1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

A heating element has a resistance of 100 $$\Omega $$ at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. what is the temperature coefficient of resistance of the heating element ?
A
0.5 $$ \times $$ 10$$-$$4 oC$$-$$1
B
5 $$ \times $$ 10$$-$$4 oC$$-$$1
C
1 $$ \times $$ 10$$-$$4 oC$$-$$1
D
2 $$ \times $$ 10$$-$$4 oC$$-$$1

Explanation

When temperature increased by 500oC then, nrw registance

Rt = $${{220} \over 2}$$ = 110 $$\Omega $$

We know,

Rt = R0 (1 + $$ \propto $$ $$\Delta $$ t)

$$ \Rightarrow $$ 110 = 100 (1 + $$ \propto $$ $$ \times $$ 500)

$$ \Rightarrow $$ $$ \propto $$ = $${{10} \over {100 \times 500}} = 2 \times {10^{ - 4}}$$ oC$$-$$1
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

When the switch S, in the circuit shown, is closed, then the value of current i will be :

A
3A
B
5A
C
4A
D
2A

Explanation



Let the voltage at C = V

$$ \therefore $$   Using KCL law,

i1 + i2 = i

$$ \Rightarrow $$   $${{20 - V} \over 2} + {{10 - V} \over 4}$$ = $${{V - 0} \over 2}$$

$$ \Rightarrow $$   V = 10 volts.

$$ \therefore $$   i = $${{10} \over 2}$$

= 5A
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is :
A
$${R \over {\sqrt 5 }}$$
B
$${R \over {\sqrt 2 }}$$
C
R
D
R$$\sqrt 2 $$

Explanation



Electric field on the axis of the ring,

$$E = {{KQh} \over {{{\left( {{R^2} + {h^2}} \right)}^{{3 \over 2}}}}}$$

For maximum electric field,

$${{dE} \over {dh}} = 0$$

$$ \Rightarrow $$   $$h = {R \over {\sqrt 2 }}$$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is $$\upsilon $$. If the electron density in copper is 9 $$ \times $$ 1028/m3 the value of $$\upsilon $$. in mm/s is close to (Take charge of electron to be = 1.6 $$ \times $$ 10$$-$$19C)
A
0.02
B
3
C
2
D
0.2

Explanation

We know,

I = neAVd

$$ \therefore $$   Vd = $${{\rm I} \over {neA}}$$

=   $${{1.5} \over {9 \times {{10}^{28}} \times 1.6{ \times ^{ - 19}} \times 5 \times {{10}^{ - 6}}}}$$

=   0.02 m/s

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