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14. A cylindrical can, whose base is horizontal andof radius 3.5 cm, contains sufficient water sothat when a sphere is placed in the can, thewater just covers the sphere. Given that thesphere just fits into the can, calculate:(i) the total surface area of the can in contactwith water when the sphere is in it;(ii) the depth of water in the can before thesphere was put into the can.

Answer»

In the cylinderof radius R, the sphere just fits. So its radius is R.

Volume of of cylinder up to the height 2R in the cylinder =πR² * 2R = 2π R³Volume of sphere = 4πR³/3

Volume of water in the gap between cylinder and sphere = 2πR³-4πR³/3 = 2πR³ /3

2) Depth of water in the can before sphere is put inside it = volume of water/area of cross section of can = (2π R³ / 3 )/π R² = 2 R/3 = 2 * 3.5 /3 = 7/3 cm

1) Surface area surface area of can in contact with water = flat surface + curved surface =πR² + 2π R * 2 R = 5π R² = 5 * 22/7 * 3.5² cm² = 110 /7 * (7/2)² cm² = 385 / 2 cm²



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