JEE Main202623 January 2026Evening ShiftPhysicsSemiconductorsActual
For the given logic gate circuit, which of the following is the correct truth table?
Options
- A( array c|c|c n & m & z 0 & 0 & 0 0 & 1 & 1 1 & 1 & 0 1 & 0 & 1 array )
- B( array c|c|c n & m & z 0 & 0 & 1 0 & 1 & 0 1 & 1 & 1 1 & 0 & 0 array )
- C( array c|c|c n & m & z 0 & 0 & 1 0 & 1 & 0 1 & 1 & 0 1 & 0 & 0 array )
- D( array c|c|c n & m & z 0 & 0 & 1 0 & 1 & 1 1 & 1 & 0 1 & 0 & 0 array )
Correct answer
D. ( array c|c|c n & m & z 0 & 0 & 1 0 & 1 & 1 1 & 1 & 0 1 & 0 & 0 array )
Step-by-step solution
The given logic circuit consists of an OR gate and a NAND gate. The output of the OR gate is y = n + m . The inputs to the NAND gate are n and y . The final output z is given by z = n y = n (n + m) . Using Boolean algebra: n (n + m) = n n + n m = n + n m = n(1 + m) = n . Therefore, z = n . Let's verify this with the truth table: n m y = n + m z = NAND(n, y) 0 0 0 1 0 1 1 1 1 1 1 0 1 0 1 0 Comparing this with the given options, option (4) matches the calculated truth table.