A cylindrical vessel whose base is horizontal and is of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the Cylinder exactly. The depth of water in the vessel before the sphere was put, is

Height of water after ball is immersed = 3.5 × 2 = 7 cm
$$\eqalign{ & \Rightarrow {\text{Volume of water}} = \pi {r^2}h - \frac{4}{3}\pi {r^3} \cr & = \pi {r^2}\left( {h - \frac{4}{3}r} \right) \cr & = \frac{{22}}{7} \times 3.5 \times 3.5\left( {7 - \frac{4}{3} \times 3.5} \right) \cr & = 11 \times 3.5\left( {\frac{7}{3}} \right) \cr & = \frac{{269.5}}{3}{\text{ c}}{{\text{m}}^3} \cr} $$
Volume of water before ball was immersed
$$\eqalign{ & \Rightarrow \pi {\left( {3.5} \right)^2} \times h = \frac{{269.5}}{3} \cr & h = \frac{{269.5 \times 7}}{{3 \times 3.5 \times 3.5 \times 22}} \cr & h = \frac{7}{3}{\text{ cm}} \cr} $$