### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

The thermo $emf$ of a thermocouple varies with temperature $\theta$ of the hot junction as $E = a\theta + b{\theta ^2}$ in volts where the ratio $a/b$ is ${700^ \circ }C.$ If the cold junction is kept at ${0^ \circ }C,$ then the neutral temperature is
A
${1400^ \circ }C$
B
${350^ \circ }C$
C
${700^ \circ }C$
D
No neutral temperature is possible for this termocouple.

## Explanation

Neutral temperature is the temperature of a hot junction at which $E$ is maximum.

$\Rightarrow {{dE} \over {d\theta }} = 0$

or $a + 2b\theta = 0 \Rightarrow \theta = {{ - a} \over {2b}} = - 350$

Neutral temperature can never be negative hence no $\theta$ is possible.
2

### AIEEE 2004

The electrochemical equivalent of a metal is ${3.35109^{ - 7}}$ $kg$ per Coulomb. The mass of the metal liberated at the cathode when a $3A$ current is passed for $2$ seconds will be
A
$6.6 \times {10^{57}}/kg$
B
$9.9 \times {10^{ - 7}}\,kg$
C
$19.8 \times {10^{ - 7}}\,kg$
D
$1.1 \times {10^{ - 7}}\,kg$

## Explanation

The mass liberated $m,$ electrochemical equivalent of a metal $Z,$ are related as $m = Zit$

$\Rightarrow m = 3.3 \times {10^{ - 7}} \times 3 \times 2$

$= 19.8 \times {10^{ - 7}}\,kg$
3

### AIEEE 2004

A
metal oxides with high temperature coefficient of resistivity
B
metals with high temperature coefficient of resistivity
C
metals with low temperature coefficient of resistivity
D
semiconducting materials having low temperature

## Explanation

Thermistors are usually made of metal-oxides with high temperature coefficient of resistivity.
4

### AIEEE 2004

In a meter bridge experiment null point is obtained at $20$ $cm$, from one end of the wire when resistance $X$ is balanced against another resistance $Y.$ If $X < Y$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $4$ $X$ against $Y$
A
$40$ $cm$
B
$80$ $cm$
C
$50$ $cm$
D
$70$ $cm$

## Explanation

In the first case ${X \over Y} = {{20} \over {80}} = {1 \over 4}$

In the second case ${{4X} \over Y} = {\ell \over {100 - \ell }} \Rightarrow \ell = 50$