### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2008

Consider a block of conducting material of resistivity $'\rho '$ shown in the figure. Current $'I'$ enters at $'A'$ and leaves from $'D'$. We apply superposition principle to find voltage $'\Delta V'$ developed between $'B'$ and $'C'$. The calculation is done in the following steps:
(i) Take current $'I'$ entering from $'A'$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $E(r)$ at distance $'r'$ from A by using Ohm's law $E = \rho j,$ where $j$ is the current per unit area at $'r'$.
(iii) From the $'r'$ dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat (i), (ii) and (iii) for current $'I'$ leaving $'D'$ and superpose results for $'A'$ and $'D'.$

For current entering at $A,$ the electric field at a distance $'r'$ from $A$ is

A
${{\rho I} \over {8\pi {r^2}}}$
B
${{\rho I} \over {{r^2}}}$
C
${{\rho I} \over {2\pi {r^2}}}$
D
${{\rho I} \over {4\pi {r^2}}}$

## Explanation

As shown above $E = {{\rho I} \over {2\pi {r^2}}}$
2

### AIEEE 2008

Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

The value of the unknown resister $R$ is

A
$13.75\Omega$
B
$220\Omega$
C
$110\Omega$
D
$55\Omega$

## Explanation

According to the condition of balancing

${{55} \over {20}} = {R \over {80}} \Rightarrow R = 220\Omega$
3

### AIEEE 2007

The resistance of a wire is $5$ ohm at ${50^ \circ }C$ and $6$ ohm at ${100^ \circ }C.$ The resistance of the wire at ${0^ \circ }C$ will be
A
$3$ ohm
B
$2$ ohm
C
$1$ ohm
D
$4$ ohm

## Explanation

KEY CONCEPT : We know that

${R_t} = R{}_0\left( {1 + \alpha t} \right),$

where ${R_t}$ is the resistance of the wire at ${t^ \circ }C,$

${R_0}$ is the resistance of the wire at ${0^ \circ }C$

and $\alpha$ is the temperature coefficient of resistance

$\Rightarrow {R_{50}} = {R_0}\left( {1 + 50\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

${R_{100}} = {R_0}\left( {1 + 100\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

From $\left( i \right),\,{R_{50}} - {R_0} = 50\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$

From $\left( {ii} \right),{R_{100}} - {R_0} = 100\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iv} \right)$

Dividing $(iii)$ by $(iv),$ we get

${{{R_{50}} - {R_0}} \over {{R_{100}} - R{}_0}} = {1 \over 2}$

Here, ${{R_{50}}}$ $= 5\Omega$ and ${{R_{100}} = 6\Omega }$

$\therefore$ ${{5 - {R_0}} \over {6 - {R_0}}} = {1 \over 2}$

or, $6 - {R_0} = 10 - 2{R_0}$

or, ${R_0} = 4\Omega .$
4

### AIEEE 2007

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be
A
$1/2$
B
$1$
C
$2$
D
$1/4$

## Explanation

Required ratio

$= {{Energy\,\,\,stored\,\,in\,\,capacitor} \over {Workdone\,\,by\,\,the\,\,battery}} = {{{1 \over 2}C{V^2}} \over {C{e^2}}}$

where $C=$ Capacitance of capacitor

$V=$ Potential difference,

$e=$ $emf$ of battery

$= {{{1 \over 2}C{e^2}} \over {C{e^2}}} = {1 \over 2}$

$\left( \, \right.$ as $V=e$ $\left. \, \right)$