Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is

$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad
\textbf{(C)}\ \sqrt{3} \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \dfrac{4}{\sqrt{3}}$
1. **Identify the given information and the goal:**
   - Two rays with a common endpoint \( O \) form a \( 30^\circ \) angle.
   - Point \( A \) lies on one ray, and point \( B \) lies on the other ray.
   - The length \( AB = 1 \).
   - We need to find the maximum possible length of \( OB \).

2. **Use the Law of Sines in triangle \( OAB \):**
   - In \(\triangle OAB\), we know \(\angle AOB = 30^\circ\).
   - Let \(\angle OAB = \theta\) and \(\angle OBA = 180^\circ - 30^\circ - \theta = 150^\circ - \theta\).

3. **Apply the Law of Sines:**
   \[
   \frac{AB}{\sin(\angle AOB)} = \frac{OB}{\sin(\angle OAB)} = \frac{OA}{\sin(\angle OBA)}
   \]
   Given \( AB = 1 \) and \(\angle AOB = 30^\circ\), we have:
   \[
   \frac{1}{\sin(30^\circ)} = \frac{OB}{\sin(\theta)}
   \]
   Since \(\sin(30^\circ) = \frac{1}{2}\), this simplifies to:
   \[
   \frac{1}{\frac{1}{2}} = \frac{OB}{\sin(\theta)}
   \]
   \[
   2 = \frac{OB}{\sin(\theta)}
   \]
   \[
   OB = 2 \sin(\theta)
   \]

4. **Maximize \( OB \):**
   - The maximum value of \(\sin(\theta)\) is 1.
   - Therefore, the maximum value of \( OB \) is:
   \[
   OB = 2 \times 1 = 2
   \]

5. **Conclusion:**
   - The maximum possible length of \( OB \) is \( 2 \).

The final answer is \(\boxed{2}\).