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1

### AIPMT 2011 Prelims

Symbolic representation of four logic gates are shown as Pick out which ones are for AND, NAND and NOT gates, respectively
A
(ii), (iii) and (iv)
B
(iii), (ii) and (i)
C
(iii), (ii) and (iv)
D
(ii), (iv) and (iii)

## Explanation

AND, NAND and NOT gates are (ii), (iv) and (iii) respectively.
2

### AIPMT 2011 Prelims

A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 $$\mu$$A to 300 $$\mu$$A produces a change in the collector current from 10 mA to 20 mA. The current gain is
A
50
B
75
C
100
D
25

## Explanation

Current gain, $$\beta$$ = $${{\Delta {I_C}} \over {\Delta {I_B}}}$$

= $${{\left( {20 - 10} \right) \times {{10}^{ - 3}}} \over {\left( {300 - 100} \right) \times {{10}^{ - 6}}}}$$ = 50
3

### AIPMT 2011 Mains

A Zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the diode is A
5 mA
B
10 mA
C
15 mA
D
20 mA

## Explanation

Voltage across 250 Ω resistance = 20 V – 15 V = 5 V

Now current through 250 $$\Omega$$ resistance: 5/250 = 20 mA

If voltage across load resistance 1 k$$\Omega$$ is 15 V, then

current through 1 kΩ is 15/1000 = 15 mA

The current through the zener diode is

= Current through 250 $$\Omega$$ resistance – Current through 1 k$$\Omega$$ resistance.

= 20 - 15

= 5 mA
4

### AIPMT 2011 Prelims

If a small amount of antimony is added to germanium crystal
A
it becomes a p-type semiconductor
B
the antimony becomes an acceptor atom
C
there will be more free electrons than holes in the semiconductor
D
its resistance is increased

## Explanation

If a small amount of antimony is added to germanium crystal, crystal becomes n-type semiconductor. Hence, there will be more free electrons than holes.

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