a trophy awarded to the best student in the class in the form of solid cylinder​ mounted on a solid hemisphere with the same radius and is made from some metal. this trophy is mounted on a wooden cuboid. the diameter of hemisphere is 21cm and the total height of the trophy is 24.5cm. find the weight of the metal used in making the trophy, if the weight of 1cm3 of the metal is 1.2g. (use pie=22/7)

Hello Tri Pragya, total height = 24.5 cm
Radius of hemi sphere = d/2 = 21/2 = 10.5 cm
Hence height of the cylinder = 24.5 - 10.5 = 14 cm
Volume of metal used = 2/3 pi r^3 + pi r^2 h = pi r^2 (2/3*r + h)
= 22/7 * (10.5)^2 (7+14) = 66 * 10.5 * 10.5= 7276.5 cm
Weight for 1 cm3 = 1.2 g
So for 7276.5 cm3 = 7276.5 * 1.2 = 8731.8 g
Hence weight of the metal used = 8 kg 731.8 g 
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Hi dude... total height = 24.5 cm Radius of hemi sphere = d/2 = 21/2 = 10.5 cm Hence height of the cylinder = 24.5 - 10.5 = 14 cm Volume of metal used = 2/3 pi r^3 + pi r^2 h = pi r^2 (2/3*r + h) = 22/7 * (10.5)^2 (7+14) = 66 * 10.5 * 10.5 = 7276.5 cm3 Weight for 1 cm3 = 1.2 g So for 7276.5 cm3 = 7276.5 * 1.2 = 8731.8 g Hence weight of the metal used = 8 kg 731.8 g
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mass=(volume of cylinder +vol of hemisphere)×1.2 total height is 24.5cm therefore height of cylinder is 24.5-10.5=14 Therefore Pi Square r h + 2 by 3 Pi R cube ×1.2 radius is 10.5cm height is 14cn ans 8kg
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