JEE Mains Previous Years Questions with Solutions

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1

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

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 $\left. {{q^2} \propto {x^3}\,\,} \right]$

$\Rightarrow v \propto {x^{ - 1/2}}$
3

AIEEE 2010

A thin semi-circular ring of radius $r$ has a positive charges $q$ distributed uniformly over it. The net field $\overrightarrow E$ at the center $O$ is
A
${q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$
B
$- {q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$
C
$- {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$
D
${q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$

Explanation

Let us consider a differential element $dl.$ charge on this element.

$dq = \left( {{q \over {\pi r}}} \right)dl$

$= {q \over {\pi r}}\left( {rd\theta } \right)\,\,\,\,\,$ (as $dl = rd\theta$)

$= \left( {{q \over \pi }} \right)d\theta$

Electric field at $O$ due to $dq$ is

$dE = {1 \over {4\pi { \in _0}}}.{{dq} \over {{r^2}}}$

$= {1 \over {4\pi { \in _0}}}.{q \over {\pi {r^2}}}d\theta$

The component $dE\cos \theta$ will be counter balanced by another element on left portion.

Hence resultant field at $O$ is the resultant of the component $dE\sin \theta$ only.

$\therefore$ $E = \int {dE\sin \theta = \int\limits_0^\pi {{q \over {4{\pi ^2}{r^2}{ \in _0}}}} } \sin \theta d\theta$

$= {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos \theta } \right]_0^\pi$

$= {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} \right)$

$= {q \over {2{\pi ^2}{r^2}{ \in _0}}}$

The directions of $E$ is towards negative $y$-axis.

$\therefore$ $\overrightarrow E = - {q \over {2{\pi ^2}{r^2}{ \in _0}}}\widehat j$
4

AIEEE 2010

Let there be a spherically symmetric charge distribution with charge density varying as $\rho \left( r \right) = {\rho _0}\left( {{5 \over 4} - {r \over R}} \right)$ upto $r=R,$ and $\rho \left( r \right) = 0$ for $r>R,$ where $r$ is the distance from the erigin. The electric field at a distance $r\left( {r < R} \right)$ from the origin is given by
A
${{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)$
B
${{4\pi {\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 3} - {r \over R}} \right)$
C
${{4{\rho _0}r} \over {4{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)$
D
${{{\rho _0}r} \over {3{\varepsilon _0}}}\left( {{5 \over 4} - {r \over R}} \right)$

Explanation

Let us consider a spherical shell of radius $x$ and thickness $dx.$

Charge on this shell

$dq = \rho .4{\pi ^2}dx = {\rho _0}\left( {{5 \over 4} - {x \over R}} \right).4\pi {x^2}dx$

$\therefore$ Total charge in the spherical region from center to $r$ $\left( {r < R} \right)$ is

$q = \int {dq = 4\pi {\rho _0}\int\limits_0^r {\left( {{5 \over 4} - {x \over R}} \right)} } {x^2}dx$

$= 4\pi {\rho _0}\left[ {{5 \over 4}.{{{r^3}} \over 3} - {1 \over R}.{{{r^4}} \over 4}} \right]$

$= \pi {\rho _0}{r^3}\left( {{5 \over 3} - {r \over R}} \right)$

$\therefore$ Electric field at $r,$ $E = {1 \over {4\pi { \in _0}}}.{q \over {{r^2}}}$

$= {1 \over {4\pi { \in _0}}}.{{\pi {\rho _0}{r^3}} \over {{r^2}}}\left( {{5 \over 3} - {r \over R}} \right)$

$= {{{\rho _0}r} \over {4{ \in _0}}}\left( {{5 \over 3} - {r \over R}} \right)$