 JEE Mains Previous Years Questions with Solutions

4.5
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1

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

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

$\therefore$ $C = 0.1 \times {10^{ - 5}}F = 1\mu F$
2

AIEEE 2004

A metal conductor of length $1$ $m$ rotates vertically about one of its ends at angular velocity $5$ radians per second. If the horizontal component of earth's magnetic field is $0.2 \times {10^{ - 4}}T,$ then the $e.m.f.$ developed between the two ends of the conductor is
A
$5mV$
B
$50\mu V$
C
$5\mu V$
D
$50mV$

Explanation

$\ell = 1m,\,\,\omega = 5\,rad/s,\,\,B = 0.2 \times {10^{ - 4}}T$

$\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V$
3

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

$\Rightarrow LC = L' \times 2C \Rightarrow L' = {L \over 2}$
4

AIEEE 2004

In a uniform magnetic field of induction $B$ a wire in the form of a semicircle of radius $r$ rotates about the diameter of the circle with an angular frequency $\omega .$ The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R,$ the mean power generated per period of rotation is
A
${{{{\left( {B\pi r\omega } \right)}^2}} \over {2R}}$
B
${{{{\left( {B\pi {r^2}\omega } \right)}^2}} \over {8R}}$
C
${{B\pi {r^2}\omega } \over {2R}}$
D
${{{{\left( {B\pi r{\omega ^2}} \right)}^2}} \over {8R}}$

Explanation

$\phi = \overrightarrow B .\overrightarrow A ;\phi = BA\cos \,\omega t$

$\varepsilon = - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,$

$i = {{\omega BA} \over R}\sin \,\omega t$

${P_{inst}} = {t^2}R = {\left( {{{\omega BA} \over R}} \right)^2} \times R{\sin ^2}\omega t$

${p_{avg}} = {{\int\limits_0^T {{P_{inst}} \times dt} } \over {\int\limits_0^T {dt} }}$

$= {{{{\left( {\omega BA} \right)}^2}} \over R}{{\int\limits_0^T {{{\sin }^2}\omega tdt} } \over {\int\limits_0^T {dt} }}$

$= {1 \over 2}{{{{\left( {\omega BA} \right)}^2}} \over R}$

$\therefore$ ${P_{abg}} = {{{{\left( {\omega B\pi {r^2}} \right)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} \right]$