Final answer:
The time to empty a cylindrical vessel filled to a height of 4H through a small hole at the bottom, according to Torricelli's theorem, will be twice the time it takes to empty the same vessel filled to a height of H.
Step-by-step explanation:
When a cylindrical vessel filled with water up to the height H becomes empty in time t0 due to a small hole at the bottom, the scenario falls under the realms of fluid dynamics in physics, specifically referring to Torricelli's theorem. This principle states that the speed of efflux from a tank depends on the height of the water column above the hole. It can be calculated that the water exits the hole with a speed that it would have if it had fallen freely from the height H without friction. Using this, the time to empty the tank can be related to the height of the water column above the hole.
The volume flow rate through the opening is the same regardless of the size of the opening, and hence the drainage time will scale with the height of the water. The relation that gives the time to empty a tank is proportional to the square root of the height of the water column. Therefore, when the vessel is filled to a height of 4H, knowing that t is proportional to the square root of H (t ∝ sqrt(H)), it will take about twice the time to empty, so the new time to empty will be 2t0.