### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

A mass $'m'$ is supported by a massless string wound around a uniform hollow cylinder of mass $m$ and radius $R.$ If the string does not slip on the cylinder, with what acceleration will the mass fall or release?
A
${{2g} \over 3}$
B
${{g} \over 2}$
C
${{5g} \over 6}$
D
$g$

## Explanation

Here string is not slipping over pulley.

From figure,

Acceleration $a = R\alpha \,\,\,\,...(i)$

Applying Newton's second law on the block we get,

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

Torque on cylinder due to tension in string about the fixed point,

$T \times R = I\alpha$

$T \times R = m{R^2}\alpha = m{R^2}\left( {{a \over R}} \right)$

[ as $I$ = MR2 for hollow cylinder]

or $T=ma$

$\Rightarrow \,\,mg - ma = ma$

$\Rightarrow \,\,\,a = {g \over 2}$
2

### JEE Main 2014 (Offline)

A bob of mass $m$ attached to an inextensible string of length $l$ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed $\omega \,rad/s$ about the vertical. About the point of suspension:
A
angular momentum is conserved
B
angular momentum changes in magnitude but not in direction.
C
angular momentum changes in direction but not in magnitude.
D
angular momentum changes both in direction and magnitude.

## Explanation

Torque working on the bob of mass $m$ is, $\tau = mg \times \ell \,\sin \,\theta .$ (Direction of torque by the weight is parallel to plane of rotation of the particle)

As $\tau$ is perpendicular to the angular momentum $\overrightarrow L$ of the bob, so the direction of $L$ changes but magnitude remains same.

3

### JEE Main 2013 (Offline)

A hoop of radius $r$ and mass $m$ rotating with an angular velocity ${\omega _0}$ is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it cases to slip?
A
${{r{\omega _0}} \over 4}$
B
${{r{\omega _0}} \over 3}$
C
${{r{\omega _0}} \over 2}$
D
${r{\omega _0}}$

## Explanation

From conservation of angular momentum at point of contact,

$m{r^2}{\omega _0} = mvr + m{r^2}\omega$

$m{r^2}{\omega _0} = mvr + m{r^2}\left( {{v \over r}} \right)$ [ as $v = r\omega$ ]

$m{r^2}{\omega _0} = mvr + mvr$

$m{r^2}{\omega _0} = 2mvr$

$v = {{{\omega _0}r} \over 2}$
4

### AIEEE 2011

A pulley of radius $2$ $m$ is rotated about its axis by a force $F = \left( {20t - 5{t^2}} \right)$ newton (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10kg$-${m^2}$ the number of rotation made by the pulley before its direction of motion is reversed, is:
A
more than $3$ but less than $6$
B
more than $6$ but less than $9$
C
more than $9$
D
less than $3$

## Explanation

Given $F = 20t - 5{t^2}$, R = 2 m and $I$ = 10 kg m2

Torque applied on pulley $\tau = FR$

$\therefore$ $\alpha = {{FR} \over I}$ [ as $\tau = I\alpha$ ]

$\Rightarrow$ $\alpha = {{\left( {20t - 5{t^2}} \right) \times 2} \over {10}}$

$\Rightarrow$ $\alpha = 4t - {t^2}$

$\Rightarrow {{d\omega } \over {dt}} = 4t - {t^2}$

$\Rightarrow \int\limits_0^\omega {d\omega } = \int\limits_0^t {\left( {4t - {t^2}} \right)} dt$

$\Rightarrow \omega = 2{t^2} - {{{r^3}} \over 3}$

( At $t = 0,6 \,s$ $\omega = 0$ )

$\omega = {{d\theta } \over {dt}} = 2{t^2} - {{{t^3}} \over 3}$

$\int\limits_0^\theta {d\theta } = \int\limits_0^6 {\left( {2{t^2} - {{{r^3}} \over 3}} \right)} dt$

$\Rightarrow \theta = 36rad\,\, \Rightarrow n = {{36} \over {2\pi }} < 6$