### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

In the given circuit, charges ${Q_2}$ on the $2\mu F$ capacitor changes as $C$ is varied from $1\,\mu F$ to $3\mu F.$ ${Q_2}$ as a function of $'C'$ is given properly by:
$\left( {figures\,\,are\,\,drawn\,\,schematically\,\,and\,\,are\,\,not\,\,to\,\,scale} \right)$
A
B
C
D

## Explanation

From figure, ${Q_2} = {2 \over {2 + 1}}Q = {2 \over 3}Q$

$Q = E\left( {{{C \times 3} \over {C + 3}}} \right)$

$\therefore$ ${Q_2} = {2 \over 3}\left( {{{3CE} \over {C + 3}}} \right) = {{2CE} \over {C + 3}}$

Therefore graph $d$ correctly dipicts.

2

### JEE Main 2014 (Offline)

A parallel plate capacitor is made of two circular plates separated by a distance $5$ $mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times {10^4}\,V/m$ the charge density of the positive plate will be close to:
A
$6 \times {10^{ - 7}}\,\,C/{m^2}$
B
$3 \times {10^{ - 7}}\,\,C/{m^2}$
C
$3 \times {10^4}\,\,C/{m^2}$
D
$6 \times {10^4}\,\,C/{m^2}$

## Explanation

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

$E = {\sigma \over {K{\varepsilon _0}}}$

Then, charge density

$\sigma = K{\varepsilon _0}E$

$= 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}$
3

### 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,

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

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

${V_A} - {V_O} = \left[ {10{x^3}} \right]_{0}^2 = - 80J/C$
4

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

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