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.

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Solution

Cylinder: $Height = 7 \ cm, Radius = 3.5 \ cm$

Sphere: $Radius = 3.5 \ cm$

Total Surface Area = Curved Surface area of Cylinder + Area of the base

$= 2 \pi r h + \pi r^2$

$= 2 \times$ $\frac{22}{7}$ $\times 3.5 \times 7 +$ $\frac{22}{7}$ $\times 3.5^2$

$= 154+38.5 = 192.5 \ cm^2$

Let the depth of the water $= h$

Therefore: $\pi \times 3.5^2 \times (7 - h) =$ $\frac{4}{3}$ $\pi \times 3.5^3$

$\Rightarrow 7 - h =$ $\frac{4}{3}$ $\times 3.5$

$\Rightarrow h = 7 -$ $\frac{4}{3}$ $\times 3.5 = 2$ $\frac{1}{3}$ $\ cm$

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Cylinder: Sphere: Total Surface Area = Curved Surface area of Cylinder + Area of the base Let the depth of the water Therefore: