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 contect with water when the sphere is in it. (ii) the depth of water in the can before the sphere was put into it. [4]
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If the can's base has a radius of 3.5 cm and the can is cylindrical, than the cylinder also has a radius of 3.5 cm. The sphere that just fits into the can must also have a radius of 3.5 cm, and therefore a diameter of 7 cm. So if the water just barely covers the sphere, assuming the sphere is touching the base of the can and not floating, that means the cylinder is 7 cm high. So the question is, what is the combined surface area of the cylindrical can's side and its base, if its radius is 3.5 cm and its height is 7 cm?
Surface area of base: A = pi r^2 = pi (3.5 cm)^2 = 12.25 pi cm^2
Surface area of cylindrical wall: A = 2 pi r h = 2 pi (3.5 cm) (7 cm) = 49 pi cm^2
Total surface area = 12.25 pi cm^2 + 49 pi cm^2 = 61.25 pi cm^2
Converted to a decimal approximation, 61.25 pi cm^2 = 192.423 cm^2
Answer: 192.423 cm^2
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Surface area of base: A = pi r^2 = pi (3.5 cm)^2 = 12.25 pi cm^2
Surface area of cylindrical wall: A = 2 pi r h = 2 pi (3.5 cm) (7 cm) = 49 pi cm^2
Total surface area = 12.25 pi cm^2 + 49 pi cm^2 = 61.25 pi cm^2
Converted to a decimal approximation, 61.25 pi cm^2 = 192.423 cm^2
Answer: 192.423 cm^2
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