JEE Mains Previous Years Questions with Solutions

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1

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.

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

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

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

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

AIEEE 2009

A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $Q/q$ equals:
A
$-1$
B
$1$
C
$- {1 \over {\sqrt 2 }}$
D
$- 2\sqrt 2$

Explanation

Let $F$ be the force between $Q$ and $Q.$ The force between $q$ and $Q$ should be attractive for net force on $Q$ to be zero. Let $F'$ be the force between $Q$ and $q.$ For equilibrium

$\sqrt 2 F' = - F$

$\sqrt 2 \times k{{Qq} \over {{\ell ^2}}} = - k{{{Q^2}} \over {{{\left( {\sqrt 2 \ell } \right)}^2}}}$

$\Rightarrow {Q \over q} = - 2\sqrt 2$
4

AIEEE 2009

Two points $P$ and $Q$ are maintained at the potentials of $10$ $V$ and $-4$ $V$, respectively. The work done in moving $100$ electrons from $P$ to $Q$ is :
A
$9.60 \times {10^{ - 17}}J$
B
$- 2.24 \times {10^{ - 16}}J$
C
$2.24 \times {10^{ - 16}}J$
D
$- 9.60 \times {10^{ - 17}}J$

Explanation

${{{W_{PQ}}} \over q} = \left( {{V_Q} - {V_P}} \right)$

$\Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)$

$= \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)$

$= + 2.24 \times {10^{ - 16}}J$