### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

A long cylindrical shell carries positives surfaces change $\sigma$ in the upper half and negative surface charge - $\sigma$ in the lower half. The electric field lines around the cylinder will look like figure given in :
(figures are schematic and not drawn to scale)
A
B
C
D

## Explanation

From the property of electric filled lines we know,

Electric filled lines stars from positive charge and ends at negative charge.

As density of cylinder is uniform so all the lines comes out of positive charge should enter to the negative charge.

From this discussion we can say (C) is correct.
2

### 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.

3

### 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

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

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

### 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,

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

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

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