1

### JEE Main 2017 (Online) 8th April Morning Slot

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :
A
${{2\,\,mg} \over {2\,m + M}}$
B
${{2\,\,Mg} \over {2\,m + M}}$
C
${{2\,\,mg} \over {2\,M + m}}$
D
${{2\,\,Mg} \over {2\,M + M}}$

## Explanation

From figure, we can say

ma = mg $-$ T . . . . . (1)

Moment of inertia of a uniform disc,

I = ${1 \over 2}$ MR2

and the acceleration of the disc, a = $\propto$ R

We know,

Torque = I$\propto$ = RT

$\therefore\,\,\,$ RT = ${1 \over 2}$ M R2 $\times$ ${a \over R}$ = ${{MaR} \over 2}$

$\Rightarrow$ $\,\,\,$ T = ${{Ma} \over 2}$

Putting the value of T in eq (1).

ma = mg $-$ ${{Ma} \over 2}$

$\Rightarrow$ $\,\,\,$ a ( m + ${{Ma} \over 2}$) = mg

$\Rightarrow$ $\,\,\,$ a = ${{2mg} \over {2m + M}}$
2

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

The machine as shown has 2 rods of length1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a :

A
Constant speed
B
decreasing speed
C
increasing speed
D
speed which is ${3 \over 4}$th of that of the roller when the weight is 0.4 m above the ground
3

### 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

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

= ${{237M{R^2}} \over {512}}$
4

### 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,

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

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

= ${{181} \over 2}M{R^2}$