 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2004

A current $i$ ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is
A
${{{\mu _0}} \over {4\pi }},{{2i} \over r}$ tesla
B
zero
C
infinite
D
${{2i} \over r}$ tesla

Explanation

Using Ampere's law at a distance $r$ from axis, $B$ is same from symmetry.

$\int {B.dl = {\mu _0}i}$

i.e., $B \times 2\pi r = {\mu _0}i$

Here $i$ is zero, for $r < R,$ whereas $R$ is the radius

$\therefore$ $B=0$
2

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.
3

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. 4

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

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

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

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

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

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