 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2009

Let $P\left( r \right) = {Q \over {\pi {R^4}}}r$ be the change density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $'p'$ inside the sphere at distance ${r_1}$ from the center of the sphere, the magnitude of electric field is :
A
${Q \over {4\pi \,{ \in _0}\,r_1^2}}$
B
${{Qr_1^2} \over {4\pi \,{ \in _0}\,{R^4}}}$
C
${{Qr_1^2} \over {3\pi \,{ \in _0}\,{R^4}}}$
D
$0$

Explanation Let us consider a spherical shell of thickness $dx$ and radius $x.$ The volume of this spherical shell $= 4\pi {r^2}dr.$

The charge enclosed within shell

$= {{{Q_r}} \over {\pi {R^4}}}\left[ {4\pi {r^2}dr} \right]$

The charge enclosed in a sphere of radius ${r_1}$ is

${{4Q} \over {{R^4}}}\int\limits_0^{{r_1}} {{r^3}} dr$

$= {{4Q} \over {{R^4}}}\left[ {{{{r^4}} \over 4}} \right]_0^{{r_1}}$

$= {Q \over {{R^4}}}r_1^4$

$\therefore$ The electric field at point $p$ inside the sphere at a distance ${r_1}$ from the center of the sphere is

$E = {1 \over {4\pi { \in _0}}}{{\left[ {{Q \over {{R^4}}}r_1^4} \right]} \over {r_1^2}}$

$= {1 \over {4\pi { \in _0}}}{Q \over {{R^4}}}r_1^2$
2

AIEEE 2008

A thin spherical shell of radius $R$ has charge $Q$ spread uniformly over its surface. Which of the following graphs most closely represents the electric field $E(r)$ produced by the shell in the range $0 \le r < \infty ,$ where $r$ is the distance from the center of the shell?
A B c
C D Explanation

The electric field inside a thin spherical shell of radius $R$ has charge $Q$ spread uniformly over its surface is zero. Outside the shell the electric field is $E = k{Q \over {{r^2}}}.$ These characteristics are represented by graph $(a)$.
3

AIEEE 2008

A parallel plate capacitor with air between the plates has capacitance of $9$ $pF.$ The separation between its plates is $'d'.$ The space between the plates has dielectric constant ${k_1}$ $=3$ and thickness ${d \over 3}$ while the other one has dielectric constant ${k_2} = 6$ and thickness ${{2d} \over 3}$. Capacitance of the capacitor is now
A
$1.8$ $pF$
B
$45$ $pF$
C
$40.5$ $pF$
D
$20.25$ $pF$

Explanation The given capacitance is equal to two capacitances connected in series where

${C_1} = {{{k_1}{ \in _0}A} \over {d/3}} = {{3{k_1}{ \in _0}A} \over d}$

$= {{3 \times 3{ \in _0}A} \over d} = {{9{ \in _0}A} \over d}$

and

${C_2} = {{{k_2}{ \in _0}A} \over {2d/3}} = {{3{k_2}{ \in _0}A} \over {2d}}$

$= {{3 \times 6{ \in _0}A} \over {2d}} = {{9{ \in _0}A} \over d}$

The equivalent capacitance ${C_{eq}}$ is

${1 \over {C{}_{eq}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$

$= {d \over {9{ \in _0}A}} + {d \over {9{ \in _0}A}}$

$= {{2d} \over {9{ \in _0}A}}$

$\therefore$ ${C_{eq}} = {9 \over 2}{{{\varepsilon _0}A} \over d} = {9 \over 2} \times 9pF = 40.5pF$
4

AIEEE 2007

A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is
A
zero
B
${1 \over 2}\,\left( {K - 1} \right)\,C{V^2}$
C
${{C{V^2}\left( {K - 1} \right)} \over K}$
D
$\left( {K - 1} \right)\,C{V^2}$

Explanation

The potential energy of a charged capacitor before removing the dielectric slat is $U = {{{Q^2}} \over {2C}}.$

The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is $U = {{{Q^2}} \over {2C}}$

There is no change in potential energy, therefore work done is zero.