### JEE Mains Previous Years Questions with Solutions

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1

### 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

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

### 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

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

$\Delta U = {{q\rho {R^2}} \over {6{ \in _0}}}$
3

### AIEEE 2011

Two identical charged spheres suspended from a common point by two massless strings of length $l$ are initially a distance $d\left( {d < < 1} \right)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $v$. Then as a function of distance $x$ between them,
A
$v\, \propto \,{x^{ - 1}}$
B
$y\, \propto \,{x^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$
C
$v\, \propto \,x$
D
$v\, \propto \,{x^{ - {\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$

## Explanation

At any instant

$T\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$T\sin \theta = {F_e}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

$\Rightarrow {{\sin \theta } \over {\cos \theta }} = {{{F_e}} \over {mg}} \Rightarrow {F_e} = mg\,\tan \theta$

$\Rightarrow {{k{q^2}} \over {{x^2}}} = mg\,\tan \theta \Rightarrow {q^2} \propto {x^2}\tan \theta$

$\sin \theta = {\textstyle{x \over {2l}}}$

For small $\theta ,\,\sin \theta \approx \tan \theta$

$\therefore$ ${q^2} \propto {x^3}$

$\Rightarrow q{{dq} \over {dt}} \propto {x^2}{{dx} \over {dt}}$

$\therefore$ ${{dq} \over {dt}} = const.$

$\therefore$ $q \propto {x^2}.v \Rightarrow {x^{3/2}}\alpha {x^2}.v\,\,$ $\,\,\,\,\,$ $\left[ {\,\,} \right.$ as $\left. {{q^2} \propto {x^3}\,\,} \right]$

$\Rightarrow v \propto {x^{ - 1/2}}$
4

### AIEEE 2011

The electrostatic potential inside a charged spherical ball is given by $\phi = a{r^2} + b$ where $r$ is the distance from the center and $a,b$ are constants. Then the charge density inside the ball is:
A
$- 6a{\varepsilon _0}r$
B
$- 24\pi a{\varepsilon _0}$
C
$- 6a{\varepsilon _0}$
D
$- 24\pi {\varepsilon _0}r$

## Explanation

Electric field

$E = {{d\phi } \over {dr}} = - 2ar\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

By Gauss's theorem

$E = {1 \over {4\theta {\varepsilon _0}{r^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

From $\left( i \right)$ and $\left( ii \right),$

$q = - 8\pi {\varepsilon _0}a{r^3}$

$\Rightarrow dq = - 24\pi {\varepsilon _0}ar{}^2dr$

Charge density, $\rho = {{dq} \over {4\pi {r^2}dr}} = - 6{\varepsilon _0}a$