IIT-JEE 2012 · Paper 1 · Physics Q18

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PhysicsInteger+4Question 18 of 60

A lamina is made by removing a small disc of diameter $2R$ from a bigger disc of uniform mass density and radius $2R$ , as shown in the figure. The moment of inertia of this lamina about axes passing through $O$ and $P$ is $I_O$ and $I_P$ , respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\dfrac{I_P}{I_O}$ to the nearest integer is

A disc of radius 2R centred at O with a circular hole of diameter 2R whose rim passes through O and touches the top of the big disc; P is on the rim of the big disc, a distance 2R to the right of O.

Your answer

Answer and worked solution

Official answer: 3

Masses. Let the full disc (radius $2R$ ) have mass $4m$ ; the removed disc (radius $R$ ) has mass $m$ . Its centre is $R$ from O along the direction perpendicular to OP, so it is $\sqrt{R^2+(2R)^2}=\sqrt5R$ from P.

About O.
$I_O=\tfrac12(4m)(2R)^2-\left[\tfrac12mR^2+mR^2\right]=8mR^2-\tfrac32mR^2=\tfrac{13}2mR^2.$

About P (parallel axis, P is $2R$ from O):
$I_P=\left[8mR^2+4m(2R)^2\right]-\left[\tfrac12mR^2+5mR^2\right]=24mR^2-\tfrac{11}2mR^2=\tfrac{37}2mR^2.$

Ratio. $\dfrac{I_P}{I_O}=\dfrac{37}{13}\approx2.85$ , nearest integer 3.

Answer: 3

Topics:Parallel and perpendicular axis theorems Moment of inertia

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IIT-JEE 2012 · Paper 1, Physics question 18