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 water in the can before the sphere was put into the can.

(i) Given,

Radius of base of cylindrical can (R) = 3.5 cm

Since,

When a sphere is placed in the can, the water just covers the sphere.

∴ Height (H) = 2R = 7 cm.

Total surface area = 2πRH + πR²

= πR(2H + R)

= $\dfrac{22}{7} \times 3.5 \times (2 \times 7 + 3.5)$

= 22 × 0.5 × (14 + 3.5)

= 192.5 cm².

Hence, the total surface area of the can in contact with water when the sphere is in it = 192.5 cm².

(ii) Let the depth of the water be h cm in the can.

Volume of water = Volume of cylinder - Volume of sphere

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

Hence, depth of water in the can before the sphere was put into the can = $2\dfrac{1}{3}$ cm.