1

### AIPMT 2008

The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be
A
1.25
B
100
C
90
D
10

## Explanation

Negative feedback is applied to reduce the output voltage of an amplifier. If there is no negative feedback, the value of output voltage could be very high. In the options given, the maximum value of voltage gain is 100. Hence it is the correct option.
2

### AIPMT 2008

A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
A
1 $\times$ 1014 Hz
B
20 $\times$ 1014 Hz
C
10 $\times$ 1014 Hz
D
5 $\times$ 1014 Hz

## Explanation

Energy gap Eg = 2.0 eV

Eg = 2.0 × 1.6 × 10–19 J

Eg = hf

$\therefore$ f = ${{{E_g}} \over h}$

= ${{3.2 \times {{10}^{ - 19}}} \over {6.625 \times {{10}^{ - 34}}}}$

= 5 $\times$ 1014 Hz
3

### AIPMT 2007

For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct ?
A
a $\ne$ b $\ne$ c  and  $\alpha$ = $\beta$ = $\gamma$ = 90o
B
a = b = c  and  $\alpha$ $\ne$ $\beta$ $\ne$ $\gamma$ = 90o
C
a = b = c  and  $\alpha$ = $\beta$ = $\gamma$ = 90o
D
a $\ne$ b $\ne$ c  and  ;$\alpha$ $\ne$ $\beta$ $\ne$  and  $\gamma$ = 90o

## Explanation

In a cubic crystal structure
a = b = c  and  $\alpha$ = $\beta$ = $\gamma$ = 90o
4

### AIPMT 2007

In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is

A
an insulator
B
a metal
C
an n-type semiconductor
D
a p-type semiconductor.

## Explanation

For a p-type semiconductor, the acceptor energy level, as shown in the diagram, is slightly above the top Ev of the valence band. With very small supply of energy an electron from the valence band can jump to the level EC and ionise acceptor negatively.