 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline)

A circular loop of radius $0.3$ $cm$ lies center of the small loop is on the axis of the bigger loop. The distance between their centers is $15$ $cm.$ If a current of $2.0$ $A$ flows through the smaller loop, than the flux linked with bigger loop is
A
$9.1 \times {10^{ - 11}}\,$ weber
B
$6 \times {10^{ - 11}}\,$ weber
C
$3.3 \times {10^{ - 11}}\,$ weber
D
$6.6 \times {10^{ - 9}}\,$ weber

Explanation

As we know, Magnetic flux, $\phi = B.A$

${{{\mu _0}\left( 2 \right){{\left( {20 \times {{10}^{ - 2}}} \right)}^2}} \over {2\left[ {{{\left( {0.2} \right)}^2} + {{\left( {0.15} \right)}^2}} \right]}} \times \pi {\left( {0.3 \times {{10}^{ - 2}}} \right)^2}$

On solving

$= 9.216 \times {10^{ - 11}} = 9.2 \times {10^{ - 11}}\,\,$ weber
2

AIEEE 2012

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :
A
development of air current when the plate is placed
B
induction of electrical charge on the plate
C
shielding of magnetic lines of force as aluminium is a para-magnetic material.
D
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Explanation

Because of the Lenz's law of conservation of energy.
3

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

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

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

$\Rightarrow R = 2.71 \times {10^6}\Omega$
4

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

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