### JEE Mains Previous Years Questions with Solutions

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1

### 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

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

$\Rightarrow I = {V \over R} = {5 \over {10}} = 0.5\,A$
2

### 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 $R = {R_0}{e^{\alpha \Delta T}}$
3

### AIEEE 2006

In a Wheatstone's bridge, three resistance $P, Q$ and $R$ connected in the three arms and the fourth arm is formed by two resistances ${S_1}$ and ${S_2}$ connected in parallel. The condition for the bridge to be balanced will be
A
${P \over Q} = {{2R} \over {{S_1} + {S_2}}}$
B
${P \over Q} = {{R\left( {{S_1} + {S_2}} \right)} \over {{S_1}{S_2}}}$
C
${P \over Q} = {{R\left( {{S_1} + {S_2}} \right)} \over {2{S_1}{S_2}}}$
D
${P \over Q} = {R \over {{S_1} + {S_2}}}$

## Explanation

${P \over Q} = {R \over S}$ where $S = {{{S_1}{S_2}} \over {{S_1} + {S_2}}}$
4

### AIEEE 2006

An electric bulb is rated $220$ volt - $100$ watt. The power consumed by it when operated on $110$ volt will be
A
$75$ watt
B
$40$ watt
C
$25$ Watt
D
$50$ Watt

## Explanation

The resistance of the bulb is $R = {{{V^2}} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}$

The power consumed when operated at $110$ $V$ is

$P = {{{{\left( {110} \right)}^2}} \over {{{\left( {220} \right)}^2}/100}} = {{100} \over 4} = 25\,W$