A right circular cylinder of height 10 cm and circular base of radius 5 cm is filled to half of its capacity by water. The cylinder is tilted so that no water goes out of the cylinder. The angle from the vertical by which the cylinder is tilted is

Given a right circular cylinder with height H = 10 cm and radius R = 5 cm, the diameter D is 10 cm. The cylinder is half-full, so the initial water height is h = 5 cm.

When the cylinder is tilted by an angle θ from the vertical, the water surface remains horizontal. The volume of water in a tilted cylinder is given by the formula V = πR² × (y₁ + y₂)/2, where y₁ and y₂ are the heights of the water levels at the opposite sides. For a half-full cylinder, (y₁ + y₂)/2 = H/2, which means y₁ + y₂ = 10 cm.

The water starts to spill when the higher level y₂ reaches the top (10 cm) and the lower level y₁ reaches the bottom (0 cm). In this limiting case, the geometry forms a triangle where tan θ = (y₂ - y₁) / D = (10 - 0) / 10 = 1. Therefore, θ = 45°. To ensure no water spills, the tilt angle must be not more than 45°.