JEE Main 2017 · 9 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

A circular hole of radius $\dfrac R4$ 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 O and perpendicular to the plane of the disc is :

A disc of radius R and centre O with a circular hole of radius R/4 whose centre O′ is 3R/4 from O, touching the rim.

\dfrac{219MR^2}{256}

\dfrac{237MR^2}{512}

\dfrac{19MR^2}{512}

\dfrac{197MR^2}{256}

Answer and worked solution

Official answer: B

The hole has mass $\frac M{16}$ and its centre is $\frac{3R}4$ from O. About O,
$I_{hole}=\frac M{16}\left(\frac{R^2}{32}+\frac{9R^2}{16}\right)=\frac{19MR^2}{512}.$
The remaining part has $I=\dfrac{MR^2}2-\dfrac{19MR^2}{512}=\dfrac{237MR^2}{512}$ .

Topics:Moment of inertia Parallel and perpendicular axis theorems

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 6