1

### JEE Main 2019 (Online) 11th January Evening Slot

The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) :
A
${\pi \over 8}$
B
${\pi \over 6}$
C
${\pi \over 4}$
D
${\pi \over 3}$

## Explanation

2.5 = 1 $\times$ 5 sin $\theta$

sin$\theta$ = 0.5 = ${1 \over 2}$

$\theta$ = ${\pi \over 6}$
2

### JEE Main 2019 (Online) 11th January Evening Slot

A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO' ,passing through the centre of D1 as shown in the figure, will be : A
3MR2
B
MR2
C
${2 \over 3}$ MR2
D
${4 \over 5}$ MR2

## Explanation

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

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

$=$ $3M{R^2}$
3

### JEE Main 2019 (Online) 11th January Evening Slot

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) : A
B
C
D

## Explanation 40 + f = m(R$\alpha$)         . . . .(i)

40 $\times$ R $-$ f $\times$ R = mR2$\alpha$

40 $-$ f = mR$\alpha$         . . . .(ii)

From (i) and (ii)

$\alpha$ = ${{40} \over {mR}}$ = 16
4

### JEE Main 2019 (Online) 12th January Morning Slot

Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is :
A
16 cm
B
12 cm
C
14 cm
D
18 cm

## Explanation

Consider an element of radius x and thickness dx Mass of element, dm = $\sigma 2\pi x\left( {dx} \right)$

Here, $\sigma$ = mass per unit area = ${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$

Moment of inertia of element, dI = (dm)x2

$\Rightarrow$ I = $\sigma 2\pi \int\limits_r^R {{x^3}dx}$

= $\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$

= ${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$

= ${m \over 2}\left( {{R^2} + {r^2}} \right)$ .....(i)

Moment of inertia of thin cylinder of same mass,

I = m$r_0^2$ ......(ii)

$\Rightarrow$ m$r_0^2$ = ${m \over 2}\left( {{R^2} + {r^2}} \right)$

$\Rightarrow$ $r_0^2$ = 250

$\Rightarrow$ r0 $\simeq$ 16 cm