Consider following program for 8085 microprocessor : MVI A , 32H The contents of A after execution of program will be :
RRC
RRC
8CH
Work in binary — rotate instructions are bit operations and hexadecimal hides what they do.
Step 1 — load the accumulator.
MVI A, 32H → A = 0011 0010
Step 2 — understand RRC. RRC is "rotate accumulator right without carry": every bit moves one place right, and the bit that falls off the right-hand end wraps around into bit 7. The carry flag receives a copy of that bit but does not take part in the rotation itself.
| Step | A (binary) | A (hex) | Bit rotated out → CY |
|---|---|---|---|
| Start | 0011 0010 | 32H | — |
| After 1st RRC | 0001 1001 | 19H | 0 |
| After 2nd RRC | 1000 1100 | 8CH | 1 |
Following the second rotation in detail: A holds 0001 1001, whose LSB is 1. Every bit shifts right, and that 1 wraps around into the MSB position, giving 1000 1100 = 8CH. The carry flag is also set to 1. So the answer is option 2.
The distinction that this question tests is between the two right-rotate instructions:
| Instruction | Bit 7 receives | Carry receives |
|---|---|---|
RRC | The old bit 0 (wrap-around) | A copy of the old bit 0 |
RAR | The old carry | The old bit 0 |
With RAR the carry is part of a 9-bit loop, so the same program would have given a different answer depending on the carry's initial value — a good reason to prefer RRC when a pure 8-bit rotation is wanted.
A useful arithmetic view. Rotating right by one is not the same as dividing by two unless the bit rotated out is zero. Here the first RRC turns 32H (50 decimal) into 19H (25 decimal) — a clean halving, because the LSB was 0. The second RRC would want to halve 25, cannot represent the remainder, and instead wraps the 1 into the MSB, producing 140 decimal. That wrap is exactly why RRC is a rotation and not a shift.
Hence, after execution A contains 8CH.
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