### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

MCQ (Single Correct Answer)
An annular ring with inner and outer radii ${R_1}$ and ${R_2}$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, ${{{F_1}} \over {{F_2}}}\,$ is
A
${\left( {{{{R_1}} \over {{R_2}}}} \right)^2}$
B
${{{{R_2}} \over {{R_1}}}}$
C
${{{{R_1}} \over {{R_2}}}}$
D
$1$

## Explanation

Let the mass of each particle is m.

Then force experienced by each particle, $F = m{\omega ^2}R$

$\therefore$ ${{{F_1}} \over {{F_2}}} = {{m{\omega ^2}{R_1}} \over {m{\omega ^2}{R_2}}}$

$\Rightarrow$ ${{{F_1}} \over {{F_2}}} = {{{R_1}} \over {{R_2}}}$
2

### AIEEE 2004

MCQ (Single Correct Answer)
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which on of the following will not be affected ?
A
Angular velocity
B
Angular momentum
C
Moment of inertia
D
Rotational kinetic energy

## Explanation

Solid sphere is rotating in free space that means no external torque is operating on the sphere.

Angular momentum will remain the same since external torque is zero.
3

### 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

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

$\therefore$ ${I_A} < {I_B}$
4

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

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

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

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

$= {1 \over {64}}$

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