### JEE Mains Previous Years Questions with Solutions

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1

### 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 $t = 0,6 \,s$ $\omega = 0$ )

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

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

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

### AIEEE 2011

A mass $m$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $m$ and radius $R.$ Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $m,$ if the string does not slip on the pulley, is:
A
$g$
B
${2 \over 3}g$
C
${g \over 3}$
D
${3 \over 2}g$

## Explanation

This is the free body diagram of pulley and mass
For translation motion of the block,

$mg - T = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right)$

For rotational motion of the pulley,

$T\times R = I\alpha = I{a \over R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

where $\alpha =$ angular acceleration of disc = ${a \over R}$

and $I = {1 \over 2}m{R^2}$ (For circular disc)

Solving $(1)$ & $(2),$

$a = {{mg} \over {\left( {m + {I \over {{R^2}}}} \right)}} = {{mg} \over {m + {{m{R^2}} \over {2{R^2}}}}}$

$= {{2mg} \over {3m}} = {{2g} \over 3}$
3

### AIEEE 2011

A thin horizontal circular disc is rotating about a vertical axis passing through its center. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc.
A
continuously decreases
B
continuously increases
C
first increases and then decreases
D
remains unchanged

## Explanation

Here no external force is applied on the disc so Torque ($\tau$) = 0.

So angular momentum is conserved.

That means ${I_1}{\omega _1} = {I_2}{\omega _2}$

$\Rightarrow$ ${\omega _2} = {{{I_1}{\omega _1}} \over {{I_2}}}$

$\therefore$ Angular speed is inversely proportional to Moment of inertia.

For disc $I = {1 \over 2}M{R^2}$

$\therefore$ Moment of Inertia is proportional to Mass.

As insect moves along a diameter, the effective mass of disc first decreases then increases and hence the moment of inertia first decreases then increases so from principle of conservation of angular momentum, angular speed, first increases then decreases.
4

### AIEEE 2009

A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its center of mass rises to a maximum height of:
A
${1 \over 6}\,\,{{l\omega } \over g}$
B
${1 \over 2}\,\,{{{l^2}{\omega ^2}} \over g}$
C
${1 \over 6}\,\,{{{l^2}{\omega ^2}} \over g}$
D
${1 \over 3}\,\,{{{l^2}{\omega ^2}} \over g}$

## Explanation

The moment of inertia of the rod about $O$ is ${1 \over 2}m{\ell ^2}.$ The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this conditions is ${1 \over 2}I{\omega ^2}$ where $I$ is the moment of inertia of the rod about $O.$ when the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the center of mass is raised through $h$, so the increase in potential energy is $mgh$. This is equal to kinetic energy ${1 \over 2}I{\omega ^2}$.

$\therefore$ $mgh = {1 \over 2}I{\omega ^2} = {1 \over 2}\left( {{1 \over 3}m{l^2}} \right){\omega ^2}$.

$\Rightarrow h = {{{\ell ^2}{\omega ^2}} \over {6g}}$