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### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
When a rubber-band is stretched by a distance $x$, it exerts restoring force of magnitude $F = ax + b{x^2}$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is :
A
$a{L^2} + b{L^3}$
B
${1 \over 2}\left( {a{L^2} + b{L^3}} \right)$
C
${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$
D
${1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)$

## Explanation

Given Restoring force, F = ax + bx2

Work done in stretching the rubber-band by a distance $dx$ is

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$

Intergrating both sides,

$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$

= $\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$

= ${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$
2

### AIEEE 2012

MCQ (Single Correct Answer)
This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements, choose the one that best describes the two Statements.

If two springs ${S_1}$ and ${S_2}$ of force constants ${k_1}$ and ${k_2}$, respectively, are stretched by the same force, it is found that more work is done on spring ${S_1}$ than on spring ${S_2}$.

STATEMENT 1: If stretched by the same amount work done on ${S_1}$, Work done on ${S_1}$ is more than ${S_2}$
STATEMENT 2: ${k_1} < {k_2}$

A
Statement 1 is false, Statement 2 is true
B
Statement 1 is true, Statement 2 is false
C
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1
D
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1

## Explanation

We know force (F) = kx

$W = {1 \over 2}k{x^2}$

$W =$ ${{{{\left( {kx} \right)}^2}} \over {2k}}$ $\,\,\,$

$\therefore$ $W = {{{F^2}} \over {2k}}$ [ as $F=kx$ ]

When force is same then,

$W \propto {1 \over k}$

Given that, ${W_1} > {W_2}$

$\therefore$ ${k_1} < {k_2}$

Statement-2 is true.

For the same extension, x1 = x2 = x

Work done on spring S1 is W1 = ${1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}$

Work done on spring S2 is W2 = ${1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}$

$\therefore$ ${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$

As ${k_1} < {k_2}$ then ${W_1} < {W_2}$

So, Statement-1 is false.
3

### AIEEE 2010

MCQ (Single Correct Answer)
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: ${{dU\left( x \right)} \over {dx}} = 0$

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

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

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

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

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

### AIEEE 2008

MCQ (Single Correct Answer)
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

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

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