JEE Physics 2026 Question with Solution

Given: A logic gate circuit with inputs $n$ and $m$ is shown.

Find: The correct truth table for the output $z$.

From the circuit working given in the solution:

• Gate 1 (Top) is NAND, so

$$X = \overline{nm}$$

• Gate 2 (Bottom) is OR, so

$$Y = n + m$$

• Gate 3 (End) is AND, so

$$z = X \cdot Y$$

Therefore,

$$z = (\overline{nm})(n+m)$$

Using De Morgan's law,

$$z = (\bar{n} + \bar{m})(n+m)$$

Expanding,

$$z = \bar{n}n + \bar{n}m + \bar{m}n + \bar{m}m$$

So,

$$z = 0 + \bar{n}m + n\bar{m} + 0$$

Hence,

$$z = \bar{n}m + n\bar{m}$$

This is the XOR function.

The truth table of XOR is:

• $n=0, m=0 \Rightarrow z=0$
• $n=0, m=1 \Rightarrow z=1$
• $n=1, m=0 \Rightarrow z=1$
• $n=1, m=1 \Rightarrow z=0$

Therefore, the circuit represents XOR, and the correct option is B.

Given: The circuit contains three gates connected in sequence.

Find: Which option matches the output truth table.

The hint states that XOR gate output is $1$ only when the two inputs are different.

From the solution explanation, the combined output becomes

$$z = \bar{n}m + n\bar{m}$$

This is exactly the Boolean expression of XOR.

So the required truth table must have output $1$ for input pairs $01$ and $10$, and output $0$ for input pairs $00$ and $11$.

Hence the matching truth table is the one corresponding to Option B.