### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

Two voltmeters, one of copper and another of silver, are joined in parallel. When a total charge $q$ flows through the voltmeters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are ${Z_1}$ and ${Z_2}$ respectively the charge which flows through the silver voltmeter is
A
${q \over {1 + {{{Z_2}} \over {{Z_1}}}}}$
B
${q \over {1 + {{{Z_1}} \over {{Z_2}}}}}$
C
$q{{{Z_2}} \over {{Z_1}}}$
D
$q{{{Z_1}} \over {{Z_2}}}$

## Explanation

Mass deposited

$m = Zq \Rightarrow Z \propto {1 \over q} \Rightarrow {{{Z_1}} \over {{Z_2}}} = {{{q_2}} \over {{q_1}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Also $q = {q_1} + {q_2}\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

$\Rightarrow {q \over {{q_2}}} = {{{q_1}} \over {{q_2}}} + 1\,\,\,\,\,\,\,\,\,\,$ $(\,Dividing\,\,\left( {ii} \right)$ by $\left. {{q_2}\,} \right)$

$\Rightarrow {q_2} = {q \over {1 + {{{q_1}} \over {{q_2}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$

From equations $(i)$ and $(iii),$ ${q_2} = {q \over {1 + {{{Z_2}} \over {{z_1}}}}}$
2

### 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

$iR = 2 = 12 - 500i$

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

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

$R = 100\Omega$
3

### 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.
4

### 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

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

${d_B} = 2{d_A}$

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

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