 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline)

A metallic rod of length $'\ell '$ is tied to a string of length $2$$\ell$ and made to rotate with angular speed $w$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field $'B'$ in the region, the $e.m.f$ induced across the ends of the rod is A
${{2B\omega \ell } \over 2}$
B
${{3B\omega \ell } \over 2}$
C
${{4B\omega {\ell ^2}} \over 2}$
D
${{5B\omega {\ell ^2}} \over 2}$

Explanation

Here, induced $e.m.f.$ $e = \int\limits_{2\ell }^{3\ell } {\left( {\omega x} \right)Bdx = B\omega } {{\left[ {{{\left( {3\ell } \right)}^2} - {{\left( {2\ell } \right)}^2}} \right]} \over 2}$

$= {{5B{\ell ^2}\omega } \over 2}$
2

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
3

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.
4

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

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

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

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