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### JEE Main 2017 (Online) 9th April Morning Slot

A circular hole of radius ${R \over 4}$ 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 O and perpendicular to the plane of the disc is :

A
${{219\,M{R^2}} \over {256}}$
B
${{237\,M{R^2}} \over {512}}$
C
${{19\,M{R^2}} \over {512}}$
D
${{197\,M{R^2}} \over {256}}$

## Explanation

Mass of removed disc = ${M \over {16}}$ Radius of removed disc = ${R \over 4}$

Moment of inertia of removed disc about it's own axis (O')

= ${1 \over 2}$ $\times$ ${M \over {16}}$ $\times$ ${\left( {{R \over 4}} \right)^2}$ = ${{M{R^2}} \over {512}}$

Moment of inertia of removed disc about O,

= Icm + mx2

= ${{M{R^2}} \over {512}}$ + ${M \over {16}}{\left( {{{3R} \over 4}} \right)^2}$

= ${{19M{R^2}} \over {512}}$

Moment of inertia of complete disc about point 'O',

I = ${{M{R^2}} \over 2}$

$\therefore\,\,\,$ Moment of Inertia of remaining disc

= ${{M{R^2}} \over 2}$ $-$ ${{19M{R^2}} \over {512}}$

= ${{237M{R^2}} \over {512}}$
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### JEE Main 2018 (Offline)

Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :
A
${{181} \over 2}M{R^2}$
B
${{55} \over 2}M{R^2}$
C
${{19} \over 2}M{R^2}$
D
${{73} \over 2}M{R^2}$

## Explanation

Moment of inertia of any disc form its center perpendicular to the plane of disc = ${1 \over 2}M{R^2}$

Moment of inertia of any one of the outer disc about an axis passing through point O and perpendicular to the plane

${I_1} = {1 \over 2}M{R^2} + M{\left( {2R} \right)^2}$ = ${9 \over 2}M{R^2}$

Moment of inertia of the entire system about point O

${I_o} = {1 \over 2}M{R^2} + 6{I_1}$

= ${1 \over 2}M{R^2} + 6 \times {9 \over 2}M{R^2}$

= ${{55} \over 2}M{R^2}$

So the moment of inertia of entire system about point P,

${I_P} = {I_O} + 7M{\left( {3R} \right)^2}$

= ${{55} \over 2}M{R^2} + 63M{R^2}$

= ${{181} \over 2}M{R^2}$
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### JEE Main 2018 (Offline)

From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :
A
${{37} \over 9}M{R^2}$
B
$4M{R^2}$
C
${{40} \over 9}M{R^2}$
D
$10M{R^2}$

## Explanation

Given that for a uniform circular disc the radius is R and mass 9M

$\therefore$ The area of uniform circular disc = $\pi {R^2}$

The radius of removed portion = ${R \over 3}$

$\therefore$ The area of removed portion = ${{\pi {R^2}} \over 9}$

So the mass of the removed portion = ${{9M} \over 9}$ = M

If the moment of inertia of removed disc about the center of the circular disc of radius R is I1 then

I1 = ${1 \over 2}M{\left( {{R \over 3}} \right)^2} + M{\left( {{{2R} \over 3}} \right)^2}$ = ${{M{R^2}} \over 2}$

Moment of inertia of the whole disc, I2 = ${{9M{R^2}} \over 2}$

Moment of inertia of the remaining disc, I = I2 - I1

$\therefore$ I = ${{9M{R^2}} \over 2} - {{M{R^2}} \over 2}$ = ${4M{R^2}}$
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