A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :

(i) the total surface area of the can in contact with water when the sphere is in it.

(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.

(i) Radius of a cylindrical can (r) = 3.5 cm

Radius of sphere (R) = r = 3.5 cm

Height of water level in can = 7 cm.

Height of cylinder (h) = 7 cm.

Total surface area of can in contact with water (T) = Curved surface area of cylinder + base area of cylinder.

$T = 2πrh + πr^2 \\[1em] = πr(2h + r) \\[1em] = \dfrac{22}{7} \times 3.5 \times (2 \times 7 + 3.5) \\[1em] = 22 \times 0.5 \times (14 + 3.5) \\[1em] = 11 \times 17.5 \\[1em] = 192.5 \text{ cm}^2.$

Hence, the surface area of can in contact with water is 192.5 cm².

(ii) Let the depth of the water before the sphere was put be d.

Volume of cylindrical can = Volume of sphere + Volume of water.

$πr^2h = \dfrac{4}{3}πR^3 + πr^2d \\[1em] πr^2h = \dfrac{4}{3}πr^3 + πr^2d \\[1em] πr^2h = πr^2\Big(\dfrac{4}{3}r + d\Big) \\[1em] h = \dfrac{4}{3}r + d \\[1em] d = h - \dfrac{4}{3}r \\[1em] d = 7 - \dfrac{4}{3} \times 3.5 \\[1em] d = 7 - \dfrac{14}{3} \\[1em] d = \dfrac{21 - 14}{3} = \dfrac{7}{3} \text{ cm}.$

Hence, the depth of water before sphere was put was $\dfrac{7}{3}$ cm.