 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2011

A resistor $'R'$ and $2\mu F$ capacitor in series is connected through a switch to $200$ $V$ direct supply. Across the capacitor is a neon bulb that lights up at $120$ $V.$ Calculate the value of $R$ to make the bulb light up $5$ $s$ after the switch has been closed. $\left( {{{\log }_{10}}2.5 = 0.4} \right)$
A
$1.7 \times {10^5}\,\Omega$
B
$2.7 \times {10^6}\,\Omega$
C
$3.3 \times {10^7}\,\Omega$
D
$1.3 \times {10^4}\,\Omega$

Explanation

We have, $V = {V_0}\left( {1 - {e^{ - t/RC}}} \right)$

$\Rightarrow 120 - 200\left( {1 - {e^{ - t/RC}}} \right)$

$\Rightarrow t = RC\,in\,\left( {2.5} \right)$

$\Rightarrow R = 2.71 \times {10^6}\Omega$
2

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

$= 15 \times {10^{ - 5}} = 0.15\,mV$
3

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 $t = {\pi \over 4}\sqrt {LC}$
4

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