# Example 25

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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. Let r be the radius of the sphere Given r = 9 cm Error in measurement of radius = ∆r ∆r = 0.03 cm Volume of the sphere = V = 43𝜋𝑟3 We need to find error in calculating the volume i.e ∆V ∆V = 𝑑𝑣𝑑𝑟 ∆r = 𝑑43 𝜋𝑟3𝑑𝑟 ∆r ∆V = 43𝜋𝑑𝑟3𝑑𝑟 ∆r ∆V = 43𝜋 3𝑟2 0.03 = 4𝜋𝑟2(0.03) = 4𝜋(9)2 (0.03) = 9.72𝜋 cm3 Hence, approximately error in calculating the volume is 9.72𝝅 cm3

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Example 25 You are here

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Chapter 6 Class 12 Application of Derivatives

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .