A hollow sphere of external and internal diameter of 8 cm and 4 cm respectively its melted and made into another solid in the shape of a right circular cone of a base diameter of 8 cm find the height of the court
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the inner radii(r) and outer radii(R)of the hollow sphere is 2cm and 4cm respectively
volume of the sphere = 4/3π(R^3–r^3)
= 4/3π(64–8)
= 4/3*22*8 = 704/3
volume of cone of height(h) = 1/3πr^2h
= 1/3π*16*h
given it is Recasted so vol of sphere = vol of cone
704/3 = 352/21 * h
therefore h = 14cm
hence the height of the cone is 14cm
volume of the sphere = 4/3π(R^3–r^3)
= 4/3π(64–8)
= 4/3*22*8 = 704/3
volume of cone of height(h) = 1/3πr^2h
= 1/3π*16*h
given it is Recasted so vol of sphere = vol of cone
704/3 = 352/21 * h
therefore h = 14cm
hence the height of the cone is 14cm
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