Question

a cylindrical can whose base is horizontal and of radius 3.5cm contains sufficient water so that if a sphere is placed in the can, the water just covers the sphere. given that the can just fits into the can, calculate:
i.) the surface area of 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.
give your answers as proper fractions and take $\pi$=22/7

Answer 1

see diagram.

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

Volume of of cylinder up to the height 2R in the cylinder = π R² * 2R = 2π R³
Volume of sphere = 4πR³/3

Volume of water in the gap between cylinder and sphere
           = 2π R³ - 4π R³/3 = 2π R³ / 3

2)   Depth of water in the can before sphere is put inside it
                     = volume of water/area of cross section of can
                     = (2 π 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
                            = π R² + 2 π R * 2 R = 5 π R²
                            = 5 * 22/7 * 3.5²  cm²
                           = 110 /7 * (7/2)²  cm²   =  385 / 2 cm²