 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

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

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

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

AIEEE 2008

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area $A=$ $10\,\,c{m^2}$ and length $=20$ $cm$ . If one of the solenoid has $300$ turns and the other $400$ turns, their mutual inductance is
$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,Tm\,{A^{ - 1}}} \right)$
A
$2.4\pi \times {10^{ - 5}}H$
B
$4.8\pi \times {10^{ - 4}}H$
C
$4.8\pi \times {10^{ - 5}}H$
D
$2.4\pi \times {10^{ - 4}}H$

Explanation

$M = {{{\mu _0}{N_1}{N_2}A} \over \ell }$

$= {{4\pi \times {{10}^{ - 7}} \times 300 \times 400 \times 100 \times {{10}^{ - 4}}} \over {0.2}}$

$= 2.4\pi \times {10^{ - 4}}H$
3

AIEEE 2007

An ideal coil of $10H$ is connected in series with a resistance of $5\Omega$ and a battery of $5V$. $2$ second after the connection is made, the current flowing in ampere in the circuit is
A
$\left( {1 - {e^{ - 1}}} \right)$
B
$\left( {1 - e} \right)$
C
$e$
D
${{e^{ - 1}}}$

Explanation

KEY CONCEPT : $I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$

(When current is in growth in $LR$ circuit)

$= {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)$

$= {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)$

$= \left( {1 - {e^{ - 1}}} \right)$
4

AIEEE 2007

In an $a.c.$ circuit the voltage applied is $E = {E_0}\,\sin \,\omega t.$ The resulting current in the circuit is $I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).$ The power consumption in the circuit is given by
A
$P = \sqrt 2 {E_0}{I_0}$
B
$P = {{{E_0}{I_0}} \over {\sqrt 2 }}$
C
$P=zero$
D
$P = {{{E_0}{I_0}} \over 2}$

Explanation

KEY CONCEPT : We know that power consumed in a.c. circuit is given by,

$P = {E_{rms}}{I_{rms}}\cos \phi$

Here, $E = {E_0}\sin \omega t$

$I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right)$

which implies that the phase difference, $\phi = {\pi \over 2}$

$\therefore$ $P = {E_{rms}}.{I_{rms}}.\cos {\pi \over 2} = 0$

$\left( {\,\,} \right.$ as $\left. {\,\,\cos {\pi \over 2} = 0\,\,} \right)$