### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii are in the ratio of ${4 \over 3}$ and ${2 \over 3}$, then the ratio of the current passing through the wires will be
A
$8/9$
B
$1/3$
C
$3$
D
$2$

## Explanation

${i_1}{R_1} = {i_2}{R_2}\,\,\,\,\,\,\,\,\,\,$ (same potential difference)

V = I1R1 = I1$\times $${{\rho {l_1}} \over {\pi r_1^2}} Also V = I2R2 = I2 \times$${{\rho {l_2}} \over {\pi r_2^2}}$

$\therefore$ I1$\times $${{\rho {l_1}} \over {\pi r_1^2}} = I2 \times$${{\rho {l_2}} \over {\pi r_2^2}}$

$\Rightarrow$ ${{{I_1}} \over {{I_2}}} = {{{\ell _1}} \over {{\ell _2}}} \times {{r_1^2} \over {r_2^2}}$

$= {3 \over 4} \times {4 \over 9} = {1 \over 3}\,\,$
2

### AIEEE 2004

The resistance of the series combination of two resistances is $S.$ When they are jointed in parallel the total resistance is $P.$ If $S = nP$ then the Minimum possible value of $n$ is
A
$2$
B
$3$
C
$4$
D
$1$

## Explanation

$S = {R_1} + {R_2}$ and $P = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$

$S = nP \Rightarrow {R_1} + {R_2} = {{n\left( {{R_1}{R_2}} \right)} \over {\left( {{R_1} + {R_2}} \right)}}$

$\Rightarrow {\left( {{R_1} + {R_2}} \right)^2} = n{R_1}{R_2}$

$\Rightarrow n = {{R_1^1 + R_2^2 + {R_1}{R_2}} \over {{R_1}{R_2}}}$

$n = {{{R_1}} \over {{R_2}}} + {{{R_2}} \over {{R_1}}} + 2$

Arithmetic mean $>$ Geometric mean

Minimum value of $n$ is $4$
3

### AIEEE 2003

A $220$ volt, $1000$ watt bulb is connected across a $110$ $volt$ mains supply. The power consumed will be
A
$750$ watt
B
$500$ watt
C
$250$ watt
D
$1000$ watt

## Explanation

We know that $R = {{V_{rated}^2} \over {{P_{rated}}}} = {{{{\left( {220} \right)}^2}} \over {1000}}$

When this bulb is connected to $110$ volt mains supply we get

$P = {{{V^2}} \over R} = {{{{\left( {110} \right)}^2} \times 1000} \over {{{\left( {220} \right)}^2}}} = {{1000} \over 4} = 250W$
4

### AIEEE 2003

A $3$ volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current ${\rm I}$, in the circuit will be
A
$1$ $A$
B
$1.5$ $A$
C
$2$ $A$
D
$1/3$ $A$

## Explanation

${R_p} = {{3 \times 6} \over {3 + 6}} = {{18} \over 9} = 2\Omega$

$\therefore$ $V = IR$

$\Rightarrow I = {V \over R} = {3 \over 2} = 1.5A$

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