JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2005

A charged ball $B$ hangs from a silk thread $S,$ which makes angle $\theta$ with a large charged conducting sheet $P,$ as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to f
A
$\cot \,\theta$
B
$\cos \,\theta$
C
$\tan \,\theta$
D
$\sin \,\theta$

Explanation

$T\sin \theta = {\sigma \over {{\varepsilon _0}K}}.q\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$T\cos \theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Dividing $(i)$ by $(ii),$

$\tan \theta = {{\sigma q} \over {{\varepsilon _0}K.mg}}$
$\therefore$ $\sigma \propto \,\tan \theta$
2

AIEEE 2005

Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+q$ and $-q.$ The potential difference between the centres of the two rings is
A
${q \over {2\pi \,{ \in _0}}}\left[ {{1 \over R} - {1 \over {\sqrt {{R^2} + {d^2}} }}} \right]$
B
${{qR} \over {4\pi \,{ \in _0}\,{d^2}}}$
C
${q \over {4\pi \,{ \in _0}}}\left[ {{1 \over R} - {1 \over {\sqrt {{R^2} + {d^2}} }}} \right]$
D
zero

Explanation

${V_A} = {V_{self}} + {V_{due}}$ to $(2)$

$\Rightarrow {V_A} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{q \over R} - {q \over {\sqrt {{R^2} + {d^2}} }}} \right]$

${V_B} = {V_{self}} + {V_{due}}$ to $(1)$

$\Rightarrow {V_B} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{ - q} \over R} + {q \over {\sqrt {{R^2} + {d^2}} }}} \right]$

$\Delta V = {V_A} - {V_B}$
$= {1 \over {4\pi {\varepsilon _0}}}\left[ {{q \over R} + {q \over R} - {q \over {\sqrt {{R^2} + {d^2}} }} - {q \over {\sqrt {{R^2} + {d^2}} }}} \right]$

$= {1 \over {2\pi {\varepsilon _0}}}\left[ {{q \over R} - {q \over {\sqrt {{R^2} + {d^2}} }}} \right]$
3

AIEEE 2005

A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $'C'$ then the resultant capacitance is
A
$\left( {n + 1} \right)C$
B
$\left( {n - 1} \right)C$
C
$nC$
D
$C$

Explanation

As $n$ plates are joined, it means $(n-1)$ capacitor joined in parallel.

$\therefore$ resultant capacitance $=(n-1)C$
4

AIEEE 2005

Two point charges $+8q$ and $-2q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$ axis at which the net electric field due to these two point charges is zero is
A
${L \over 4}$
B
$2$ $L$
C
$4$ $L$
D
$8$ $L$

Explanation

${{ - K2q} \over {{{\left( {x - L} \right)}^2}}} + {{K8q} \over {{x^2}}} = 0 \Rightarrow {1 \over {{{\left( {x - L} \right)}^2}}} = {4 \over {{x^2}}}$

or, ${1 \over {x - L}} = {2 \over x} \Rightarrow x = 2x - 2L$

or, $x=2L$