1

### 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

### 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

### 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

### 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