### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

An inductor $(L=0.03$ $H)$ and a resistor $\left( {R = 0.15\,k\Omega } \right)$ are connected in series to a battery of $15V$ $EMF$ in a circuit shown below. The key ${K_1}$ has been kept closed for a long time. Then at $t=0$, ${K_1}$ is opened and key ${K_2}$ is closed simultaneously. At $t=1$ $ms,$ the current in the circuit will be : $\left( {{e^5} \cong 150} \right)$
A
$6.7$ $mA$
B
$0.67$ $mA$
C
$100$ $mA$
D
$67$ $mA$

## Explanation

${\rm I}\left( 0 \right) = {{15 \times 100} \over {0.15 \times {{10}^3}}} = 0.1A$

${\rm I}\left( \infty \right) = 0$

${\rm I}\left( t \right) = \left[ {{\rm I}\left( 0 \right) - {\rm I}\left( \infty \right)} \right]\,{e^{{{ - t} \over {L/R}}}} + i\left( \infty \right)$

${\rm I}\left( t \right) = 0.1\,{e^{{{ - t} \over {L/R}}}} = 0.1{e^{{R \over L}}}$

${\rm I}\left( t \right) = 0.1\,\,{e^{{{0.15 \times 1000} \over {0.03}}}} = 0.67mA$
2

### JEE Main 2015 (Offline)

An $LCR$ circuit is equivalent to a damped pendulum. In an $LCR$ circuit the capacitor is charged to ${Q_0}$ and then connected to the $L$ and $R$ as shown below :

If a student plots graphs of the square of maximum charge $\left( {Q_{Max}^2} \right)$ on the capacitor with time$(t)$ for two different values ${L_1}$ and ${L_2}$ $\left( {{L_1} > {L_2}} \right)$ of $L$ then which of the following represents this graph correctly ?
$\left( {plots\,\,are\,\,schematic\,\,and\,\,niot\,\,drawn\,\,to\,\,scale} \right)$

A
B
C
D

## Explanation

From $KVL$ at any time $t$

${q \over c} - iR - L{{di} \over {dt}} = 0$

$i = - {{dq} \over {dt}} \Rightarrow {q \over c} + {{dq} \over {dt}}R + {{L{d^2}q} \over {d{t^2}}} = 0$

${{{d^2}q} \over {d{t^2}}} + {R \over L}{{dq} \over {dt}} + {q \over {Lc}} = 0$

From damped harmonic oscillator, the amplitude is

given by $A = {A_0}e - {{dt} \over {2m}}$

Double differential equation

${{{d^2}x} \over {d{t^2}}} + {b \over m}{{dx} \over {dt}} + {k \over m}x = 0$

${Q_{\max }} = {Q_0}e - {{Rt} \over {2L}} \Rightarrow Q_{\max }^2 = Q_0^2e - {{Rt} \over L}$

Hence damping will be faster for lesser self inductance.
3

### JEE Main 2015 (Offline)

An arc lamp requires a direct current of $10$ $A$ at $80$ $V$ to function. If it is connected to a $220$ $V$ $(rms),$ $50$ $Hz$ $AC$ supply, the series inductor needed for it to work is close to :
A
$0.044$ $H$
B
$0.065$ $H$
C
$80$ $H$
D
$0.08$ $H$

## Explanation

Here

$i = {e \over {\sqrt {{R^2} + X_L^2} }}$

$= {e \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}$

$= {e \over {\sqrt {{R^2} + 4{\pi ^2}{v^2}{L^2}} }}$

$10 = {{220} \over {\sqrt {64 + 4{\pi ^2}{{\left( {50} \right)}^2}L} }}$

$\left[ {\,\,\,} \right.$ as $\left. {R = {V \over {\rm I}} = {{80} \over {10}} = 8\,\,\,} \right]$

On solving we get

$L = 0.065\,H$
4

### JEE Main 2015 (Offline)

Two coaxial solenoids of different radius carry current $I$ in the same direction. $\overrightarrow {{F_1}}$ be the magnetic force on the inner solenoid due to the outer one and $\overrightarrow {{F_2}}$ be the magnetic force on the outer solenoid due to the inner one. Then :
A
$\overrightarrow {{F_1}}$ is radially in wards and $\overrightarrow {{F_2}} = 0$
B
$\overrightarrow {{F_1}}$ is radially outwards and $\overrightarrow {{F_2}} = 0$
C
$\overrightarrow {{F_1}} = \overrightarrow {{F_2}} = 0$
D
$\overrightarrow {{F_1}}$ is radially inwards and $\overrightarrow {{F_2}}$ is radially outards

## Explanation

$\mathop {F{}_1}\limits^ \to = \mathop {F{}_2}\limits^ \to = 0$

because of action and reaction pair