JEE Main 2026 Physics Solution | 23 Jan Shift 2 | Logic Gates & Truth Tables (Q.43) скачать в хорошем качестве

JEE Main 2026 Physics Solution | 23 Jan Shift 2 | Logic Gates & Truth Tables (Q.43)

Welcome to Physics Unlocked! In this video, we provide a detailed, step-by-step solution for a Logic Gates question from the JEE Main 2026 Physics paper (23 January, Shift 2). Question (Q.43): For the given logic gate circuit, which of the following is the correct truth table? (The circuit features inputs 'n' and 'm' passing through an OR gate and a NAND gate to produce output 'z'.) Step-by-Step Mathematical Solution & Derivation: 1. Identify the logic gates in the circuit: We have one OR gate and one NAND gate. 2. Determine the output of the first gate (OR gate): The inputs to the OR gate are n and m. Output of OR gate = n + m 3. Determine the inputs to the second gate (NAND gate): One input is directly connected to n. The second input is the output of the OR gate, which is (n + m). 4. Calculate the final output z (NAND gate): A NAND gate performs an AND operation followed by a NOT operation (inversion). z = (n • (n + m))' 5. Simplify the Boolean expression using the Distributive Law: z = ((n • n) + (n • m))' 6. Apply Boolean algebra rules: Since a variable ANDed with itself remains unchanged (n • n = n), we can write: z = (n + (n • m))' 7. Factor out n: z = (n • (1 + m))' 8. Apply the Boolean identity 1 + m = 1 (ORing any variable with 1 always results in 1): z = (n • 1)' 9. Final simplification: z = n' (or NOT n) 10. Conclusion: The final output 'z' depends solely on input 'n' and will always be its exact opposite (complement). The value of input 'm' does not affect the output at all. Looking at the given options, Option 2 is the correct truth table where z is the complement of n. Timestamps: 0:00 - Introduction and Reading the Question 0:19 - Formulating the Initial Boolean Expression 0:49 - Step-by-Step Boolean Algebra Simplification 1:36 - Matching the Final Truth Table and Conclusion Don't forget to like, subscribe, and hit the notification bell for more comprehensive JEE Main physics solutions without skipping any steps!