 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

A circuit has a resistance of $12$ $ohm$ and an impedance of $15$ $ohm$. The power factor of the circuit will be
A
$0.4$
B
$0.8$
C
$0.125$
D
$1.25$

Explanation

Power factor $= \cos \phi = {R \over Z} = {{12} \over {15}} = {4 \over 5} = 0.8$
2

AIEEE 2005

A coil of inductance $300$ $mH$ and resistance $2\,\Omega$ is connected to a source of voltage $2$ $V$. The current reaches half of its steady state value in
A
$0.1$ $s$
B
$0.05$ $s$
C
$0.3$ $s$
D
$0.15$ $s$

Explanation

KEY CONCEPT : The charging of inductance given

by, $i = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right)$

${{{i_0}} \over 2} = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right) \Rightarrow {e^{ - {{Rt} \over L}}} = {1 \over 2}$

Taking log on both the sides,

$- {{Rt} \over L} = \log 1 - \log 2$

$\Rightarrow t = {L \over R}\log 2 = {{300 \times {{10}^{ - 3}}} \over 2} \times 0.69$

$\Rightarrow t - 0.1\,\sec .$
3

AIEEE 2005

The self inductance of the motor of an electric fan is $10$ $H$. In order to impart maximum power at $50$ $Hz$, it should be connected to a capacitance of
A
$8\mu F$
B
$4\mu F$
C
$2\mu F$
D
$1\mu F$

Explanation

For maximum power, ${X_L} = X{}_C,$ which yields

$C = {1 \over {{{\left( {2\pi n} \right)}^2}L}} = {1 \over {4{\pi ^2} \times 50 \times 50 \times 10}}$

$\therefore$ $C = 0.1 \times {10^{ - 5}}F = 1\mu F$
4

AIEEE 2004

In a $LCR$ circuit capacitance is changed from $C$ to $2$ $C$. For the resonant frequency to remain unchaged, the inductance should be changed from $L$ to
A
$L/2$
B
$2L$
C
$4L$
D
$L/4$

Explanation

For resonant frequency to remain same $LC$ should be const. $LC=$ const

$\Rightarrow LC = L' \times 2C \Rightarrow L' = {L \over 2}$