### JEE Mains Previous Years Questions with Solutions

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1

### 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

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

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

$= 2.4\pi \times {10^{ - 4}}H$
2

### 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)

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

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

$= \left( {1 - {e^{ - 1}}} \right)$
3

### 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 $\left. {\,\,\cos {\pi \over 2} = 0\,\,} \right)$
4

### 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

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

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

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