1

### JEE Main 2016 (Online) 10th April Morning Slot

To get an output of 1 from the circuit shown in figure the input must be : A
a = 0, b = 1, c = 0
B
a = 1, b = 0, c = 0
C
a = 1, b = 0, c = 1
D
a = 0, b = 0, c = 1

## Explanation

Here,

Y = c . (a + b)

Now  Y = 1   when,

c = 1 and (a = 1, b = 0 or a = 0, b = 1 or a = 1, b = 1)
2

### JEE Main 2016 (Online) 10th April Morning Slot

A realistic graph depicting the variation of the reciprocal of input resistance in an input characteristics measurement in a commonemitter transistor configuration is :
A B C D ## Explanation For common emitter configuration, the input characteristic graph is shown above,

ri $=$ ${{\Delta {V_{BE}}} \over {\Delta {{\rm I}_B}}}$

$\Rightarrow$   ${1 \over {{r_i}}} = {{d{{\rm I}_B}} \over {d{V_{BE}}}}$ $=$ shope of this curve.

Upto knee voltage 0.7 V the shope is almost constant. Then it increases sharply.

So, option (c) is the correct choice.
3

### JEE Main 2017 (Offline)

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be:
A
180°
B
45°
C
90°
D
135°

## Explanation

In common emitter configuration for n-p-n transistor input and output signals are 180° out of phase i.e., phase difference between output and input voltage is 180°.
4

### JEE Main 2017 (Online) 8th April Morning Slot

What is the conductivity of a semiconductor sale having electron concentration of $5 \times {10^{18}}\,\,{m^{ - 3}},$ hole concentration of $5 \times {10^{19}}\,\,{m^{ - 3}},$ electron mobility of 2.0 m2 V$-$1 s-1 and hole mobility of 0.01 m2 V$-$1 s$-$1 ?

(Take charge of electronas 1.6 $\times$ 10 $-$19 c)
A
1.68 ($\Omega$-m)$-$1
B
1.83 ($\Omega$-m)$-$1
C
0.59 ($\Omega$-m)$-$1
D
1.20 ($\Omega$-m)$-$1

## Explanation

Conductivity of semiconductor,

$\sigma$ = e$\left( {{\eta _e}{\mu _e} + \eta '{\mu _h}} \right)$

= 1.6 $\times$ 10$-$19 (5 $\times$ 1018 $\times$ 2 + 5 $\times$ 1019 $\times$ 0.01)

= 1.6 $\times$ 1.05

= 1.68