### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

The thermo $e.m.f.$ of a thermo -couple is $25$ $\mu V/{}^ \circ C$ at room temperature. A galvanometer of $40$ $ohm$ resistance, capable of detecting current as low as ${10^{ - 5}}\,A,$ is connected with the thermo couple. The smallest temperature difference that can be detected by this system is
A
${16^0}C$
B
${12^0}C$
C
${8^0}C$
D
${20^0}C$

## Explanation

Let $\theta$ be the smallest temperature difference that can be detected by the thermocouple, then

$I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta$

where ${\rm I}$ is the smallest current which can be detected by the galvanometer of resistance $R.$

$\therefore$ ${10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta$

$\therefore$ $\theta = {16^ \circ }C.$
2

### AIEEE 2003

The length of a wire of a potentiometer is $100$ $cm$, and the $e.$ $m.$ $f.$ of its standard cell is $E$ volt. It is employed to measure the $e.m.f.$ of a battery whose internal resistance in $0.5\Omega .$ If the balance point is obtained at $1=30$ $cm$ from the positive end, the $e.m.f.$ of the battery is

where $i$ is the current in the potentiometer wire.

A
${{30E} \over {100.5}}$
B
${{30E} \over {\left( {100 - 0.5} \right)}}$
C
${{30\left( {E - 0.5i} \right)} \over {100}}$
D
${{30E} \over {100}} - 0.5i$, where i is the current in the potentiometer wire

## Explanation

Potential gradient along wire, K = ${E \over {100}}$ volt/cm

For battery V = E' – ir, where E' is emf of battery.

or K × 30 = E' – ir, where current i is drawn from battery

or ${{E \times 30} \over {100}}$ = E' + 0.5i

or E' = ${{30E} \over {100}} - 0.5i$
3

### AIEEE 2002

If in the circuit, power dissipation is $150W,$ then $R$ is
A
$2\,\Omega$
B
$6\,\Omega$
C
$5\,\Omega$
D
$4\,\Omega$

## Explanation

The equivalent resistance is ${{\mathop{\rm R}\nolimits} _{eq}} = {{2 \times R} \over {2 + R}}$

$\therefore$ Powder dissipation $P = {{{V^2}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

$\therefore$ $150 = {{15 \times 15} \over {{R_{eq}}}}$

$\therefore$ ${{\mathop{\rm R}\nolimits} _{eq}} = {{15} \over {10}} = {3 \over 2}$

$\Rightarrow {{2R} \over {2 + R}} = {3 \over 2}$

$\Rightarrow 4R = 6 + 3R$

$\Rightarrow R = 6\Omega$
4

### AIEEE 2002

If a current is passed through a spring then the spring will
A
expand
B
compress
C
remains same
D
none of these

## Explanation

When current is passed through a spring then current flows parallel in the adjacent turns.

NOTE : When two wires are placed parallel to each other and current flows in the same direction, the wires attract each other.

Similarly here the various turns attract each other and the spring will compress.