 JEE Mains Previous Years Questions with Solutions

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1

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, $F \propto y$
2

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.
3

AIEEE 2012

This question has statement- $1$ and statement- $2.$ Of the four choices given after the statements, choose the one that best describe the two statements.
An insulating solid sphere of radius $R$ has a uniformly positive charge density $\rho$. As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero.

Statement- $1:$ When a charge $q$ is take from the centre of the surface of the sphere its potential energy changes by ${{q\rho } \over {3{\varepsilon _0}}}$
Statement- $2:$ The electric field at a distance $r\left( {r < R} \right)$ from the center of the sphere is ${{\rho r} \over {3{\varepsilon _0}}}.$

A
Statement- $1$ is true, Statement- $2$ is true; Statement- $2$ is not the correct explanation of Statement- $1$.
B
Statement $1$ is true, Statement $2$ is false.
C
Statement $1$ is false, Statement $2$ is true.
D
Statement- $1$ is true, Statement- $2$ is true; Statement- $2$ is the correct explanation of Statement- $1$.

Explanation

The electric field inside a uniformly charged sphere is

= ${{\rho .r} \over {3{ \in _0}}}$

The electric potential inside a uniformly charged sphere

$= {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - {{{r^2}} \over {{R^2}}}} \right]$

$\therefore$ Potential difference between center and surface

$= {{\rho {R^2}} \over {6{ \in _0}}}\left[ {3 - 2} \right] = {{\rho {R^2}} \over {6{ \in _0}}}$

$\Delta U = {{q\rho {R^2}} \over {6{ \in _0}}}$
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

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