 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2006

An inductor $(L=100$ $mH)$, a resistor $\left( {R = 100\,\Omega } \right)$ and a battery $\left( {E = 100V} \right)$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points $A$ and $B$. The current in the circuit $1$ $ms$ after the short circuit is A
$1/eA$
B
$eA$
C
$0.1$ $A$
D
$1$ $A$

Explanation

Initially, when steady state is achieved,

$i = {E \over R}$

Let $E$ is short circuited at a $t=0.$ Then

At $t=0,$ ${i_0} = {E \over R}$

Let during decay of current at any time the current

flowing is $- L{{di} \over {dt}} - iR = 0$

$\Rightarrow {{di} \over i} = - {R \over L}dt$

$\Rightarrow \int\limits_{{i_0}}^i {{{di} \over i}} = \int\limits_0^t { - {R \over L}dt}$

$\Rightarrow {\log _e}{i \over {{i_0}}} = - {R \over L}t$

$\Rightarrow i = {i_0}\,{e^{ - {R \over L}t}}$

$\Rightarrow i = {E \over R}{e^{{R \over L}t}} = {{100} \over {100}}{e^{{{ - 100 \times {{10}^{ - 3}}} \over {100 \times {{10}^{ - 3}}}}}} = {1 \over e}$
2

AIEEE 2006

The flux linked with a coil at any instant $'t'$ is given by
$\phi = 10{t^2} - 50t + 250$
The induced $emf$ at $t=3s$ is
A
$-190$ $V$
B
$-10$ $V$
C
$10$ $V$
D
$190$ $V$

Explanation

$\phi = 10{t^2} - 50t + 250$

$e = - {{d\phi } \over {dt}} = - \left( {20t - 50} \right)$

${e_{t = 3}} = - 10\,V$
3

AIEEE 2006

In an $AC$ generator, a coil with $N$ turns, all of the same area $A$ and total resistance $R,$ rotates with frequency $\omega$ in a magnetic field $B.$ The maximum value of $emf$ generated in the coil is
A
$N.A.B.R.$$\omega B N.A.B C N.A.B.R. D N.A.B.$$\omega$

Explanation

$e = - {{d\phi } \over {dt}} = - {{d\left( {N\overrightarrow B .\overrightarrow A } \right)} \over {dt}}$

$= - N{d \over {dt}}\left( {BA\,\cos \,\omega t} \right) = NBA\omega \sin \,\omega t$

$\Rightarrow {e_{\max }} = NBA\omega$
4

AIEEE 2006

In a series resonant $LCR$ circuit, the voltage across $R$ is $100$ volts and $R = 1\,k\Omega$ with $C = 2\mu F.$ The resonant frequency $\omega$ is $200$ $rad/s$. At resonance the voltage across $L$ is
A
$2.5 \times {10^{ - 2}}V$
B
$40$ $V$
C
$250$ $V$
D
$4 \times {10^{ - 3}}V$

Explanation

Across resistor, $I = {V \over R} = {{100} \over {1000}} = 0.1A$

At resonance,

${X_L} = {X_C} = {1 \over {\omega C}}$

$= {1 \over {200 \times 2 \times {{10}^{ - 6}}}} = 2500$

Voltage across $L$ is

$I{X_L} = 0.1 \times 2500 = 250V$