A uniform rod is partially in water with one end suspended, as shown in figure. The density of the rod is $5/9$ that of water. At equilibrium, what portion of the rod is above water?

$\textbf{(A) } 0.25\\
\textbf{(B) }  0.33\\
\textbf{(C) } 0.5\\
\textbf{(D) } 0.67\\
\textbf{(E) } 0.75$
1. **Identify the forces and torques acting on the rod:**
   - The torque due to gravity ($T_g$) acts at the center of mass of the rod.
   - The torque due to buoyancy ($T_b$) acts at the center of buoyancy, which is the midpoint of the submerged part of the rod.

2. **Set up the torque due to gravity:**
   - The center of mass of the rod is at $\frac{L}{2}$ from the pivot.
   - The mass of the rod is $p_r \cdot L$, where $p_r$ is the density of the rod.
   - The torque due to gravity is:
     \[
     T_g = \frac{L}{2} \cdot p_r \cdot L \cdot g = \frac{L^2 \cdot p_r \cdot g}{2}
     \]

3. **Set up the torque due to buoyancy:**
   - The buoyant force acts at the center of the submerged part of the rod, which is at $\frac{L + \ell}{2}$ from the pivot.
   - The buoyant force is $p_w \cdot (L - \ell) \cdot g$, where $p_w$ is the density of water.
   - The torque due to buoyancy is:
     \[
     T_b = p_w \cdot (L - \ell) \cdot g \cdot \frac{L + \ell}{2}
     \]

4. **Set the torques equal to each other for equilibrium:**
   \[
   \frac{L^2 \cdot p_r \cdot g}{2} = p_w \cdot (L - \ell) \cdot g \cdot \frac{L + \ell}{2}
   \]

5. **Simplify the equation:**
   \[
   L^2 \cdot p_r = p_w \cdot (L - \ell) \cdot \frac{L + \ell}{2}
   \]

6. **Substitute the given density ratio $\frac{p_r}{p_w} = \frac{5}{9}$:**
   \[
   \frac{5}{9} L^2 = (L - \ell)(L + \ell)
   \]

7. **Simplify the right-hand side:**
   \[
   \frac{5}{9} L^2 = L^2 - \ell^2
   \]

8. **Solve for $\ell$:**
   \[
   1 - \frac{\ell^2}{L^2} = \frac{5}{9}
   \]
   \[
   \frac{\ell^2}{L^2} = 1 - \frac{5}{9} = \frac{4}{9}
   \]
   \[
   \frac{\ell}{L} = \frac{2}{3}
   \]

9. **Determine the portion of the rod above water:**
   - The length of the rod above water is $L - \ell$.
   - The portion of the rod above water is:
     \[
     \frac{L - \ell}{L} = 1 - \frac{\ell}{L} = 1 - \frac{2}{3} = \frac{1}{3}
     \]

The final answer is $\boxed{0.33}$