1

### JEE Main 2018 (Online) 16th April Morning Slot

In the given circuit, the current through zener diode is : A
5.5 mA
B
6.7 mA
C
2.5 mA
D
3.3 mA
2

### JEE Main 2019 (Online) 9th January Morning Slot

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m$-$3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :
A
$2\,\Omega$m
B
4$\,\Omega$m
C
0.4 $\,\Omega$m
D
0.2 $\,\Omega$m

## Explanation

For semiconductor,

Conductivity, $\sigma$ = ne q $\mu$e + nh q $\mu$h

given that semiconductor is n-type. So contribution of holes is ignored.

$\therefore$   nh q $\mu$h = 0

$\therefore$   $\sigma$ = ne q $\mu$e

Resistivity, $\rho$ = ${1 \over \sigma }$

= ${1 \over {{n_e}q{\mu _e}}}$

= ${1 \over {{{10}^{19}} \times 1.6 \times {{10}^{ - 19}} \times 1.6}}$

= ${1 \over {1.6 \times 1.6}}$

= 0.4 $\Omega$ m
3

### JEE Main 2019 (Online) 9th January Evening Slot

Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of V0 changes by : (assume that the Ge diode has large breakdown voltage) A
0.8 V
B
0.6 V
C
0.2 V
D
0.4 V

## Explanation

Case 1 : Ge need 0.3 V for start conducting and Si need 0.7 V for start conducting.

So, conduction will be start by Ge.

Out of 12 V, 0.3V will be consumed by Ge.

$\therefore$  V01 = 12 $-$ 0.3 = 11.7 V

Case 2 : Here as Ge diode is in reverse bias so no conduction will happen through Ge diode

As Si diode is in forward bias so conduction will happen through Si diode.

Here Si diode consume 0.7 V.

So, V02 = 12 $-$ 0.7 = 11.3 V

$\therefore$  Value of V0 changes by

= 11.7 $-$ 11.3 = 0.4 V
4

### JEE Main 2019 (Online) 10th January Morning Slot

To get output 1 at R, for the given logic gate circuit the input values must be A
x = 0,  y = 0
B
x = 1,  y = 0
C
x = 0,  y = 1
D
x = 1,  y = 1

## Explanation

P = $\overline x +$ y

Q = $\overline {\overline y .x}$ = y + $\overline x$

Output, R = $\overline {P + Q}$

To make R = 1

$\overline {P + Q}$ should be = 1

$\therefore$ P + Q must be 0

$\therefore$ ($\overline x +$ y) + (y + $\overline x$) = 0

Now by checking each option you can see

When y = 0 and x = 1 then

$\overline x +$ y = 0

$\therefore$ ($\overline x +$ y) + (y + $\overline x$) = 0