A hollow sphere is filled with water through a small hole in it. It is hung by a long thread and, as the water flows out of the hole at the bottom, one finds that the period of oscillation first increases and then decreases. Explain.

Hollow sphere acts as a physical pendulum as the sphere is filled with water. The time period T for a physical pendulum is defined as,
T = 2π√I/mgd
Here, I is the rotational inertia of the sphere, m is the mass of the sphere which is filled with water, g is the acceleration due to gravity and d is the distance from the pivot to the center of mass.
In the above equation T = 2π√I/mgd,
I/mgd = mk²/mgd (Since, I = mk²)
= k²/gd
So, k²/d is the deciding factor here.
As the water flow out from the distance of center of mass from axis, that increases up to a certain limit and radius of gyration increases in square. So there is net increase in period. But after certain limit d decreases, because, center of mass shifted toward heavy portion of sphere, and rate in decrease in d is increases. But the increment in radius of gyration remains constant. Thus after a certain limit k²/d will decrease. Therefore the period of oscillation first increases and then decreases