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### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Two long current carrying thin wires, both with current $I,$ are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $'\theta '$ with the vertical. If wires have mass $\lambda$ per unit-length then the value of $I$ is :
($g=$ $gravitational$ $acceleration$ )
A
$2\sqrt {{{\pi gL} \over {{\mu _0}}}\tan \theta }$
B
$\sqrt {{{\pi \lambda gL} \over {{\mu _0}}}\tan \theta }$
C
$\sin \theta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos \theta }}}$
D
$2\sin \theta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos \theta }}}$

## Explanation

Let us consider $'\ell '$ length of current carrying wire,

At equilibrium

$T\cos \theta = \lambda g\ell$

and $T\sin \theta = {{{\mu _0}} \over {2\pi }}{{I \times Il} \over {2L\sin \theta }}$

$\left[ {\,\,} \right.$ as $\left. {{{{F_B}} \over \ell } = {{{\mu _0}} \over {4\pi }}{{2I \times I} \over {2\ell \sin \theta }}\,\,} \right]$

Therefore, $I = 2\sin \theta \sqrt {{{\pi \lambda gL} \over {{u_0}\cos \theta }}}$
2

### JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
A conductor lies along the $z$-axis at $- 1.5 \le z < 1.5\,m$ and carries a fixed current of $10.0$ $A$ in $- {\widehat a_z}$ direction (see figure). For a field $\overrightarrow B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}\,\,{\widehat a_y}\,\,T,$ find the power required to move the conductor at constant speed to $x=2.0$ $m$, $y=0$ $m$ in $5 \times {10^{ - 3}}s.$ Assume parallel motion along the $x$-axis.
A
$1.57W$
B
$2.97W$
C
$14.85$ $W$
D
$29.7W$

## Explanation

Work done in moving the conductor is,

$W = \int_0^2 {Fdx}$

$= \int_0^2 {3.0 \times {{10}^{ - 4}}\,{e^{ - 02x}} \times 10 \times 3dx}$

$= 9 \times {10^{ - 3}}\int_0^2 {{e^{ - 0.2x}}\,dx}$

$= {{9 \times {{10}^{ - 3}}} \over {0.2}}\left[ { - {e^{ - 0.2 \times 2}} + 1} \right]\,\,\,$

$B = 3.0 \times {10^{ - 4}}\,{e^{ - 0.2x}}$

(By exponential function)

$= {{9 \times {{10}^{ - 3}}} \over {0.2}} \times \left[ {1 - {e^{ - 0.4}}} \right]$

$= 9 \times {10^{ - 3}} \times \left( {0.33} \right) = 2.97 \times {10^{ - 3}}J$

Power required to move the conductor is,

$P = {W \over t}$

$P = {{2.97 \times {{10}^{ - 3}}} \over {\left( {0.2} \right) \times 5 \times {{10}^{ - 3}}}} = 2.97W$
3

### JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
The coercivity of a small magnet where the ferromagnet gets demagnetized is $3 \times {10^3}\,A{m^{ - 1}}.$ The current required to be passed in a solenoid of length $10$ $cm$ and number of turns $100,$ so that the magnet gets demagnetized when inside the solenoid, is :
A
$30$ $mA$
B
$60$ $mA$
C
$3$ $A$
D
$6A$

## Explanation

Magnetic field in solenoid $B = {\mu _0}ni$

$\Rightarrow {B \over {{\mu _0}}} = ni$

(Where $n=$ number of turns per unit length)

$\Rightarrow {B \over {{\mu _0}}} = {{Ni} \over L}$

$\Rightarrow 3 \times {10^3} = {{100i} \over {10 \times {{10}^{ - 2}}}}$

$\Rightarrow i = 3A$
4

### JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
Two short bar magnets of length $1$ $cm$ each have magnetic moments $1.20$ $A{m^2}$ and $1.00$ $A{m^2}$ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $20.0$ $cm.$ The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to $\left( \, \right.$ Horizontal com-ponent of earth. $s$ magnetic induction is $3.6 \times 10.5Wb/{m^2})$
A
$3.6 \times 10.5\,\,Wb/{m^2}$
B
$2.56 \times 10.4\,\,Wb/{m^2}$
C
$3.50 \times 10.4\,\,Wb/{m^2}$
D
$5.80 \times 10.4\,Wb/{m^2}$

## Explanation

Given: ${M_1} = 1.20A{m^2}\,\,\,$ and $\,\,\,{M_2} = 1.00A{m^2}$

$r = {{20} \over 2}cm = 0.1m$

${B_{net}} = {B_1} + {B_2} + {B_H}$

${B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}$

$= {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}$

$= 2.56 \times {10^{ - 4}}\,\,wb/{m^2}$

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