 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

A circular disc $X$ of radius $R$ is made from an iron plate of thickness $t,$ and another disc $Y$ of radius $4$ $R$ is made from an iron plate of thickness ${t \over 4}.$ Then the relation between the moment of inertia ${I_X}$ and ${I_Y}$ is
A
${I_Y} = 32{I_X}$
B
${I_Y} = 16{I_X}$
C
${I_Y} = {I_X}$
D
${I_Y} = 64{I_X}$

Explanation

We know that density $\left( d \right) = {{mass\left( M \right)} \over {volume\left( V \right)}}$

$\therefore$ Mass of disc $M = d \times V = d \times \left( {\pi {R^2} \times t} \right).$

The moment of inertia of any disc is $I = {1 \over 2}M{R^2}$

$\therefore$ $I = {1 \over 2}\left( {d \times \pi {R^2} \times t} \right){R^2} = {{\pi d} \over 2}t \times {R^4}$

$\therefore$ ${{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}$

$= {{t \times {R^4}} \over {{t \over 4} \times {{\left( {4R} \right)}^4}}}$

$= {1 \over {64}}$
2

AIEEE 2003

Let $\overrightarrow F$ be the force acting on a particle having position vector $\overrightarrow r ,$ and $\overrightarrow \tau$ be the torque of this force about the origin. Then
A
$\overrightarrow {r.} \overrightarrow \tau = 0\,\,$ and $\overrightarrow {F.} \overrightarrow \tau \ne 0\,\,$
B
$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu} {\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$
C
$\overrightarrow {r.} \vec \tau \ne 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau \ne 0$
D
$\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$

Explanation As we know $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F$

So the angle between $\overrightarrow \tau$ and $\overrightarrow r$ is ${90^ \circ }$ and the angle between $\overrightarrow t$ and $\overrightarrow F$ is also ${90^ \circ }.$

We also know that the dot product of two vectors which have an angle of ${90^ \circ }$ between them is zero.

$\therefore$ $\overrightarrow {r.} \vec \tau = 0{\mkern 1mu}$ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$

Therefore $(d)$ is the correct option.
3

AIEEE 2003

A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is
A
${L \over 4}$
B
$2L$
C
$4L$
D
${L \over 2}$

Explanation

We know Rotational Kinetic Energy$={1 \over 2}I{\omega ^2},$

Angular Momentum $L = I\omega \Rightarrow I = {L \over \omega }$

$\therefore$ Initial $K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega$

Final $K.E'$ = ${{K.E} \over 2}$ = ${1 \over 2}{L'} \times 2\omega$

$\therefore$ ${{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}}$

$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}$

$\therefore$ $L' = {L \over 4}$
4

AIEEE 2002

Moment of inertia of a circular wire of mass $M$ and radius $R$ about its diameter is
A
${{M{R^2}} \over 2}$
B
$M{R^2}$
C
$2M{R^2}$
D
${{M{R^2}} \over 4}$

Explanation

Moment of Inertia of a circular wire about an axis $nn'$ passing through the centre of the circle and perpendicular to the plane of the circle $= M{R^2}$ As shown in the figure, $X$-axis and $Y$-axis lies in the plane of the ring. Then by perpendicular axis theorem

${I_X} + {I_Y} = {I_Z}$

$\Rightarrow 2{I_X} = M{R^2}\,$ $\left[ \, \right.$ as ${I_X} - {I_Y}$ (by symmetry) and ${I_Z} = M{R^2}$ $\left. \, \right]$

$\therefore$ ${I_X} = {1 \over 2}M{R^2}$