JEE Mains Previous Years Questions with Solutions

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1

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,

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

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

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

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, $F \propto y$
3

AIEEE 2012

The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant $\tau$ of this circuit lies between
A
100 sec and 150 sec
B
0 and 50 sec
C
50 sec and 100 sec
D
150 sec and 200 sec

Explanation

During discharging of a capacitor, V = V0${e^{ - {t \over \tau }}}$

At t = $\tau$,

V = ${{{V_0}} \over e}$ = 0.37V0

From the graph, t = 0, V0 = 25 V

$\therefore$ V = 0.37 $\times$ 25 = 9.25 V

This voltage will occur at time between 100 sec and 150 sec. Hence, time constant $\tau$ of this circuit lies between 100 sec and 150 sec.
4

AIEEE 2012

In a uniformly charged sphere of total charge $Q$ and radius $R,$ the electric field $E$ is plotted as function of distance from the center. The graph which would correspond to the above will be:
A
B
C
D

Explanation

${E_{in}} \propto r$
${E_{out}} \propto {1 \over {{r^2}}}$