 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

Two electric bulbs marked $25W$ $-$ $220$ $V$ and $100W$ $-$ $220V$ are connected in series to a $440$ $V$ supply. Which of the bulbs will fuse?
A
Both
B
$100$ $W$
C
$25$ $W$
D
Neither

Explanation

The current upto which bulb of marked $25W$-$220V,$ will

not fuse ${I_1} = {{{W_1}} \over {{V_1}}} = {{25} \over {220}}Amp$

Similarly, ${I_2} = {{{W_2}} \over {{V_2}}} = {{100} \over {220}}\,Amp$

The current flowing through the circuit $I = {{440} \over {{{\mathop{\rm R}\nolimits} _{eff}}}},\,\,{{\mathop{\rm R}\nolimits} _{eff}} = {R_1} + {R_2}$

${R_1} = {{V_1^2} \over {{P_1}}} = {{{{\left( {220} \right)}^2}} \over {25}};\,\,\,$

${R_2} = {{V_2^2} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}$

$I = {{440} \over {{{{{\left( {220} \right)}^2}} \over {25}} + {{{{\left( {220} \right)}^2}} \over {100}}}}$

$= {{440} \over {{{\left( {220} \right)}^2}\left[ {{1 \over {25}} + {1 \over {100}}} \right]}}$

$I = {{40} \over {220}}\,\,Amp$

as ${I_1}\left( { = {{25} \over {220}}A} \right) < I\left( { = {{40} \over {220}}A} \right) < {I_2}\left( { = {{100} \over {200}}A} \right)$

Thus the bulb marked $25W$-$220$ will fuse.
2

AIEEE 2011

If a wire is stretched to make it $0.1\%$ longer, its resistance will:
A
increase by $0.2\%$
B
decrease by $0.2\%$
C
decrease by $0.05\%$
D
increase by $0.05\%$

Explanation

Resistance of wire

$R = {{\rho l} \over A} = {{\rho {l^2}} \over C}$ (where $Al=C$ )

$\therefore$ Fractional charge in resistance

${{\Delta R} \over R} = 2{{\Delta l} \over l}$

$\therefore$ Resistance will increase by $0.2\%$
3

AIEEE 2010

Two conductors have the same resistance at ${0^ \circ }C$ but their temperature coefficients of resistance are ${\alpha _1}$ and ${\alpha _2}.$ The respective temperature coefficients of their series and parallel combinations are nearly
A
${{{\alpha _1} + {\alpha _2}} \over 2},\,{\alpha _1} + {\alpha _2}$
B
${\alpha _1} + {\alpha _2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$
C
${\alpha _1} + {\alpha _2},\,{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}$
D
${{{\alpha _1} + {\alpha _2}} \over 2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$

Explanation

${R_1} = {R_0}\left[ {1 + {\alpha _1}\Delta t} \right];$

${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$

$R = {R_1} + {R_2}$

$= {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$

$= 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$

${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$

In Parallel, ${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$

$= {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$

$\Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$

$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$

$\therefore$ ${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$
4

AIEEE 2010

Let $C$ be the capacitance of a capacitor discharging through a resistor $R.$ Suppose ${t_1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and ${t_2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio ${t_1}/{t_2}$ will be
A
$1$
B
${1 \over 2}$
C
${1 \over 4}$
D
$2$

Explanation

Initial energy of capacitor, ${E_1} = {{q_1^2} \over {2C}}$

Final energy of capacitor, ${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}$

$\therefore$ ${t_1}=$ time for the charge to reduce to ${1 \over {\sqrt 2 }}$ of its initial value

and ${t_2} =$ time for the charge to reduce to ${1 \over 4}$ of its initial value

We have, ${q_2} = {q_1}{e^{ - t/CR}}$

$\Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}$

$\therefore$$\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)$

and $\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)$

By $(1)$ and $(2),$ ${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}$

$= {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}$