Mass of the cup G Quantity of water the cup will hold L The

The aluminum cup shown has a uniform thickness of 0.1 cm. Knowing that the density ofaluminum is 2.8 g/cm, determine the surface area and mass of the cup. When the cup is turned vertically and ignoring the cup thickness, what is the quantity of water (liters) that the cup will hold? Note: 1 liter 1000 cm3 50 26cm 20cm t-- -- 58 cm 22.4 cm 19.8 cm 9 cm

Solution

For frustum of a cone, Slant height s = sqrt ((r₁ - r₂)² + h²)

For smallest frustum, we get s1 = sqrt ((13-10)² + 22.4²) = 22.6 cm

For middle frustum, we get s2 = sqrt ((25-13)² + 9²) = 15 cm

For biggest frustum, we get s3 = sqrt ((29-25)² + 19.8²) = 20.2 cm

For frustum of a cone, Lateral surface area L = pi * (r₁ + r₂) * s

For smallest frustum, we get L1 = pi* (13+10) * 22.6 = 519.8*pi cm²

For middle frustum, we get L2 = pi* (25+13) * 15 = 570*pi cm²

For biggest frustum, we get L3 = pi* (29+25) * 20.2 = 1090.8*pi cm²

For base, surface area L4 = pi*10² = 100*pi cm²

Total surface area L = 519.8*pi + 570*pi + 1090.8*pi + 100*pi

L = 2280.6*pi cm²

L = 2280.6*3.14

Surface area L = 7161.084 cm²

Volume of material used to make the cup = Surface area * thickness

= 7161.084 * 0.1

= 716.108 cm³

Mass of cup = Density*Volume of material

= 2.8 * 716.108

Mass of cup m = 2005.1 grams

For frustum of a cone, Volume = (1/3) pi * h * (r₁² + r₂² + r₁r₂)

For smallest frustum, V1 = (1/3) pi * 22.4 * (13² + 10² + 13*10) = 2979.2*pi cm³

For middle frustum, V2 = (1/3) pi * 9 * (25² + 13² + 25*13) = 3357*pi cm³

For biggest frustum, V3 = (1/3) pi * 19.8 * (29² + 25² + 29*25) = 14460.6*pi cm³

Total volume of cup V = 2979.2*pi + 3357*pi + 14460.6*pi

V = 20796.8*pi cm³

V = 20796.8 * 3.14 cm³

V = 65301.952 cm³

Volume of cup V = 65.302 L