JEE Physics 2024 Question with Solution

Given: The circuit includes a NOT gate on input $A$, then an OR gate with inputs $\text{NOT}(A)$ and $B$, and finally an AND gate with this OR output and $B$.

Find: Which truth table matches the circuit output $Y$.

From the circuit description in the solution:

$$\text{NOT}(A)$$

The OR gate output is

$$\text{NOT}(A) + B$$

The final output is

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

Now evaluate for all input combinations:

$$\begin{aligned} A=0,\; B=0 &\Rightarrow Y=(1+0)\cdot 0 = 0 \\ A=0,\; B=1 &\Rightarrow Y=(1+1)\cdot 1 = 1 \\ A=1,\; B=0 &\Rightarrow Y=(0+0)\cdot 0 = 0 \\ A=1,\; B=1 &\Rightarrow Y=(0+1)\cdot 1 = 1 \end{aligned}$$

So the truth table is:

$$\begin{array}{c|c|c} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}$$

This matches option B. Therefore, the correct option is B.

Given: $Y = (\text{NOT}(A) + B) \cdot B$

Find: The matching truth table.

Use Boolean algebra:

$$\begin{aligned} Y &= (\overline{A} + B)B \\ &= \overline{A}B + B\cdot B \\ &= \overline{A}B + B \\ &= B \end{aligned}$$

So the output depends only on $B$. Hence whenever $B=0$, output is $0$, and whenever $B=1$, output is $1$.

Therefore, the correct truth table is the one corresponding to option B.