Question:
Assume that a 4-bit serial in/serial out shift register is initially clear. Bits are shifted in from left. We wish to store the nibble 1100. What will be the 4-bit pattern after the second clock pulse?

Answer:
To determine the 4-bit pattern after the second clock pulse, we need to understand how a 4-bit serial in/serial out shift register works and how data is shifted in and out.

Serial In/Serial Out Shift Register:
A serial in/serial out shift register is a type of digital circuit that can store and shift data in a sequential manner. It consists of a chain of flip-flops connected in series, with each flip-flop storing one bit of data. The data can be shifted in and out of the shift register one bit at a time.

Initial State:
In this question, the 4-bit serial in/serial out shift register is initially clear. This means that all the flip-flops in the shift register are reset to 0.

Shifting in Data:
To store the nibble 1100 in the shift register, we need to shift in the bits one by one from left to right.

- First Clock Pulse: When the first clock pulse is applied, the first bit of the nibble (1) is shifted into the first flip-flop of the shift register. The remaining three flip-flops are still in the reset state (0).

- Second Clock Pulse: After the second clock pulse is applied, the second bit of the nibble (1) is shifted into the second flip-flop, while the first bit (1) is shifted from the first flip-flop to the second flip-flop. The remaining two flip-flops are still in the reset state (0).

Pattern after the Second Clock Pulse:
After the second clock pulse, the 4-bit pattern in the shift register will be as follows:

- Fourth Flip-Flop: The fourth flip-flop is still in the reset state (0) as no bits have been shifted into it yet.

- Third Flip-Flop: The third flip-flop contains the first bit of the nibble (1) after the second clock pulse.

- Second Flip-Flop: The second flip-flop contains the second bit of the nibble (1) after the second clock pulse.

- First Flip-Flop: The first flip-flop contains the third bit of the nibble (0) after the second clock pulse.

Therefore, the 4-bit pattern after the second clock pulse will be 0011, which corresponds to the nibble 1100.

Answer:
The correct answer is option 'D' - 0000.