Question

A solid sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?

Answer 1

Diameter of the sphere = 6 cm

Radius of the sphere, r = 3 cm

∴ Volume of sphere =

\[\frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left( 3 \right)^3 = 36\pi {cm}^3\]

Diameter of the cylinderical vessel = 12 cm

Radius of the cylinderical vessel, R = 6 cm

Let the rise in the water level in the cylinder vassel be h cm.

∴ Volume of the water displaced in the cylindrical vassel =

\[\pi\]R²h =\[\pi\](6)²h = 36 \[\pi\]h cm³

 

It is given that the sphere is completely submerged in the water in the vassel.

∴ Volume of water displaced = Volume of the sphere

⇒ 36 \[\pi\]h = 36\[\pi\]

⇒ h = 1 cm

Thus, the rise in the water level in the cylindrical vassel is 1 cm.