 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

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

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

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

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

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

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

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

$= {1 \over {64}}$