Time Delay Calculation:

The time delay (td) of a shift register can be calculated using the formula:

td = N/f

Where:
- td is the time delay
- N is the number of bits in the shift register
- f is the clock frequency

Given information:
- An 8-bit serial in/serial out shift register is used
- The clock frequency is 2 MHz

Substituting these values into the formula:

td = 8 / (2 * 10^6) = 8 / 2,000,000 = 4 us

Therefore, the time delay achieved by the 8-bit serial in/serial out shift register with a clock frequency of 2 MHz is 4 microseconds (us).

Explanation:

1. Understanding the Components:
- A shift register is a sequential logic circuit that can store and shift data.
- An 8-bit shift register has 8 flip-flops, allowing it to store 8 bits of data.
- Serial in/serial out means that data is input one bit at a time and output one bit at a time, in a serial manner.
- The clock frequency (f) represents the rate at which the shift register operates.

2. Time Delay Calculation:
- The time delay (td) represents the time it takes for the data to propagate through all the stages of the shift register.
- It can be calculated by dividing the number of bits (N) in the shift register by the clock frequency (f).
- In this case, since we have an 8-bit shift register and a clock frequency of 2 MHz, the time delay is calculated as td = 8 / (2 * 10^6).
- Simplifying the expression gives us td = 8 / 2,000,000 = 4 us.

3. Interpretation of the Result:
- The calculated time delay of 4 us means that it takes 4 microseconds for the data to propagate through all the stages of the shift register.
- This time delay is independent of the data being shifted, as it only depends on the number of bits and the clock frequency.
- A shorter time delay would allow for faster data transfer, while a longer time delay would result in slower data transfer.

In conclusion, the correct answer is option 'C' - 4 us, as it accurately represents the time delay achieved by the 8-bit serial in/serial out shift register with a clock frequency of 2 MHz.