Not quite. But before getting into the math, you'd need to fix up the problem a bit: if the string and the ball are both massless, then there will be nothing left to oscillate once the water drains out. It probably makes the most sense to assume that the ball is a uniform spherical shell with a mass $m$. For simplicity, I'll make two more assumptions:
- The rate at which the water drains is slow, so that the change in mass over each oscillation is negligible (i.e. the process is adiabatic)
- The momentum/energy of the stream of water escaping is also negligible
I'm also using the result which Georg brought up, that the surface of the water remains perpendicular to the pendulum's axis even as it swings.
First I'll compute the volume of water in the ball when it's filled up to a height $h$,
$$V = \int_{-r}^{h-r} \pi(r^2 - z^2)\mathrm{d}z = \pi h^2\biggl(r - \frac{h}{3}\biggr)$$
and then the moment of that water,
$$M z_{w} = \int_{-r}^{h-r} zd\pi (r^2 - z^2)\mathrm{d}z = \pi d h^2 \biggl(hr - r^2 - \frac{h^2}{4}\biggr)$$
To find the overall center of mass, you add this to the moment of the sphere (which is zero since its center of mass is at $z = 0$) and divide by the total mass. Since $M = dV$, we get
$$z_{CM} = \frac{\pi d h^2\bigl(hr - r^2 - h^2/4\bigr)}{(d) \pi h^2\bigl(r - h/3\bigr) + m} = \frac{hr - r^2 - \frac{h^2}{4}}{r - \frac{h}{3} + \frac{m}{\pi h^2 d}}$$
The thing that you're calling effective length in the problem is $l + r - z_{CM}$. Using $m = 1$, $r = 1$, and $d = 1$ as sample values (e.g. if you are using CGS units with lengths in centimeters and masses in grams), a plot looks like this:
The graph does start and end at zero, which means that at the beginning and at the end, the center of mass is at the center of the sphere at those times, as you said. But in the middle, it never goes below about -0.24, which means that the radius of the swing of the center of mass never gets longer than $l + 1.24 r$ (in this example). It won't reach all the way out to $l + 2r$.
However, as Georg also mentioned, approximating the pendulum as a point mass on a string of varying length isn't quite accurate. To do better, we can use the rotational equation $\vec\tau = I\vec\alpha$ to account for the pendulum's finite extent perpendicular to the axis. The torque is produced by the force of gravity acting downward through the center of mass, and after a little rearranging the equation becomes
$$\ddot\theta + \frac{mg r_g}{I}\sin\theta = 0$$
where $I$ is the moment of inertia of the pendulum and $r_g$ is the radius at which the gravitational force acts.
Notice that the behavior of the pendulum in this model is determined entirely by the value of the constant $\omega^2 \equiv \frac{mg r_g}{I}$. For a point pendulum of length $l$, you have $r_g = l$ and $I = ml^2$, so $\omega^2 = \frac{g}{l}$. We can therefore define the "effective length" of a more complex pendulum as the length that gives the same value of $\omega^2$,
$$l_\text{eff} = \frac{g}{\omega^2} = \frac{I}{m r_g}$$
For the dripping pendulum, we already have $r_g$: it's just the radius of the swing of the center of mass, which is what you were calling "effective length" in the problem. But the new definition also incorporates the moment of inertia of the pendulum bob around the pivot point.
$$I = \bigl(I_\text{shell,CM} + m(l + r)^2\bigr) + \bigl(I_\text{water,CM} + M(l + r - z_w)^2\bigr)$$
The first term is the moment of inertia of the spherical shell around its own center of mass, $I_\text{shell,CM} = \frac{2}{3}mr^2$. The third term is the moment of inertia of the water around its own center of mass. It takes the shape of a "chopped" sphere so we have to do the integral to calculate it:
$$I_\text{water,CM} = \iiint x^2\,\mathrm{d}^3 M = \int_{-r}^{h-r}\int_{0}^{2\pi}\int_{0}^{\sqrt{r^2 - z^2}} \bigl((s\cos\phi)^2 + (z - z_w)^2\bigr) d\,s\,\mathrm{d}s\mathrm{d}\phi\mathrm{d}z$$
The second and fourth terms in $I$ come from the parallel axis theorem.
Putting it all together, we get a long but straightforward expression which can be written as:
$$l_\text{eff} = l + \frac{12mrl + 20mr^2 +\pi dh^2\bigl[-\frac{9}{5}h^3 + 24r^2(l+2r)+3h^2(l+5r)-4hr(4l+11r)\bigr]}{12m(l+r) + \pi dh^2\bigl[4l(3r-h)+3h^2-16hr+24r^2\bigr]}$$
(yes, I used Mathematica for the algebra :-P) The graph for $l=10$ looks like this:
The red line is $l_{eff}$, the blue line is the swing radius of the center of mass (the same thing that was in the last graph) just for comparison. Notice that for these values they're pretty similar, and as $r$ shrinks relative to $l$ the curves get even closer together.
In fact, you can show this from the analytical analysis as well. I won't show the complete details, but if you eliminate $h$ and $r$ in favor of $\epsilon = \frac{h}{r}$ and $\beta = \frac{r}{l}$, and do a series expansion in $\beta$, if Mathematica is to be trusted, you wind up with
$$\frac{l_{eff}}{l} = 1 + \beta + \frac{2}{3}(\beta^2 - \beta^3) + P^4(\epsilon)\mathcal{O}(\beta^4)$$
In other words, the amount by which the sphere is filled doesn't even matter until you reach the fourth-order correction in $\frac{r}{l}$.
