### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

A material $'B'$ has twice the specific resistance of $'A'.$ A circular wire made of $'B'$ has twice the diameter of a wire made of $'A'$. Then for the two wires to have the same resistance, the ratio ${l \over B}/{l \over A}$ of their respective lengths must be
A
$1$
B
${l \over 2}$
C
${l \over 4}$
D
$2$

## Explanation

${\rho _B} = 2{\rho _A}$

${d_B} = 2{d_A}$

${R_B} = {R_A} \Rightarrow {{{\rho _B}{\ell _B}} \over {{A_B}}} = {{{P_A}{\ell _A}} \over {{A_A}}}$

$\therefore$ ${{{\ell _B}} \over {{\ell _A}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{d_B^2} \over {d_A^2}}$ $= {{{\rho _A}} \over {2{\rho _A}}} \times {{4d_d^2} \over {d_A^2}} = 2$
2

### AIEEE 2004

The Kirchhoff's first law $\left( {\sum i = 0} \right)$ and second law $\left( {\sum iR = \sum E} \right),$ where the symbols have their usual meanings, are respectively based on
A
conservation of charge, conservation of momentum
B
conservation of energy, conservation of charge
C
conservation of momentum, conservation of charge
D
conservation of charge, conservation of energy

## Explanation

NOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoffs second law is based on conservation of energy.
3

### 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.
4

### 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 $m = Zit$

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

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