Question

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:

1. the total surface area of the can in contact with water when the sphere is in it;
2. the depth of water in the can before the sphere was put into the can.

Answer 1

 
Radius of the base of the cylindrical can = 3.5 cm

i. When the sphere is in can, then total surface area of the can = Base area + Curved surface area 

= `pir^2 + 2pirh` 

= `(22/7 xx 3.5 xx 3.5) + (2 xx 22/7 xx 3.5 xx 7)` 

= `77/2+154` 

= 38.5 + 154 

= 192.5 cm² 

ii. Let depth of water = x cm

When sphere is not in the can, then volume of the can = Volume of water + Volume of sphere 

`=> pir^2h + pir^2x xx + 4/3pir^3` 

`=> pir^2h + pir^2(x + 4/3r)` 

`=> h = x + 4/3r` 

`=> x = h - 4/3r` 

`=> x = 7 - 4/3 xx 7/2` 

`=> x = 7 - 14/3` 

`=> x = (21 - 14)/3` 

`=> x = 7/3` 

`=> x = 2 1/3 cm`