 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2010

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$ where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$ is
A
${{{b^2}} \over {2a}}$
B
${{{b^2}} \over {12a}}$
C
${{{b^2}} \over {4a}}$
D
${{{b^2}} \over {6a}}$

Explanation

Given $U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$

${U\left( {x = \infty } \right)}$ = 0

We know $F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$

At equilibrium: ${{dU\left( x \right)} \over {dx}} = 0$

$\Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}}$

$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$

$\therefore$ ${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$

$= - {{{b^2}} \over {4a}}$

$\therefore$ $D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$
2

AIEEE 2008

An athlete in the olympic games covers a distance of $100$ $m$ in $10$ $s.$ His kinetic energy can be estimated to be in the range
A
$200J-500J$
B
$2 \times {10^5}J - 3 \times {10^5}J$
C
$20,000J - 50,000J$
D
$2,000J - 5,000J$

Explanation

The average speed of the athelete

$v = {{100} \over {10}} = 10m/s\,\,\,\,$ $\therefore$ $K.E. = {1 \over 2}m{v^2}$

If mass of athlete is $40$ $kg$ then, $K.E.$ $= {1 \over 2} \times 40 \times {\left( {10} \right)^2} = 2000J$

If mass of athlete is $100$ $kg$ then, $K.E.$ $= {1 \over 2} \times 100 \times {\left( {10} \right)^2} = 5000J$

His kinetic energy can be in the range = 2000 J to 5000 J.
3

AIEEE 2008

A block of mass $0.50$ $kg$ is moving with a speed of $2.00$ $m{s^{ - 1}}$ on a smooth surface. It strike another mass of $1.0$ $kg$ and then they move together as a simple body. The energy loss during the collision is
A
$0.16J$
B
$1.00J$
C
$0.67J$
D
$0.34$ $J$

Explanation

Let $m$ = 0.50 kg and $M$ = 1.0 kg

Initial kinetic energy of the system when 1 kg mass is at rest,

$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$

$= {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$

For collision, applying conservation of linear momentum

$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$

$\therefore$ $0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$

Final kinetic energy of the system is

$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$

$= {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$

$\therefore$ Energy loss during collision

$= \left( {1 - {1 \over 3}} \right)J = 0.67J$
4

AIEEE 2007

A $2$ $kg$ block slides on a horizontal floor with a speed of $4m/s.$ It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is $15N$ and spring constant is $10, 000$ $N/m.$ The spring compresses by
A
$8.5cm$
B
$5.5cm$
C
$2.5cm$
D
$11.0cm$

Explanation

Let the block compress the spring by $x$ before coming to rest.

Initial kinetic energy of the block $=$ (potential energy of compressed spring) $+$ work done due to friction.

${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$

$10,000{x^2} + 30x - 32 = 0$

$\Rightarrow 5000{x^2} + 15x - 16 = 0$

$\therefore$ $x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$

$\,\,\,\,\, = 0.055m = 5.5cm.$