### JEE Mains Previous Years Questions with Solutions

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1

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

### AIEEE 2006

The $rms$ value of the electric field of the light coming from the Sun is $720$ $N/C.$ The average total energy density of the electromagnetic wave is
A
$4.58 \times {10^{ - 6}}\,J/{m^3}$
B
$6.37 \times {10^{ - 9}}\,J/{m^3}$
C
$81.35 \times {10^{ - 12}}\,J/{m^3}$
D
$3.3 \times {10^{ - 3}}\,J/{m^3}$

## Explanation

${E_{rms}} = 720$

The average total energy density

$= {1 \over 2}{ \in _0}\,E_0^2 = {1 \over 2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}\,E_{rms}^2$

$= 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2}$

$= 4.58 \times {10^{ - 6}}\,J/{m^3}$
3

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

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

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

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

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

$\Rightarrow i = {E \over R}{e^{{R \over L}t}} = {{100} \over {100}}{e^{{{ - 100 \times {{10}^{ - 3}}} \over {100 \times {{10}^{ - 3}}}}}} = {1 \over e}$
4

### 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

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

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

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

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