The Correct Answer is Option (D)

$Y=\overline{\stackrel{―}{A}+\stackrel{―}{B}}=\overline{\stackrel{―}{A}}.\overline{\stackrel{―}{B}}=A.B$

AND gate

for AND gate

The Correct Answer is (A)

$y=\overline{\stackrel{―}{\mathrm{A}}\cdot \stackrel{―}{\mathrm{B}}}=\overline{\stackrel{―}{\mathrm{A}}}+\overline{\stackrel{―}{\mathrm{B}}}$

 $=(\mathrm{A}+\mathrm{B})\mathrm{OR}$ Gate

The Correct Answer is (B)

The given circuit clearly indicates that the bulb will glow when either of the switches are closed.

The corresponding truth table will be

This suggests that it will correspond to OR gate.

The Correct Answer is (C)

$\Rightarrow C=\overline{A.B+\stackrel{―}{A}.B}$

$\Rightarrow C=\overline{\left(A+\stackrel{―}{A}\right)B}$

$\Rightarrow C=\bar{B}$

The truth table would be

The Correct Answer is (C)

$Y=AB+\overline{BC}$

The Correct Answer is (B)

From the given logic circuit LED will glow, when voltage across LED in high. This can be verified by the given truth table.

Boolean expression Y = $\overline{A.B}$

This is output of NAND gate.

The Correct Answer is ()

The Correct Answer is (B)

y₁ = $\overline{A+B}$

y₂ = $\overline{y_1+y_1}$

= $A+B$

y = $\overline{y_2}$ = $\overline{A+B}$

So it is NOR gate.

The Correct Answer is (B)

Voltage gain, A = $\beta \frac{R_C}{R_B}$ = $100\times \frac{2000}{1000}$ = 200

Also, A =$\frac{V_0}{V_i}$

$\Rightarrow$$V_i=\frac{V_0}{A}$ = $\frac{4}{200}$ = 20 mA

The Correct Answer is (C)

Applying De Morgan’s law:

Output Y = [(A ⋅ B) ⋅ C]' = A' + B' + C'

When A, B, C are 0$\Rightarrow$ Y = 1

When A, B, C are 1$\Rightarrow$ Y = 0

The Correct Answer is (B)

Output of the circuit, Y = (A + B)·C

Y = 1 if C = 1 and A = 0, B = 1

or A = 1, B = 0 or A = B = 1

The Correct Answer is (A)

The Boolean expression of this arrangement is

Y = $\overline{\stackrel{―}{A}+\stackrel{―}{B}}$ = $\overline{\stackrel{―}{A}}+\overline{\stackrel{―}{B}}$ = A.B

Thus, the combination represents AND gate.

The Correct Answer is (A)

C' = $\overline{A.B}$

$\Rightarrow$ C = $\overline{\stackrel{―}{A.B}}$ = A.B

Hence the resultant gate is AND gate.

The Correct Answer is (C)

The output of the given logic circuit is

X = $\overline{\stackrel{―}{A.B}}=\; A.B$

The Correct Answer is (B)

The Boolen expression for the given combination is

output Y = (A + B) · C

So it is clear that Y = 1, when A = 1, B = 0 and C = 1

The Correct Answer is (A)

The truth table is
The logic circuit is OR gate.

The Correct Answer is (C)

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

The Correct Answer is (D)

It is clear from given logic circuit, that out put Y is low when both the inputs are high, otherwise it is high. Thus logic circuit is NAND gate.

The Correct Answer is (B)

The Boolen expression for the given combination is

output Y = (A + B) · C

So it is clear that Y = 1, when A = 1, B = 0 and C = 1

The Correct Answer is (B)

Integrated circuit can act as a complete electronic circuit.

The Correct Answer is (B)

OR gate, NOT gate and NAND gates are (iv), (ii) and (i) respectively.

The Correct Answer is (C)

Current gain,$\beta$ = $\frac{\mathrm{\Delta }{I}_C}{\mathrm{\Delta }{I}_B}$

= $\frac{\left(10-5\right)\times {10}^{-3}}{\left(200-100\right)\times {10}^{-6}}$ = 50

The Correct Answer is (C)

Y' = $\overline{A+B}$

Y = $\overline{Y^{\prime }}$

Y = $\overline{\stackrel{―}{A+B}}$

Y = A + B which is OR-gate.

The Correct Answer is (B)

The truth table corresponding to waveform is given by

$\begin{array}{ccc}A& B& C\\ 1& 1& 1\\ 0& 1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}$

$\therefore$ The given logic circuit gate is AND gate.

The Correct Answer is Option (B)

The truth table of OR gate is

$\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$

It is seen that output of OR gate is Y = A + B.
If both or any one has 1 input then output will be 1.

The Correct Answer is Option (A)

This truth table represents NAND gate.

The Correct Answer is Option (A)

This truth table represents NAND gate.