JEE Physics 2025 Question with Solution

Given: Inputs are $A$ and $B$ for the shown logic circuit.

Find: The input combination for which $Y = 0$.

From the solution working, the circuit is analyzed as follows:

• Let the output of the first AND gate be $X$.
• The NOT gate gives $\overline{B}$.
• Let the OR gate output be $Z$.
• The final gate is a NAND gate producing $Y$.

Thus,

$$X = A \cdot B$$ $$\overline{B} = \text{NOT}(B)$$ $$Z = A + \overline{B}$$ $$Y = \overline{X \cdot Z} = \overline{(A \cdot B) \cdot (A + \overline{B})}$$

Now evaluate the options.

For $A = 1$ and $B = 1$:

$$X = 1 \cdot 1 = 1$$ $$\overline{B} = 0$$ $$Z = 1 + 0 = 1$$ $$Y = \overline{1 \cdot 1} = \overline{1} = 0$$

For $A = 0$ and $B = 1$:

$$X = 0 \cdot 1 = 0$$ $$\overline{B} = 0$$ $$Z = 0 + 0 = 0$$ $$Y = \overline{0 \cdot 0} = \overline{0} = 1$$

For $A = 1$ and $B = 0$:

$$X = 1 \cdot 0 = 0$$ $$\overline{B} = 1$$ $$Z = 1 + 1 = 1$$ $$Y = \overline{0 \cdot 1} = \overline{0} = 1$$

For $A = 0$ and $B = 0$:

$$X = 0 \cdot 0 = 0$$ $$\overline{B} = 1$$ $$Z = 0 + 1 = 1$$ $$Y = \overline{0 \cdot 1} = \overline{0} = 1$$

Therefore, the output is $Y = 0$ only for $A = 1$ and $B = 1$. The correct option is A.

Given: A logic circuit with inputs $A$ and $B$.

Find: Which input pair makes $Y = 0$.

A direct way is to compute the output for each option using the Boolean expression obtained in the solution:

$$Y = \overline{(A \cdot B) \cdot (A + \overline{B})}$$

A NAND gate gives output $0$ only when its input product is $1$. So we need

$$(A \cdot B) \cdot (A + \overline{B}) = 1$$

This requires both factors to be $1$.

First,

$$A \cdot B = 1 \Rightarrow A = 1,\; B = 1$$

Now check the second factor for $A = 1, B = 1$:

$$A + \overline{B} = 1 + 0 = 1$$

So the NAND input becomes $1$, hence

$$Y = \overline{1} = 0$$

No other input pair can make $A \cdot B = 1$, so no other option can give $Y = 0$.

Therefore, the correct option is A.