Discussion
Consider a 8-bit serial input-parallel output shift registrar operated at clock frequency of 10 MHz. The time taken by the registrar to shift 8 bit number is ..........
Answer: A
The time taken is 800 ns.
A serial-in shift register accepts one bit per clock pulse, so loading an 8-bit number takes 8 clock periods. At 10 MHz each period is
\[ \begin{aligned} T &= \frac{1}{10 \times 10^{6}\ \text{Hz}} = 100\ \text{ns} \\ t &= 8 \times 100\ \text{ns} = 800\ \text{ns} \end{aligned} \]
100 ns is a single clock period, enough for only one bit. 10 µs and 100 ms are far longer than 8 periods of 100 ns and do not follow from the clock frequency.
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