Matching electronic devices with their primary applications:

• Photodiode: Operated in reverse bias to detect optical signals as a Photo detector (Matches IV).
• Zener diode: Operates in reverse breakdown region to provide constant output voltage as a Voltage regulator (Matches III).
• Light Emitting Diode (LED): Emits spontaneous radiation in forward bias, widely used in optical displays and infrared Remote controls (Matches I).
• Transistor: Used in active region to amplify small alternating signals as an Amplifier (Matches II).

Correct matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II). Hence, option 2 is the correct answer.

Evaluation of the given statements:

• Statement A is incorrect: A p-n junction diode is used as a rectifier, not a transistor.
• Statement B is correct: A Zener diode operates in its reverse breakdown region as a voltage regulator.
• Statement C is incorrect: NAND and NOR gates are universal gates; NOT is a basic gate.
• Statement D is correct: A junction transistor in common-emitter configuration is widely used as an amplifier.
• Statement E is incorrect: A transistor with positive feedback is used as an oscillator, not a photodiode.

Analysis of the four I-V curves shown in the diagram:

• Curve (A): Lies in the fourth quadrant ($V \gt 0, I \lt 0$) where the device delivers electrical power to an external load. This is the characteristic curve of a Solar Cell (Matches I).
• Curve (B): Operates in forward bias with a knee voltage (typically $1.5\,\text{V} - 3.0\,\text{V}$). This represents a Light Emitting Diode (LED) (Matches II).
• Curve (C): Features a sharp, vertical reverse breakdown at Zener voltage $V_Z$. This is a Zener Diode (Matches III).
• Curve (D): Operates in the third quadrant (reverse bias) where reverse saturation current increases linearly with incident light intensity. This is a Photodiode (Matches IV).

Analysis of the logic gate combination in the diagram:

• The inputs $A$ and $B$ are each inverted using NOT gates (or single-input NOR gates) to obtain $\overline{A}$ and $\overline{B}$.
• These inverted inputs are fed into a NOR gate:

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

Applying De Morgan's theorem:

$$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$$

The Boolean output $Y = A \cdot B$ corresponds precisely to an AND gate.