 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

Two sources of equal $emf$ are connected to an external resistance $R.$ The internal resistance of the two sources are ${R_1}$ and ${R_2}\left( {{R_1} > {R_1}} \right).$ If the potential difference across the source having internal resistance ${R_2}$ is zero, then
A
$R = {R_2} - {R_1}$
B
$R = {R_2} \times \left( {{R_1} + {R_2}} \right)/\left( {{R_2} - {R_1}} \right)$
C
$R = {R_1}{R_2}/\left( {{R_2} - {R_1}} \right)$
D
$R = {R_1}{R_2}/\left( {{R_1} - {R_2}} \right)$

Explanation ${\rm I} = {{2\varepsilon } \over {R + {R_1} + {R_2}}}$

Potential difference across second cell

$= V = \varepsilon - {\rm I}{R_2} = 0$

$\varepsilon - {{2\varepsilon } \over {R + {R_1} + {R_2}}}.{R_2} = 0$

$R + {R_1} + {R_2} - 2{R_2} = 0$

$R + {R_1} - {R_2} = 0$

$\therefore$ $R = {R_2} - {R_1}$
2

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 $(iii),$ ${q_2} = {q \over {1 + {{{Z_2}} \over {{z_1}}}}}$
3

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 $iR = 2 = 12 - 500i$

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

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

$R = 100\Omega$
4

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.