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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:
- the total surface area of the can in contact with water when the sphere is in it;
- the depth of water in the can before the sphere was put into the can.
Solution
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`
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