JEE Physics 2026 Question with Solution

Given: The circuit has inputs $A, B, C, D$. Inputs $A$ and $B$ pass through an AND gate followed by a NOT gate, while inputs $C$ and $D$ pass through an OR gate. These two outputs then enter an AND gate followed by a NOT gate.

Find: The correct truth table for output $Y$.

Step 1: Identify individual logic operations.

From the given circuit, the upper branch gives NAND of $A$ and $B$, so its output is $\overline{A \cdot B}$. The lower branch gives OR of $C$ and $D$, so its output is $C + D$.

Step 2: Write the Boolean expression.

The final gate is an AND followed by NOT, therefore

Using De Morgan's theorem,

Step 3: Evaluate the given rows.

Checking the listed input combinations row-wise, the resulting output values match the truth table shown in option D.

Conclude: The correct option is D.

Given: The circuit is a combination of NAND, OR, and final NAND stages.

Find: Which option matches the truth table.

A quick approach is to simplify the circuit before checking rows. First write

Now apply De Morgan's theorem directly:

This works because the complement of a product becomes the sum of complements.

Now compare this simplified form with the rows in the options instead of tracing the entire circuit gate by gate each time. The matching table is option D.

Therefore, the correct option is D.