### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

Curie temperature is the temperature above which
A
a ferromagnetic material becomes paramagnetic
B
a paramagnetic material becomes diamagnetic
C
a ferromagnetic material becomes diamagnetic
D
a paramagnetic material becomes ferromagnetic

## Explanation

KEY CONCEPT : The temperature above which a ferromagnetic substance becomes para-magnetic is called Curie's temperature.
2

### AIEEE 2003

The magnetic lines of force inside a bar magnet
A
are from north-pole to south-pole of the magnet
B
do not exist
C
depend upon the area of cross-section of the bar magnet
D
are from south-pole to north-pole of the Magnet

## Explanation

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.

3

### AIEEE 2003

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through ${60^ \circ }.$ The torque needed to maintain the needle in this position will be
A
$\sqrt 3 \,W$
B
$W$
C
${{\sqrt 3 } \over 2}W$
D
$2W$

## Explanation

$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$

$= MB\left( {\cos {\theta ^ \circ } - \cos {{60}^ \circ }} \right)$

$= MB\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}$

$\therefore$ $\tau = MB\,\sin \theta = MB\,\sin \,{60^ \circ }$

$= \sqrt 3 {{MB} \over 2} = \sqrt 3 W$
4

### AIEEE 2003

A particle of charge $- 16 \times {10^{ - 18}}$ coulomb moving with velocity $10m{s^{ - 1}}$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis, and an electric field of magnitude ${10^4}V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis, the magnitude of $B$ is
A
${10^3}Wb/{m^2}$
B
${10^5}Wb/{m^2}$
C
${10^{16}}Wb/{m^2}$
D
${10^{ - 3}}Wb/{m^2}$

## Explanation

The situation is shown in the figure.

${F_E} =$ Force due to electric field

${F_B} =$ Force due to magnetic field

It is given that the charged particle remains moving along $X$-axis (i.e. undeviated).

Therefore ${F_B} = {F_E}$

$\Rightarrow qvB = qE$

$\Rightarrow B = {E \over v} = {{{{10}^4}} \over {10}}$

$= {10^3}\,\,weber/{m^2}$