
The given circuit consists of two NOT gates and one OR gate. To identify the logic operation, let's analyze the circuit step-by-step:
Each input (A and B) passes through a NOT gate. The function of a NOT gate is to invert the input signal.
The outputs of the NOT gates are then fed into an OR gate. The OR gate outputs 1 if at least one of its inputs is 1.
Combining these, we realize that the entire circuit performs an OR operation on the outputs of the NOT gates. Since the NOT gates invert A and B before passing to the OR gate, the final operation resembles an OR operation.
Therefore, the correct answer is OR.
The circuit simplifies to:
\[ Y = \overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B \]
(De-Morgan’s law)
Match List-I with List-II:
| List-I (Amplifiers) | List-II (Characteristics) |
|---|---|
| (A) CE Amplifier | (I) Current buffer circuit |
| (B) CB Amplifier | (II) Voltage buffer circuit |
| (C) CC Amplifier | (III) High current gain |
| (D) Darlington Amplifier | (IV) High power gain |
Choose the correct answer:
Match List-I with List-II:
| List-I (Effects) | List-II (Electronic Devices) |
|---|---|
| (A) Channel length modulation | (I) Zener diode |
| (B) Channel width modulation | (II) BJTs |
| (C) Early effect | (III) JFETs |
| (D) Tunneling effect | (IV) MOSFETs |
Choose the correct answer:
Match List-I with List-II
| List-I (Instructions) | List-II (Addressing Mode) |
|---|---|
| (A) LDA 2100 H | (I) Immediate |
| (B) RAL | (II) Register |
| (C) ADD C | (III) Direct |
| (D) ANI 08 H | (IV) Implied |
Match List-I with List-II
| List-I (Data Bus Status Output) | List-II (Status Signals) |
|---|---|
| (A) Memory read | (I) 0, 1, 1 |
| (B) Op-code fetch | (II) 0, 1, 0 |
| (C) INTR acknowledge | (III) 0, 0, 1 |
| (D) Memory write | (IV) 1, 1, 1 |

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.