 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

The length of a given cylindrical wire is increased by $100\%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be
A
$200\%$
B
$100\%$
C
$50\%$
D
$300\%$

Explanation

${R_f} = {n^2}{R_1}$

Here $n=2$ (length becomes twice)

$\therefore$ ${R_f} = 4{R_i}$

New resistance $=400$ of ${R_i}$

$\therefore$ Increase $= 300\%$
2

AIEEE 2003

Three charges $- {q_1}, + {q_2}$ and $- {q_3}$ are placed as shown in the figure. The $x$-component of the force on $- {q_1}$ is proportional to A
${{{q_2}} \over {{b^2}}} - {{{q_3}} \over {{a^2}}}\cos \theta$
B
${{{q_2}} \over {{b^2}}} + {{{q_3}} \over {{a^2}}}\sin \theta$
C
${{{q_2}} \over {{b^2}}} + {{{q_3}} \over {{a^2}}}\cos \theta$
D
${{{q_2}} \over {{b^2}}} - {{{q_3}} \over {{a^2}}}sin\theta$

Explanation Force on charge ${q_1}$ due to ${q_2}$ is ${F_{12}} = k{{{q_1}{q_2}} \over {{b^2}}}$

Force on charge ${q_1}$ due to ${q_3}$ is ${F_{13}} = k{{{q_1}{q_3}} \over {{a^2}}}$

The $X$-component of the force $\left( {{F_x}} \right)$ on

$q{}_1$ is ${F_{12}} + {F_{13}}$ $\sin \theta$

$\therefore$ ${F_x} = k{{{q_1}{q_2}} \over {{b^2}}} + k{{{q_1}{q_2}} \over {{a^2}}}\sin \theta$

$\therefore$ ${F_x} \propto {{{q_2}} \over {{b^2}}} + {{{q_3}} \over {{a^2}}}\sin \theta$
3

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

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

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

$= 32 \times {10^{ - 32}}J$
4

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

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