### JEE Mains Previous Years Questions with Solutions

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### AIEEE 2011

A boat is moving due east in a region where the earth's magnetic fields is $5.0 \times {10^{ - 5}}$ $N{A^{ - 1}}\,{m^{ - 1}}$ due north and horizontal. The best carries a vertical aerial $2$ $m$ long. If the speed of the boat is $1.50\,m{s^{ - 1}},$ the magnitude of the induced $emf$ in the wire of aerial is :
A
$0.75$ $mV$
B
$0.50$ $mV$
C
$0.15$ $mV$
D
$1$ $mV$

## Explanation

Induced $emf$ $= v{B_H}l = 1.5 \times 5 \times {10^{ - 5}} \times 2$

$= 15 \times {10^{ - 5}} = 0.15\,mV$
2

### AIEEE 2011

A fully charged capacitor $C$ with initial charge ${q_0}$ is connected to a coil of self inductance $L$ at $t=0.$ The time at which the energy is stored equally between the electric and the magnetic fields is :
A
${\pi \over 4}\sqrt {LC}$
B
$2\pi \sqrt {LC}$
C
$\sqrt {LC}$
D
$\pi \sqrt {LC}$

## Explanation

Energy stored in magnetic field $= {1 \over 2}L{i^2}$

Energy stored in electric field $= {1 \over 2}{{{q^2}} \over C}$

$\therefore$ ${1 \over 2}L{i^2} = {1 \over 2}{{{q^2}} \over C}$

Also $q = {q_0}\,\cos \,\omega t$ and $\omega = {1 \over {\sqrt {LC} }}$

On solving $t = {\pi \over 4}\sqrt {LC}$
3

### AIEEE 2010

In the circuit shown below, the key $K$ is closed at $t=0.$ The current through the battery is
A
${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$ at $t=0$ and ${V \over {{R_2}}}$ at $t = \infty$
B
${V \over {{R_2}}}$ at $\,t = 0$ and ${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$ at $t = \infty$
C
${V \over {{R_2}}}$ at $\,t = 0$ and ${{V{R_1}{R_2}} \over {\sqrt {R_1^2 + R_2^2} }}$ at $t = \infty$
D
${{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$ at $t=0$ and ${V \over {{R_2}}}$ at $t = \infty$

## Explanation

At $t=0,$ no current will flow through $L$ and ${R_1}$

$\therefore$ Current through battery $= {V \over {{R_2}}}$

At $t = \infty ,$

effective resistance, ${{\mathop{\rm R}\nolimits} _{eff}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$

$\therefore$ Current through battery $= {V \over {{{\mathop{\rm R}\nolimits} _{eff}}}} = {{V\left( {{R_1} + {R_2}} \right)} \over {{R_1}{R_2}}}$
4

### AIEEE 2010

A rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R$ $\Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents ${I_1},{I_2}$ and $I$ are
A
${I_1} = - {I_2} = {{Blv} \over {6R}},\,\,I = {{2Blv} \over {6R}}$
B
${I_1} = {I_2} = {{Blv} \over {3R}},\,\,I = {{2Blv} \over {3R}}$
C
${I_1} = {I_2} = I = {{Blv} \over R}$
D
${I_1} = {I_2} = {{Blv} \over {6R}},I = {{Blv} \over {3R}}$

## Explanation

Due to the movement of resistor $R,$ an $emf$ equal to $Blv$ will be induced in it as shown in figure clearly,

$I = {I_1} + {I_2}$
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Also, ${I_1} = {I_2}$

Solving the circuit, we get

${I_1} = {I_2} = {{Blv} \over {3R}}$

and $I = 2{I_1} = {{2Blv} \over {3R}}$