### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $R$ will be -
A
$100\Omega$
B
$200\Omega$
C
$1000\Omega$
D
$500\Omega$

## Explanation

$iR = 2 = 12 - 500i$

$\therefore$ $i = {1 \over {50}}$

$\therefore$ ${1 \over {50}} \times R = 2$

$R = 100\Omega$
2

### AIEEE 2005

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be
A
four times
B
doubled
C
halved
D
one fourth

## Explanation

$H = {{{V^2}t} \over R}$

Resistance of half the coil $= {R \over 2}$

$\therefore$ As $R$ reduces to half, $'H'$ will be doubled.
3

### 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

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

${d_B} = 2{d_A}$

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

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

### 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.