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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


Transcript

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 = (𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤𝑛 𝑜𝑢𝑡)/(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 1 𝑙𝑒𝑎𝑑 𝑠ℎ𝑜𝑡) 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 𝜋𝑟^2 ℎ = 1/3×22/7×5×5×8 = 4400/21 cm3 Volume of water flown out = 1/4 × Volume of cone = 1/4 × 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 πr3 = 4/3×22/7×(0.5)3 = 11/21 cm3 Now, Number of lead shots = (𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑓𝑙𝑜𝑤𝑛 𝑜𝑢𝑡)/(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 1 𝑙𝑒𝑎𝑑 𝑠ℎ𝑜𝑡) = (1100/21)/(11/21) = 1100/21 × 21/11 = 100 Hence there are 100 such lead shots.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.