8051 logical AND practice — accumulator result Compute the accumulator after executing these two instructions (showing real newlines): MOV A, #0BH ANL A, #2CH

Introduction / Context:This exercise checks understanding of bitwise AND on hexadecimal values in 8051 assembly. Being fluent in hex-to-binary conversion and bitwise operations is crucial for low-level I/O and mask manipulation.

Given Data / Assumptions:

• MOV A, #0BH loads A with 0x0B.
• ANL A, #2CH computes A = A AND 0x2C.
• Binary conversions are performed to visualize the result.

Concept / Approach:0x0B = 0000 1011, 0x2C = 0010 1100. The AND operator keeps only positions where both operands have 1s. The outcome is 0000 1000 (0x08).

Step-by-Step Solution:

Verification / Alternative check:Compute in hex: 0x0B AND 0x2C = 0x08, which equals binary 00001000.

Why Other Options Are Wrong:

• 11010111, 11011010, 00101000: do not match the correct AND result of the given operands.

Common Pitfalls:Misaligning bits during conversion or confusing AND with OR/XOR. Always write both operands in full 8-bit binary before applying the operator.

Final Answer:00001000