Rotational Motion

JEE Main level, single correct, medium

86 of 280

Question 86 of 280

A circular hole of radius R4\dfrac{R}{4} is made in a thin uniform disc having mass and radius RR, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point OO and perpendicular to the plane of the disc is-

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Answer: (B) 237MR2512\dfrac{237M{R}^{2}}{512}

ID=mr22{I}_{D}=\dfrac{m{r}^{2}}{2}

Moment of inertia of removed portion about the axis.

Iremoved=12(m16)(r216)+(m16)(9r216){I}_{removed}=\dfrac{1}{2}\left(\dfrac{m}{16}\right)\left(\dfrac{{r}^{2}}{16}\right)+\left(\dfrac{m}{16}\right)\left(\dfrac{9{r}^{2}}{16}\right)(by parallel axis theorem)

=mr2+18mr2512=\dfrac{m{r}^{2}+18m{r}^{2}}{512}

=19mr2512=\dfrac{19m{r}^{2}}{512}

Iremaining=mr22−19512mr2{I}_{remaining}=\dfrac{m{r}^{2}}{2}-\dfrac{19}{512}m{r}^{2}

=237512mr2=\dfrac{237}{512}m{r}^{2}

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