### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

Assume that an electric field $\overrightarrow E = 30{x^2}\widehat i$ exists in space. Then the potential difference ${V_A} - {V_O},$ where ${V_O}$ is the potential at the origin and ${V_A}$ the potential at $x=2$ $m$ is :
A
$120$ $J/C$
B
$-120$ $J/C$
C
$-80$ $J/C$
D
$80$ $J/C$

## Explanation

Potential difference between any two points in an electric field is given by,

$dV = - \overrightarrow E .\overrightarrow {dx}$

$\int\limits_{{V_O}}^{{V_A}} {dV = - \int\limits_0^2 {30{x^2}} } dx$

${V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C$
2

### JEE Main 2013 (Offline)

Two capacitors ${C_1}$ and ${C_2}$ are charged to $120$ $V$ and $200$ $V$ respectively. It is found that connecting them together the potential on each one can be made zero. Then
A
$5{C_1} = 3{C_2}$
B
$3{C_1} = 5{C_2}$
C
$3{C_1} + 5{C_2} = 0$
D
$9{C_1} = 4{C_2}$

## Explanation

For potential to be made zero, after connection

$120{C_1} = 200{C_2}$

$\left[ {\,\,} \right.$ as $\left. {C = {q \over v}\,\,} \right]$

$\Rightarrow 3{C_1} = 5{C_2}$
3

### JEE Main 2013 (Offline)

A charge $Q$ is uniformly distributed over a long rod $AB$ of length $L$ as shown in the figure. The electric potential at the point $O$ lying at distance $L$ from the end $A$ is
A
${Q \over {8\pi {\varepsilon _0}L}}$
B
${{3Q} \over {4\pi {\varepsilon _0}L}}$
C
${Q \over {4\pi {\varepsilon _0}L\,\ln \,2}}$
D
${{Q\ln \,2} \over {4\pi {\varepsilon _0}L\,{}^s}}$

## Explanation

Electric potential is given by,

$V = \int\limits_L^{2L} {{{kdq} \over x}}$

$= {{2L} \over L}{1 \over {4\pi {\varepsilon _0}}}{{\left( {{q \over L}} \right)dx} \over x}$

$= {q \over {4\pi {\varepsilon _0}L}}\ln \left( 2 \right)$
4

### JEE Main 2013 (Offline)

Two charges, each equals to $q,$ are kept at $x=-a$ and $x=a$ on the $x$-axis. A particle of mass $m$ and charge ${q_0} = {q \over 2}$ is placed at the origin. If charge ${q_0}$ is given a small displacement $\left( {y < < a} \right)$ along the $y$-axis, the net force acting on the particle is proportional to
A
$y$
B
$-y$
C
${1 \over y}$
D
$-{1 \over y}$

## Explanation

$\Rightarrow {F_{net}} = 2F\,\cos \theta$

${F_{net}} = {{2kq\left( {{q \over 2}} \right)} \over {{{\left( {\sqrt {{y^2} + {a^2}} } \right)}^2}}}.{y \over {\sqrt {{y^2} + {a^2}} }}$

${F_{net}} = {{2kq\left( {{q \over 2}} \right)y} \over {{{\left( {{y^2} + {a^2}} \right)}^{3/2}}}} \Rightarrow {{k{q^2}y} \over {{a^3}}}$

S0, $F \propto y$