 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

A thin rectangular magnet suspended freely has a period of oscillation equal to $T.$ Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T',$ the ratio ${{T'} \over T}$ is
A
${1 \over {2\sqrt 2 }}$
B
${1 \over 2}$
C
$2$
D
${1 \over 4}$

Explanation

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

$T = 2\pi \sqrt {{I \over {\mu {B_H}}}}$

where $I=$ Moment of inertia of the rectangular magnet

$\mu =$ Magnetic moment

${B_H} =$ Horizontal component of the earth's magnetic field

Case 1 : $T = 2\pi \sqrt {{I \over {\mu {B_H}}}}$ where $I = {1 \over {12}}M{\ell ^2}$

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

$T' = 2\pi \sqrt {{{I'} \over {\mu '{B_H}}}}$

where $I' = {1 \over {12}}\left( {{M \over 2}} \right){\left( {{\ell \over 2}} \right)^2} = {I \over 8}$ and $\mu ' = {\mu \over 2}$

$\therefore$ ${{T'} \over T} = \sqrt {{{I'} \over {\mu '}} \times {\mu \over I}}$

$= \sqrt {{{I/8} \over {\mu /2}} \times {\mu \over I}}$

$= \sqrt {{1 \over 4}} = {1 \over 2}$
2

AIEEE 2002

Wires $1$ and $2$ carrying currents $i{}_1$ and $i{}_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$ (as shown in figure) due to the magnetic field of wire $1$? A
${{{\mu _0}} \over {2\pi r}}{i_1}{i_2}\,dl\,\tan \,\theta$
B
${{{\mu _0}} \over {2\pi r}}{i_1}{i_2}\,dl\,\sin \,\theta$
C
${{{\mu _0}} \over {2\pi r}}{i_1}{i_2}\,dl\,\cos \,\theta$
D
${{{\mu _0}} \over {4\pi r}}{i_1}{i_2}\,dl\,\sin \,\theta$

Explanation

Magnetic field due to current in wire $1$ at point $P$ distant $r$ from the wire is $B = {{{\mu _0}} \over {4\pi }}{{{i_1}} \over r}\left[ {\cos \theta + \cos \theta } \right]$

$B = {{{\mu _0}} \over {2\pi }}{{{i_1}\cos \theta } \over r}$ (directed perpendicular to the plane of paper, inwards)

The force exerted due to this magnetic field on current element ${i_2}\,dl$ is

$dF = {i_2}\,dl\,B\,\sin \,{90^ \circ }$

$\therefore$ $dF = {i_2}dl\left[ {{{{\mu _0}} \over {2\pi }}{{{i_1}\cos \theta } \over r}} \right]$

$= {{{\mu _0}} \over {2\pi r}}{i_1}\,{i_2}dl\,\cos \,\theta$
3

AIEEE 2002

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its
A
speed
B
mass
C
charge
D
magnetic induction

Explanation

KEY CONCEPT : The time period of a charged particle

$\left( {m,q} \right)$ moving in a magnetic field $(B)$ is $T = {{2\pi m} \over {qB}}$

The time period does not depend on the speed of the particle.
4

AIEEE 2002

If an electron and a proton having same momentum enter perpendicular to a magnetic field, then
A
curved path of electron and proton will be same (ignoring the sense of revolution)
B
they will move undeflected
C
curved path of electron is more curved than that of the proton
D
path of proton is more curved.

Explanation

KEY CONCEPT : When a charged particle enters perpendicular to a magnetic field,

then it moves in a circular path of radius.

$r = {p \over {qB}}$

where $q=$ Charge of the particle

$p=$ Momentum of the particle

$B=$ Magnetic field

Here $p,q$ and $B$ are constant for electron and proton, therefore the radius will be

same.