### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

Four charges equal to -$Q$ are placed at the four corners of a square and a charge $q$ is at its center. If the system is in equilibrium the value of $q$ is
A
$- {Q \over 2}\left( {1 + 2\sqrt 2 } \right)$
B
${Q \over 4}\left( {1 + 2\sqrt 2 } \right)$
C
$- {Q \over 4}\left( {1 + 2\sqrt 2 } \right)$
D
${Q \over 2}\left( {1 + 2\sqrt 2 } \right)$

## Explanation

Net field at A should be zero

$\sqrt 2 \,{E_1} + {E_2} = {E_3}$

$\therefore$ ${{kQ \times \sqrt 2 } \over {{a^2}}} + {{kQ} \over {\left( {\sqrt 2 a} \right)}} = {{kq} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}}$

$\Rightarrow {{Q\sqrt 2 } \over 1} + {Q \over 2} = 2q$

$\Rightarrow q = {Q \over 4}\left( {2\sqrt 2 + 1} \right).$
2

### AIEEE 2004

A charge particle $'q'$ is shot towards another charged particle $'Q'$ which is fixed, with a speed $'v'$. It approaches $'Q'$ upto a closest distance $r$ and then returns. If $q$ were given a speed of $'2v'$ the closest distances of approaches would be
A
$r/2$
B
$2r$
C
$r$
D
$r/4$

## Explanation

${1 \over 2}m{v^2} = {{kQq} \over r}$

$\Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}$

$\Rightarrow r' = {r \over 4}$
3

### AIEEE 2004

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges on them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that $B$ but uncharged is brought in contact with $B,$ then brought in correct with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is
A
$F/8$
B
$3$ $F/4$
C
$F/4$
D
$3$ $F/8$

## Explanation

$F \propto {{{Q_A}{Q_C}} \over {{x^2}}}$

$x$ is distance between the spheres. After first operation charge on $B$ is halved i.e ${Q \over 2}.$

and charge on third sphere becomes ${Q \over 2}.$ Now it is touched to $C$, charge then equally

distributes them selves to make potential same, hence charge on $C$ becomes

$\left( {Q + {Q \over 2}} \right){1 \over 2} = {{3Q} \over 4}.$

$\therefore$ ${F_{new}} \propto {{{Q_C}{Q_B}} \over {{x^2}}}$

$= {{\left( {{{3Q} \over 4}} \right)\left( {{Q \over 2}} \right)} \over {{x^2}}}$

$= {3 \over 8}{{{Q^2}} \over {{x^2}}}$

or, ${F_{new}} = {3 \over 8}F$
4

### AIEEE 2003

The length of a given cylindrical wire is increased by $100\%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be
A
$200\%$
B
$100\%$
C
$50\%$
D
$300\%$

## Explanation

${R_f} = {n^2}{R_1}$

Here $n=2$ (length becomes twice)

$\therefore$ ${R_f} = 4{R_i}$

New resistance $=400$ of ${R_i}$

$\therefore$ Increase $= 300\%$