### JEE Mains Previous Years Questions with Solutions

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1

### 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, ${R_0} = 4\Omega .$
2

### 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 $V=e$ $\left. \, \right)$
3

### AIEEE 2006

The current ${\rm I}$ drawn from the $5$ volt source will be
A
$0.33$ $A$
B
$0.5$ $A$
C
$0.67$ $A$
D
$0.17$ $A$

## Explanation

The network of resistors is a balanced wheatstone bridge. The equivalent circuit is

${{\mathop{\rm R}\nolimits} _{eq}} = {{15 \times 30} \over {15 + 30}} = 10\Omega$

$\Rightarrow I = {V \over R} = {5 \over {10}} = 0.5\,A$
4

### AIEEE 2006

The resistance of bulb filmanet is $100\Omega$ at a temperature of ${100^ \circ }C.$ If its temperature coefficient of resistance be $0.005$ per $^ \circ C$, its resistance will become $200\,\Omega$ at a temperature of
A
${300^ \circ }C$
B
${400^ \circ }C$
C
${500^ \circ }C$
D
${200^ \circ }C$

## Explanation

$R{}_1 = {R_0}\left[ {1 + \alpha \times 100} \right] = 100\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

${R_2} = {R_0}\left[ {1 + \alpha \times T} \right] = 200\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

On dividing we get

${{200} \over {100}} = {{1 + \alpha T} \over {1 + 100\alpha }}$

$\Rightarrow 2 = {{1 + 0.005T} \over {1 + 100 \times 0.005}}$

$\Rightarrow T = {400^ \circ }C$

NOTE : We may use this expression as an approximation because the difference in the answers is appreciable. For accurate results one should use $R = {R_0}{e^{\alpha \Delta T}}$