This question was asked inGPSC AE (Electrical) Class-2 R&B Department Paper 2025 – Series MQuestion 52 of 179 published · Gujarat Public Service CommissionOpen paper
Q161ModerateSequential Circuits

The group of bits 10110111 is serially shifted (rightmost bit first) into an 8-bit parallel output shift register with an initial state 11110000. After two clock pulses, the register contains ____________

Answer & step-by-step solution
Correct answer: D. 11111100
Quick explanation

A serial-in/parallel-out register accepts one incoming bit per clock. With the rightmost bit transmitted first, perform two serial shifts from the stated initial register contents; the answer follows from those two state transitions.

new register = shifted(old register) + next serial bit

Concept & Reasoning

The phrase 'rightmost bit first' fixes the transmission order. Each clock causes the existing register contents to move one position while one new serial bit enters. Only two of the eight incoming bits have entered after two pulses, so most of the initial state is still present. This is why stepwise register tracing is essential.

How to Solve It in the Exam

Write the first two transmitted bits explicitly before shifting.

Important Exam Point

Check whether the serial stream is LSB-first or MSB-first before the first clock.

Quick Trick

Write the first two transmitted bits explicitly before shifting.

Common Mistake

Reading 10110111 left-to-right despite the instruction 'rightmost bit first'.

Solution diagram – The group of bits 10110111 is serially shifted (rightmost bit first) into an…
Exam tipCheck whether the serial stream is LSB-first or MSB-first before the first clock.
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