### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

The temperature dependence of resistance of $Cu$ and undoped $Si$ in the temperature range $300-400$ $K,$ is best described by :
A
Linear increases for $Cu,$ exponential decrease of $Si.$
B
Linear decrease for $Cu,$ linear decrease for $Si$
C
Linear increase for $Cu,$ linear increase for $Si.$
D
Linear increase for $Cu,$ exponential increase for $Si$

## Explanation

2

### JEE Main 2015 (Offline)

In the circuit shown, the current in the $1\Omega$ resistor is :
A
$0.13$ $A,$ from $Q$ to $P$
B
$0.13$ $A$, from $P$ to $Q$
C
$1.3A$ from $P$ to $Q$
D
$0A$

## Explanation

From $KVL$

$- 6 + 3{{\rm I}_1} + {\rm I}\left( {{{\rm I}_1} - {{\rm I}_2}} \right) = 0$

$6 = 3{{\rm I}_1} + {{\rm I}_1} - {{\rm I}_2}$

$4{{\rm I}_1} - {{\rm I}_2} = 6\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$- 9 + 2{{\rm I}_2} - \left( {{{\rm I}_1} - {{\rm I}_2}} \right) + 3{{\rm I}_2} = 0$

$- {{\rm I}_1} + 6{{\rm I}_2} = 9\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

On solving $\left( 1 \right)$ and $\left( 2 \right)$

${{\rm I}_1} = 0.13A$

Direction $Q$ to $P,$ since ${{\rm I}_1} > {{\rm I}_2}.$

Alternatively

$Eq = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}}}}$

$= {{{6 \over 3} - {9 \over 5}} \over {{1 \over 3} + {1 \over 5}}} = {3 \over {8V}}$

$\therefore$ ${\rm I} = {{{3 \over 8}} \over {{{15} \over 8} + 1}} = {3 \over {23}} = 0.13A$

Considering potential at $P$ as $0V$ and at $Q$ as $x$ volt, then

${{x - 6} \over 3} + {{x - 0} \over 1} + {{x + 9} \over 5} = 0$

$\therefore$ $x = {2 \over {23}}$

$\therefore$ $i = {{x - 0} \over 1} = {2 \over {23}} = 0.13A$

From $Q$ to $P$
3

### JEE Main 2015 (Offline)

When $5V$ potential difference is applied across a wire of length $0.1$ $m,$ the drift speed of electrons is $2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}.$ If the electron density in the wire is $8 \times {10^{28}}\,\,{m^{ - 3}},$ the resistivity of the material is close to :
A
$1.6 \times {10^{ - 6}}\Omega m$
B
$1.6 \times {10^{ - 5}}\Omega m$
C
$1.6 \times {10^{ - 8}}\Omega m$
D
$1.6 \times {10^{ - 7}}\Omega m$

## Explanation

$V = IR = \left( {neA{v_d}} \right)\rho {\ell \over A}$

$\therefore$ $\rho = {V \over {{V_d}\ln e}}$

Here $V=$ potential difference

$l =$ length of wire

$n=$ no. of electrons per unit volume of conductor.

$e=$ no. of electrons

Placing the value of above parameters we get resistivity

$\rho = {5 \over {8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}}$

$= 1.6 \times {10^{ - 5}}\Omega m$
4

### 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)