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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

Questions Asked from Current Electricity

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