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Example 7 - A solid toy is in form of a hemisphere surmounted - Volume - Added

Example 7 - Chapter 13 Class 10 Surface Areas and Volumes - Part 2
Example 7 - Chapter 13 Class 10 Surface Areas and Volumes - Part 3 Example 7 - Chapter 13 Class 10 Surface Areas and Volumes - Part 4

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Example 7 A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy.(Take = 3.14) Volume of the toy = Volume of cone + Volume of hemisphere Volume of cone Height of cone = OA = h = 2 cm Diameter of cone = BC = 4 cm So, radius = r = /2 "=" 4/2 = 2 cm Volume of cone = 1/3 2 = 1/3 (2)2 (2) = 8 /3 cm3 Volume of hemisphere Diameter of hemisphere = BC = 4 cm So, radius = r = /2 "=" 4/2 = 2 cm Volume of hemisphere = 2/3 3 = 2/3 (2)3 = 2/3 2 2 2 = 16 /3 cm3 Volume of the toy = Volume of cone + Volume of hemisphere = 8 /3+16 /3 = (8 + 16 )/3 = 24 /3 = (24 3.14)/3 = 8 3.14 = 25.12 cm3 Now, we need to find the difference of the volumes of the cylinder and the toy. Volume of cylinder Cylinder circumscribes the toy. So, Diameter of cylinder = HG = BC = 4 cm So, radius = r = /2 "=" 4/2 = 2 cm Height of cylinder = OA + OP = Height of cone + Radius of hemisphere = 2 + 2 = 4 cm Volume of cylinder = 2 = 3.14 (2)2 (4) = 3.14 4 4 = 50.24 Therefore, Difference of the volume = Volume of cylinder Volume of toy = 50.24 25.12 = 25.12 cm3 Hence, difference of the volume = 25.12 cm3

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.