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1

### JEE Main 2021 (Online) 27th August Evening Shift

For a transistor $$\alpha$$ and $$\beta$$ are given as $$\alpha = {{{I_C}} \over {{I_E}}}$$ and $$\beta = {{{I_C}} \over {{I_B}}}$$. Then the correct relation between $$\alpha$$ and $$\beta$$ will be :
A
$$\alpha = {{1 - \beta } \over \beta }$$
B
$$\beta = {\alpha \over {1 - \alpha }}$$
C
$$\alpha \beta = 1$$
D
$$\alpha = {\beta \over {1 - \beta }}$$

## Explanation

$$\alpha = {{{I_C}} \over {{I_E}}}$$, $$\beta = {{{I_C}} \over {{I_B}}}$$; $${I_E} = {I_C} + {I_B}$$

$$\alpha = {{{I_C}} \over {{I_C} + {I_B}}} = {{{I_C}/{I_B}} \over {{{{I_C}} \over {{I_B}}} + 1}} = {\beta \over {\beta + 1}}$$

$$1 + {1 \over \beta } = {1 \over \alpha }$$

$${1 \over \beta } = {{1 - \alpha } \over \alpha }$$

$$\beta = {\alpha \over {1 - \alpha }}$$
2

### JEE Main 2021 (Online) 27th August Morning Shift

For a transistor in CE mode to be used as an amplifier, it must be operated in :
A
Both cut-off and Saturation
B
Saturation region only
C
Cut-off region only
D
The active region only

## Explanation

Active region of the CE transistor is linear region and is best suited for its use as an amplifier.
3

### JEE Main 2021 (Online) 26th August Evening Shift

Four NOR gates are connected as shown in figure. The truth table for the given figure is : A B C D ## Explanation $$y = \overline {(\overline {A + \overline {A + B} } ) + (\overline {B + \overline {A + B} } )}$$

$$y = (A + \overline {A + B} ).(B + \overline {A + B} )$$

A B y
0 0 1
0 1 0
1 0 0
1 1 1
4

### JEE Main 2021 (Online) 26th August Morning Shift

Statement I : By doping silicon semiconductor with pentavalent material, the electrons density increases.

Statement II : The n-type semiconductor has net negative charge.

In the light of the above statements, choose the most appropriate answer from the options given below :
A
Statement - I is true but Statement - II is false.
B
Statement - I is false but Statement - II is true.
C
Both Statement I and Statement II are true.
D
Both Statement I and Statement II are false.

## Explanation

Pentavalent activities have excess free e$$-$$, so e$$-$$ density increases but overall semiconductor is neutral.

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