 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline)

The supply voltage to room is $120V.$ The resistance of the lead wires is $6\Omega$. A $60$ $W$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $240$ $W$ heater is switched on in parallel to the bulb?
A
zero
B
$2.9$ Volt
C
$13.3$ Volt
D
$10.04$ Volt

Explanation Power of bulb $=60W$ $\left( {given} \right)$

Resistance of bulb $= {{120 \times 120} \over {60}} = 240\Omega$

$\left[ {\,\,} \right.$ $\left. {\,P = {{{V^2}} \over R}\,} \right]$

Power of heater $=240W$ (given)

Resistance of heater $= {{120 \times 120} \over {240}} = 60\Omega$

Voltage across bulb before heater is switched on,

${V_1} = {{240} \over {246}} \times 120 = 117.73\,\,$ volt

Voltage across bulb after heater is switched on,

${V_2} = {{48} \over {54}} \times 120 = 106.66$ volt

Hence decrease in voltage

${V_1} - {V_2} = 117.073 - 106.66 = 10.04$ Volt (approximately)
2

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.
3

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 $0.2\%$
4

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

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

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

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

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