### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2014 (Offline)

In a large building, three are $15$ bulbs of $40$ $W$, $5$ bulbs of $100$ $W$, $5$ fans of $80$ $W$ and $1$ heater of $1$ $kW.$ The voltage of electric mains is $220$ $V.$ The minimum capacity of the main fuse of the building will be:
A
$8$ $A$
B
$10$ $A$
C
$12$ $A$
D
$14$ $A$

## Explanation

Total power consumed by electrical appliances in the building, ${P_{total}} = 2500W$

Watt $=$ Volt $\times$ ampere

$\Rightarrow 2500 = V \times {\rm I}$

$\Rightarrow 2500 = 220$ ${\rm I}$

$\Rightarrow I = {{2500} \over {220}} = 11.36 \approx 12A$

(Minimum capacity of main fuse)
2

### JEE Main 2013 (Offline)

This questions has Statement - ${\rm I}$ and Statement - ${\rm I}$${\rm I}. Of the four choices given after the Statements, choose the one that best describes into two Statements. Statement - {\rm I} : Higher the range, greater is the resistance of ammeter. Statement - {\rm I}$${\rm I}$ : To increase the range of ammeter, additional shunt needs to be used across it.

A
Statement - ${\rm I}$ is true, Statement - ${\rm II}$ is true, Statement - ${\rm II}$ is the correct explanation of statement - ${\rm I}$.
B
Statement - ${\rm I}$ is true, Statement - ${\rm II}$ is true, Statement - ${\rm II}$ is not the correct explanation of statement - ${\rm I}$.
C
Statement - ${\rm I}$ is true, Statement - ${\rm II}$ is false
D
Statement - ${\rm I}$ is false, Statement - ${\rm II}$ is true

## Explanation

Statements ${\rm I}$ is false and Statement ${\rm I}$${\rm I}$ is true

For ammeter, shunt resistance, $S = {{{{\rm I}_g}G} \over {{\rm I} - {{\rm I}_g}}}$

Therefore for ${\rm I}$ to increase, $S$ should decrease, So additional $S$ can be connected across it.
3

### 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)
4

### 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.