## Download our App

### JEE Mains Previous Years Questions with Solutions

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

### AIEEE 2003

MCQ (Single Correct Answer)
In an oscillating $LC$ circuit the maximum charge on the capacitor is $Q$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is
A
${Q \over 2}$
B
${Q \over {\sqrt 3 }}$
C
${Q \over {\sqrt 2 }}$
D
$Q$

## Explanation

When the capacitor is completely charged, the total energy in the $L.C$ circuit is with the capacitor and that energy is $E = {1 \over 2}{{{Q^2}} \over C}$

When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get ${E \over 2} = {1 \over 2}{{Q{'^2}} \over C}$ where $Q'$ is the charge on plate of the capacitor

$\therefore$ ${1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}$

$\Rightarrow Q' = {Q \over {\sqrt 2 }}$
2

### AIEEE 2003

MCQ (Single Correct Answer)
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon
A
the rates at which currents are changing in the two coils
B
relative position and orientation of the two coils
C
the currents in the two coils
D
the materials of the wires of the coils

## Explanation

Mutual conductance depends on the relative position and orientation of the two coils.
3

### AIEEE 2002

MCQ (Single Correct Answer)
Which of the following are not electromagnetic waves?
A
cosmic rays
B
gamma rays
C
$\beta$-rays
D
$X$-rays

## Explanation

$\beta$ -rays are fast moving beam of electrons.
4

### AIEEE 2002

MCQ (Single Correct Answer)
A conducting square loop of side $L$ and resistance $R$ moves in its plane with a uniform velocity $v$ perpendicular to one of its sides. A magnetic induction $B$ constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced $emf$ is
A
zero
B
$RvB$
C
$vBL/R$
D
$vBL$

## Explanation

The induced $emf$ is

$e = {{ - d\phi } \over {dt}} = - {{d\left( {\overrightarrow B .\overrightarrow A } \right)} \over {dt}}$

$= {{ - d\left( {BA\cos {0^ \circ }} \right)} \over {dt}}$

$\therefore$ $e = - B{{dA} \over {dt}} = - B{{d\left( {\ell \times x} \right)} \over {dt}}$

$= - B\ell {{dx} \over {dt}} = - B\ell v$

### 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