## Download our App

### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
(100k+ download)
1

### AIEEE 2004

MCQ (Single Correct Answer)
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

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

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

### AIEEE 2004

MCQ (Single Correct Answer)
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

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

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

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

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

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

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

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

$\therefore$ ${P_{abg}} = {{{{\left( {\omega B\pi {r^2}} \right)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} \right]$
3

### AIEEE 2004

MCQ (Single Correct Answer)
A coil having $n$ turns and resistance $R\Omega$ is connected with a galvanometer of resistance $4R\Omega .$ This combination is moved in time $t$ seconds from a magnetic field ${W_1}$ weber to ${W_2}$ weber. The induced current in the circuit is
A
${{\left( {{W_2} - {W_1}} \right)} \over {Rnt}}$
B
$- {{n\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rt}}$
C
$- {{\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rnt}}$
D
$- {{n\left( {{W_2} - {W_1}} \right)} \over {Rt}}$

## Explanation

${{\Delta \phi } \over {\Delta t}} = {{\left( {{W_2} - {W_1}} \right)} \over t}$

${R_{tot}} = \left( {R + 4R} \right)\Omega = 5R\Omega$

$i = {{nd\phi } \over {{R_{tot}}dt}} = {{ - n\left( {{W_2} - {W_1}} \right)} \over {5Rt}}$

( as ${W_2}\,\,\& \,\,{W_1}$ are magnetic flux )
4

### AIEEE 2004

MCQ (Single Correct Answer)
In an $LCR$ series $a.c.$ circuit, the voltage across each of the components, $L,C$ and $R$ is $50V$. The voltage across the $L.C$ combination will be
A
$100V$
B
$50\sqrt 2$
C
$50$ $V$
D
$0$ $V$ (zero)

## Explanation

Since the phase difference between $L$ & $C$ is $\pi ,$

$\therefore$ net voltage difference across $LC=50-50=0$

### Joint Entrance Examination

JEE Advanced JEE Main

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE ME GATE PI GATE EE GATE CE GATE IN

NEET

Class 12