Serial load into an 8-bit parallel-out register (LSB first) Bits 10110111 are shifted in right-most bit first to an 8-bit parallel-out shift register with initial state 11110000. What is the register state after two clock pulses?

Introduction / Context:
Intermediate states in serial loading depend on the initial contents, the input bit order, and the shift direction. Recognizing where the serial input enters helps avoid common off-by-one mistakes.

Given Data / Assumptions:

• Initial register: 11110000 (MSB→LSB).
• Input pattern: 10110111, right-most bit first → input sequence = 1, 1, 1, 0, 1, 1, 0, 1.
• On each clock the register shifts right; the serial input enters the MSB.

Concept / Approach:
At each clock, the current MSB receives the next serial bit, and all other bits shift one place toward the LSB. With the first two inputs both equal to 1, the upper two bits become 1 while the lower bits are previous contents shifted down.

Step-by-Step Solution:
Initial: 1 1 1 1 0 0 0 0.Clock 1 (in=1): 1 1 1 1 1 0 0 0.Clock 2 (in=1): 1 1 1 1 1 1 0 0 → 11111100.

Verification / Alternative check:
Quick simulation or a state table confirms that two leading 1s populate the top two positions while lower positions are the earlier upper bits shifting down.

Why Other Options Are Wrong:
11111000: corresponds to a different intermediate count of clocks.Other options either show no change, the final fully loaded pattern, or a wrong shift direction.

Common Pitfalls:
Misinterpreting where the serial input enters or counting the number of applied clocks incorrectly.

Final Answer:
11111100