 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

The resistance of hot tungsten filament is about $10$ times the cold resistance. What will be resistance of $100$ $W$ and $200$ $V$ lamp when not in use ?
A
$20\Omega$
B
$40\Omega$
C
$200\Omega$
D
$400\Omega$

Explanation

$P = Vi = {{{V_2}} \over R}$

${R_{hot}} = {{{V^2}} \over P} = {{200 \times 200} \over {100}} = 400\Omega$

${R_{cold}} = {{400} \over {10}} = 40\Omega$
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

A moving coil galvanometer has $150$ equal divisions. Its current sensitivity is $10$- divisions per milliampere and voltage sensitivity is $2$ divisions per millivolt. In order that each division reads $1$ volt, the resistance in $ohms$ needed to be connected in series with the coil will be -
A
${10^5}$
B
${10^3}$
C
$9995$
D
$99995$

Explanation

KEY CONCEPT : Resistance of Galvanometer,

$G = {{Current\,\,\,sensitivity} \over {Voltage\,\,\,sensityvity}} \Rightarrow G = {{10} \over 2} = 5\Omega$

Here ${i_g} =$ Full scale deflection current $= {{150} \over {10}} = 15\,\,mA$

$V=$ voltage to be measured $=150$ volts

(such that each division reads $1$ volt)

$\Rightarrow R = {{150} \over {15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega$
4

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

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

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

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

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

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