A 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 just completely submerged in water, then the rise of water level in the cylindrical vessel is:

$$\eqalign{ & {\text{Volume of sphere}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3}\pi \times 3 \times 3 \times 3 \cr & = 36\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{If the water level rises by H cm}} \cr & \pi {R^2}H = 36\pi \cr & 6 \times 6 \times H = 36 \cr & H = 1{\text{ cm}} \cr} $$