \[{{I}_{remaining}}\,=\frac{m{{r}^{2}}}{2}-\frac{19}{512}\,m{{r}^{2}}\] \[=\frac{237}{512}\,m{{r}^{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner