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

Ex 13.2, 5 - Chapter 13 Class 10 Surface Areas and Volumes - Part 2
Ex 13.2, 5 - Chapter 13 Class 10 Surface Areas and Volumes - Part 3

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Ex 12.2, 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 = (π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘“π‘™π‘œπ‘€π‘› π‘œπ‘’π‘‘)/(π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 1 π‘™π‘’π‘Žπ‘‘ π‘ β„Žπ‘œπ‘‘) 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 πœ‹π‘Ÿ^2 β„Ž = 1/3Γ—22/7Γ—5Γ—5Γ—8 = 4400/21 cm3 Volume of water flown out = 1/4 Γ— Volume of cone = 1/4 Γ— 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 Ο€r3 = 4/3Γ—22/7Γ—(0.5)3 = 11/21 cm3 Now, Number of lead shots = (π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘“π‘™π‘œπ‘€π‘› π‘œπ‘’π‘‘)/(π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 1 π‘™π‘’π‘Žπ‘‘ π‘ β„Žπ‘œπ‘‘) = (1100/21)/(11/21) = 1100/21 Γ— 21/11 = 100 Hence there are 100 such lead shots.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.