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Example 25 - If radius of a sphere is 9 cm with error 0.03 cm

Example 25 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Example 25 If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.Given Radius of sphere = r = 9 cm Error in radius = βˆ†r = 0.03 cm We need to find Error in calculating Volume Let Volume of sphere = V = πŸ’/πŸ‘ 𝝅𝒓^πŸ‘ ∴ We need to find βˆ†V Now, βˆ†V = 𝑑𝑉/π‘‘π‘Ÿ βˆ†r = 𝑑(4/3 πœ‹π‘Ÿ^3 )/π‘‘π‘Ÿ βˆ†r = 4/3 πœ‹ (π‘‘π‘Ÿ^3)/π‘‘π‘Ÿ βˆ†r = 4/3 πœ‹ (3π‘Ÿ^2 )(0.03 ) = 4πœ‹π‘Ÿ^2(0.03) = 4πœ‹(9)2 (0.03) = 9.72𝝅 cm3 Hence, approximate error in calculating volume is 9.72πœ‹ 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.