### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2004

MCQ (Single Correct Answer)
One solid sphere $A$ and another hollow sphere $B$ are of same mass and same outer radii. Their moment of inertia about their diameters are respectively ${I_A}$ and ${I_B}$ such that
A
${I_A} < {I_B}$
B
${I_A} > {I_B}$
C
${I_A} = {I_B}$
D
${{{I_A}} \over {{I_B}}} = {{{d_A}} \over {{d_B}}}$ where ${d_A}$ and ${d_B}$ are their densities.

## Explanation

For solid sphere the moment of inertia of $A$ about its diameter

${I_A} = {2 \over 5}M{R^2}.$

The moment of inertia of a hollow sphere $B$ about its diameter

${I_B} = {2 \over 3}M{R^2}.$

$\therefore$ ${I_A} < {I_B}$
2

### AIEEE 2003

MCQ (Single Correct Answer)
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 $I = {1 \over 2}M{R^2}$

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

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

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

$= {1 \over {64}}$
3

### AIEEE 2003

MCQ (Single Correct Answer)
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.
4

### AIEEE 2003

MCQ (Single Correct Answer)
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 $K.E'$ = ${{K.E} \over 2}$ = ${1 \over 2}{L'} \times 2\omega$

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

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

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

### Joint Entrance Examination

JEE Advanced JEE Main

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET

Class 12