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Ex 13.2, 5 - A vessel is in form of an inverted cone - Ex 13.2

  1. Chapter 13 Class 10 Surface Areas and Volumes
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Ex 13.2, 5 A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. Number of lead shots = (๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ ๐‘“๐‘™๐‘œ๐‘ค๐‘› ๐‘œ๐‘ข๐‘ก)/(๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘œ๐‘“ 1 ๐‘™๐‘’๐‘Ž๐‘‘ ๐‘ โ„Ž๐‘œ๐‘ก) Volume of water flown out Volume of water flown out = 1/4 of volume of cone Given radius of cone = r = 5 cm height of cone = h = 8 cm Volume of cone = 1/3 ๐œ‹๐‘Ÿ^2 โ„Ž = 1/3ร—22/7ร—5ร—5ร—8 = 4400/21 cm3 Volume of water flown out = 1/4 ร— Volume of cone = 1/4 ร— 4400/21 = 1100/21 cm3 Volume of 1 lead shot Lead shot is in from of sphere with Radius = r = 0.5 cm Volume of 1 lead shot = Volume of sphere = 4/3 ฯ€r3 = 4/3ร—22/7ร—(0.5)3 = 11/21 cm3 Now, Number of lead shots = (๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘œ๐‘“ ๐‘ค๐‘Ž๐‘ก๐‘’๐‘Ÿ ๐‘“๐‘™๐‘œ๐‘ค๐‘› ๐‘œ๐‘ข๐‘ก)/(๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘œ๐‘“ 1 ๐‘™๐‘’๐‘Ž๐‘‘ ๐‘ โ„Ž๐‘œ๐‘ก) = (1100/21)/(11/21) = 1100/21 ร— 21/11 = 100 Hence there are 100 such lead shots.

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