Solution

Let $x$ be the length of the vertical part of the rope. (We may think of $x$ as the distance of the right endpoint of the rope from where it will be when the rope is completely horizontal.) The mass of the vertical part of the rope is $mx/l$, where $m$ is the mass of the entire rope, and its center of mass is $x/2$ high. Consequently, the gravitational potential energy of the rope is

$$\frac{m\omega^2x^2}{2}=\frac{mgx^2}{2l}\qquad\Rightarrow\qquad\omege=\sqrt{\frac{g}{l}}.$$

The motion of the rope from when it is released to when it is completely horizontal is a quarter of a period of this motion, so

$$\frac{T}{4}=\frac{\pi}{2\omega}=\frac{\pi}{2}\sqrt{l}{g}.$$

Note that the motion of the rope remains simple harmonic withe the same (quarter) period as long as there is a point on the rope at the bend of the pipe. Thus, releasing the rope when a part of it is already horizontal does not change the time it takes to become completely horizontal, because the (quarter) period of simple harmonic motion is independent of its amplitude.