A cylindrical vessel 60 cm in diameter is partially filled with water. A sphere of diameter 36 cm is dropped into it and is fully submerged in water. Find the increase in the level of water in the vessel.

Radius of the sphere, r = $\dfrac{\text{diameter}}{2} = \dfrac{36}{2}$ = 18 cm

Radius of cylinder, R = $\dfrac{\text{diameter}}{2} = \dfrac{60}{2}$ = 30 cm

Let height of water raised be h cm.

Volume of water rise in cylinder = Volume of sphere

$\Rightarrow π\text{R}^2\text{h} = \dfrac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \dfrac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{h} = \dfrac{4}{3} \times \dfrac{\text{r}^3}{\text{R}^2} \\[1em] \Rightarrow \text{h} = \dfrac{4}{3} \times \dfrac{18^3}{30^2} \\[1em] \Rightarrow \text{h} = \dfrac{4}{3} \times \dfrac{5832}{900} \\[1em] \Rightarrow \text{h} = \dfrac{23328}{2700} \\[1em] \Rightarrow \text{h} = 8.64 \text{ cm.}$

Hence, the height by which water level raised is 8.64 cm.