### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2008

A parallel plate capacitor with air between the plates has capacitance of $9$ $pF.$ The separation between its plates is $'d'.$ The space between the plates has dielectric constant ${k_1}$ $=3$ and thickness ${d \over 3}$ while the other one has dielectric constant ${k_2} = 6$ and thickness ${{2d} \over 3}$. Capacitance of the capacitor is now
A
$1.8$ $pF$
B
$45$ $pF$
C
$40.5$ $pF$
D
$20.25$ $pF$

## Explanation

The given capacitance is equal to two capacitances connected in series where

${C_1} = {{{k_1}{ \in _0}A} \over {d/3}} = {{3{k_1}{ \in _0}A} \over d}$

$= {{3 \times 3{ \in _0}A} \over d} = {{9{ \in _0}A} \over d}$

and

${C_2} = {{{k_2}{ \in _0}A} \over {2d/3}} = {{3{k_2}{ \in _0}A} \over {2d}}$

$= {{3 \times 6{ \in _0}A} \over {2d}} = {{9{ \in _0}A} \over d}$

The equivalent capacitance ${C_{eq}}$ is

${1 \over {C{}_{eq}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$

$= {d \over {9{ \in _0}A}} + {d \over {9{ \in _0}A}}$

$= {{2d} \over {9{ \in _0}A}}$

$\therefore$ ${C_{eq}} = {9 \over 2}{{{\varepsilon _0}A} \over d} = {9 \over 2} \times 9pF = 40.5pF$
2

### AIEEE 2007

A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is
A
zero
B
${1 \over 2}\,\left( {K - 1} \right)\,C{V^2}$
C
${{C{V^2}\left( {K - 1} \right)} \over K}$
D
$\left( {K - 1} \right)\,C{V^2}$

## Explanation

The potential energy of a charged capacitor before removing the dielectric slat is $U = {{{Q^2}} \over {2C}}.$

The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is $U = {{{Q^2}} \over {2C}}$

There is no change in potential energy, therefore work done is zero.
3

### AIEEE 2007

The potential at a point $x$ (measured in $\mu \,m$) due to some charges situated on the $x$-axis is given by $V\left( x \right) = 20/\left( {{x^2} - 4} \right)$ volt
The electric field $E$ at $x = 4\,\mu \,m$ is given by
A
$(10/9)$ volt / $\mu$ $m$ and in the $+ ve$ $x$ direction
B
$\left( {5/3} \right)$ volt/ $\mu$ $m$ and in the $-ve$ $x$ direction
C
$\left( {5/3} \right)$ volt/$\mu$ $m$ and in the $+ve$ $x$ direction
D
$\left( {10/9} \right)$ volt/ $\mu \,m$ and in the $-ve$ $x$ direction

## Explanation

Here, $V\left( x \right) = {{20} \over {{x^2} - 4}}volt$

We know that $E = - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} \right)$
or, $E = + {{40x} \over {{{\left( {{x^2} - 4} \right)}^2}}}$

At $x = 4\mu m,$

$E = + {{40 \times 4} \over {{{\left( {{4^2} - 4} \right)}^2}}}$

$= + {{160} \over {144}} = + {{10} \over 9}volt/\mu m.$

Positive sign indicates that $\overrightarrow E$ is in $+ve$ $x$-direction.
4

### AIEEE 2007

Charges are placed on the vertices of a square as shown. Let $\overrightarrow E$ be the electric field and $V$ the potential at the center. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then
A
$\overrightarrow E$ changes, $V$ remains unchanged
B
$\overrightarrow E$ remains unchanged, $V$ changes
C
both $\overrightarrow E$ and $V$ change
D
$\overrightarrow E$ and $V$ remain unchanged

## Explanation

As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions.

Also, the potential will be same in both cases as it is a scalar quantity.