### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

The work done in placing a charge of $8 \times {10^{ - 18}}$ coulomb on a condenser of capacity $100$ micro-farad is
A
$16 \times {10^{ - 32}}\,\,joule$
B
$3.1 \times {10^{ - 26}}\,\,joule$
C
$4 \times {10^{ - 10}}\,\,joule$
D
$32 \times {10^{ - 32}}\,\,joule$

## Explanation

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by

$U = {1 \over 2}{{{Q^2}} \over C}$

$= {1 \over 2} \times {{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}} \over {100 \times {{10}^{ - 6}}}}$

$= 32 \times {10^{ - 32}}J$
2

### AIEEE 2003

A thin spherical conducting shell of radius $R$ has a charge $q.$ Another charge $Q$ is placed at the center of the shell. The electrostatic potential at a point $P$ a distance ${R \over 2}$ from the center of the shell is
A
${{2Q} \over {4\pi {\varepsilon _0}R}}$
B
${{2Q} \over {4\pi {\varepsilon _0}R}} - {{2q} \over {4\pi {\varepsilon _0}R}}$
C
${{2Q} \over {4\pi {\varepsilon _0}R}} + {q \over {4\pi {\varepsilon _0}R}}$
D
${{\left( {q + Q} \right)2} \over {4\pi {\varepsilon _0}R}}$

## Explanation

Electric potential due to charge $Q$ placed at the center of spherical shell at point $P$ is

${V_1} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {R/2}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over R}$

Electric potential due to charge $q$ on the surface of the spherical shell at any point inside the shell is

${V_2} = {1 \over {4\pi {\varepsilon _0}}}{q \over R}$
3

### AIEEE 2003

If the electric flux entering and leaving an enclosed surface respectively is ${\phi _1}$ and ${\phi _2},$ the electric charge inside the surface will be
A
$\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}$
B
$\left( {{\phi _2} + {\phi _1}} \right)/{\varepsilon _0}$
C
$\left( {{\phi _2} - {\phi _1}} \right)/{\varepsilon _0}$
D
$\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}$

## Explanation

The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface $= {\phi _2} - {\phi _1}$

$\therefore$ the change enclosed in the surface by Gauss's law is $q = { \varepsilon _0}\,\left( {{\phi _2} - {\phi _1}} \right)$
4

### AIEEE 2003

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor
A
decreases
B
remains unchanged
C
becomes infinite
D
increases

## Explanation

The capacitancce of parallel plate capacitor in which a metal plate of thickness $t$ is inserted is given by

$C = {{{\varepsilon _0}A} \over {d - t}}.\,\,\,\,\,$

Here $t \to 0\,\,\,\,\,\,$ $\therefore$ $C = {{{\varepsilon _0}A} \over d}$