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Ex 13.3, 6 - How many silver coins, 1.75 cm in diameter - Conversion of one shape to another

Ex 13.3, 6 - Chapter 13 Class 10 Surface Areas and Volumes - Part 2
Ex 13.3, 6 - Chapter 13 Class 10 Surface Areas and Volumes - Part 3

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Question 6 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm 10 cm 3.5 cm? Number of coins = ( )/( 1 ) Volume of cuboid Length (l) = 5.5 cm Breadth (b) = 10 cm Height (h) = 3.5 cm Volume of cuboid = = 5.5 10 3.5 = 192.5 cm3 Volume of 1 coin Coin is in shape of cylinder with Diameter = 1.75 cm Radius = Diamete /2 = 1.75/2 cm= 0.875 cm = 875/1000 cm Height = 2 mm = 2 1/10 cm = 2/10 cm Volume of 1 coin = 2 = 22/7 (875/1000)^2 2/10 = 22/7 875/1000 875/1000 0.2 = 0.48125 cm3 Volume of cuboid = ๐‘™๐‘โ„Ž = 5.5 ร—10ร—3.5 = 192.5 cm3 Volume of 1 coin Coin is in shape of cylinder with Diameter = 1.75 cm Radius = Diamete๐‘Ÿ/2 = 1.75/2 cm= 0.875 cm = 875/1000 cm Height = 2 mm = 2 ร— 1/10 cm = 2/10 cm Volume of 1 coin = ๐œ‹๐‘Ÿ2โ„Ž = 22/7ร—(875/1000)^2ร— 2/10 = 22/7ร—875/1000 ร—875/1000ร—0.2 = 0.48125 cm3 Volume of cuboid = ๐‘™๐‘โ„Ž = 5.5 ร—10ร—3.5 = 192.5 cm3 Volume of 1 coin Coin is in shape of cylinder with Diameter = 1.75 cm Radius = Diamete๐‘Ÿ/2 = 1.75/2 cm= 0.875 cm = 875/1000 cm Height = 2 mm = 2 ร— 1/10 cm = 2/10 cm Volume of 1 coin = ๐œ‹๐‘Ÿ2โ„Ž = 22/7ร—(875/1000)^2ร— 2/10 = 22/7ร—875/1000 ร—875/1000ร—0.2 = 0.48125 cm3 Number of coins = ( )/( 1 ) = 192.5/0.48125 = 1925/4.8125 = 19250000/48125 = 400 Hence, number of coins = 400

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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.