### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2010

In the circuit shown below, the key $K$ is closed at $t=0.$ The current through the battery is
A
${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$ at $t=0$ and ${V \over {{R_2}}}$ at $t = \infty$
B
${V \over {{R_2}}}$ at $\,t = 0$ and ${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$ at $t = \infty$
C
${V \over {{R_2}}}$ at $\,t = 0$ and ${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$ at $t = \infty$
D
${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$ at $t=0$ and ${V \over {{R_2}}}$ at $t = \infty$

## Explanation

At $t=0,$ no current will flow through $L$ and ${R_1}$

$\therefore$ Current through battery $= {V \over {{R_2}}}$

At $t = \infty ,$

effective resistance, ${{\mathop{\rm R}\nolimits} _{eff}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$

$\therefore$ Current through battery $= {V \over {{{\mathop{\rm R}\nolimits} _{eff}}}} = {{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$
2

### AIEEE 2010

A rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R$ $\Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents ${I_1},{I_2}$ and $I$ are
A
${I_1} = - {I_2} = {{Blv} \over {6R}},\,\,I = {{2Blv} \over {6R}}$
B
${I_1} = {I_2} = {{Blv} \over {3R}},\,\,I = {{2Blv} \over {3R}}$
C
${I_1} = {I_2} = I = {{Blv} \over R}$
D
${I_1} = {I_2} = {{Blv} \over {6R}},I = {{Blv} \over {3R}}$

## Explanation

Due to the movement of resistor $R,$ an $emf$ equal to $Blv$ will be induced in it as shown in figure clearly,

$I = {I_1} + {I_2}$
<
Also, ${I_1} = {I_2}$

Solving the circuit, we get

${I_1} = {I_2} = {{Blv} \over {3R}}$

and $I = 2{I_1} = {{2Blv} \over {3R}}$
3

### AIEEE 2010

In a series $LCR$ circuit $R = 200\Omega$ and the voltage and the frequency of the main supply is $220V$ and $50$ $Hz$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by ${30^ \circ }.$ On taking out the inductor from the circuit the current leads the voltage by ${30^ \circ }.$ The power dissipated in the $LCR$ circuit is
A
$305$ $W$
B
$210$ $W$
C
$zero$ $W$
D
$242$ $W$

## Explanation

When capacitance is taken out, the circular is $LR.$

$\therefore$ $\tan \phi = {{\omega L} \over R}$

$\Rightarrow \omega L = R\,\tan \phi$

$= 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$

Again, when inductor is taken out, the circuit is $CR.$

$\therefore$ $\tan \phi = {1 \over {\omega CR}}$

$\Rightarrow {1 \over {\omega c}} = R\tan \phi$

$= 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}$

Now, $Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}}$

$= \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega$

Power dissipated $= {V_{rms}}{I_{rms}}\cos \phi$

$= {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}$

$\left( {\,\,\,} \right.$ as $\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)$

$= {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}$

$= {{220 \times 220} \over {200}} = 242\,W$
4

### AIEEE 2009

An inductor of inductance $L=400$ $mH$ and resistors of resistance ${R_1} = 2\Omega$ and ${R_2} = 2\Omega$ are connected to a battery of $emf$ $12$ $V$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0.$ The potential drop across $L$ as a function of time is :
A
${{12} \over t}{e^{ - 3t}}V$
B
$6\left( {1 - {e^{ - t/0.2}}} \right)V$
C
$12{e^{ - 5t}}V$
D
$6{e^{ - 5t}}V$

## Explanation

Growth in current in $L{R_2}$ branch when switch is closed is given by

$i = {E \over {{R_2}}}\left[ {1 - {e^{ - {R_2}t/L}}} \right]$

$\Rightarrow {{di} \over {dt}} = {E \over {{R_2}}}.{{{R_2}} \over L}.{e^{ - {R_{2t/L}}}}$

$= {E \over L}{e^{{{{R_2}t} \over L}}}$

Hence, potential drop across

$L = \left( {{E \over L}{e^{ - {R_2}t/L}}} \right)L = E{e^{ - {R_2}t/L}}$

$= 12{e^{ - {{2t} \over {400 \times {{10}^{ - 3}}}}}} = 12{e^{ - 5t}}V$