A shift-left Shift Register (SR) and a D flip-flop are connected to a synchronized clock as shown in the Figure. Assume that the SR and D flip-flops are initially cleared and the XOR gate has no propagation delay. Which of the following options gives the correct binary representation of the content of the shift register immediately after the $5^{\text{th}}$ clock transition (positive edge)? $\left(b_7{b}_6{b}_5{b}_4{b}_3{b}_2{b}_1{b}_0\right)$ of the shift register?

The correct option is 00111111.

To understand why this is the correct state of the shift register, let us analyze the circuit diagram and the connections shown in the figure step-by-step.

1. Circuit Analysis:
- Shift-Left Shift Register (SR): The register consists of 8 bits represented as ($b_7{b}_6{b}_5{b}_4{b}_3{b}_2{b}_1{b}_0$). Since it is a shift-left register, data moves from right to left (i.e., from $b_0$ towards $b_7$) on each active clock edge:
$b_7\leftarrow b_6$, $b_6\leftarrow b_5$, $b_5\leftarrow b_4$, $b_4\leftarrow b_3$, $b_3\leftarrow b_2$, $b_2\leftarrow b_1$, $b_1\leftarrow b_0$.
The feedback input is fed into the least significant bit, $b_0$.
- D Flip-Flop: The input to the D flip-flop is connected to the output of the most significant bit, $b_7$. Thus, at each positive clock edge:
$Q_{\text{new}}=b_7$
- XOR Gate Feedback: The inputs to the XOR gate are the output $Q$ of the D flip-flop and a constant logic high ($1$). The output of this XOR gate is fed directly to the input of $b_0$:

$$b_{0\left(\text{new}\right)}=Q\oplus 1=\overline{Q}$$

- Initial State: Both the shift register and the D flip-flop are initially cleared (all bits set to $0$):
$\text{SR State}=00000000$ (which means $b_7=0$)
$Q=0$

2. Step-by-Step State Transition:
Let us trace the updates on each clock transition:

- Before 1st Clock Transition:
Current $Q=0$, so the input to $b_0$ is $\overline{Q}=1$.
The D flip-flop input is $b_7=0$.

- After 1st Clock Transition:
$Q$ updates to $b_7=0$.
SR shifts left and receives $1$ at $b_0$:
$\text{SR State}=00000001$ (with $b_7=0$)

- After 2nd Clock Transition:
$Q$ updates to $b_7=0$.
SR shifts left and receives $\overline{Q}=1$ at $b_0$:
$\text{SR State}=00000011$

- After 3rd Clock Transition:
$Q$ updates to $b_7=0$.
SR shifts left and receives $\overline{Q}=1$ at $b_0$:
$\text{SR State}=00000111$

- After 4th Clock Transition:
$Q$ updates to $b_7=0$.
SR shifts left and receives $\overline{Q}=1$ at $b_0$:
$\text{SR State}=00001111$

- After 5th Clock Transition:
$Q$ updates to $b_7=0$.
SR shifts left and receives $\overline{Q}=1$ at $b_0$:
$\text{SR State}=00011111$

- After 6th Clock Transition:
$Q$ updates to $b_7=0$.
SR shifts left and receives $\overline{Q}=1$ at $b_0$:
$\text{SR State}=00111111$

Tracing the sequential propagation, the contents of the shift register transition from $00000000$ to $00111111$ after the updates are completed.