Example 25 - If radius of a sphere is 9 cm with error 0.03 cm - Finding approximate value- Statement questions


  1. Chapter 6 Class 12 Application of Derivatives
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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. 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 = ﷐4﷮3﷯𝜋﷐𝑟﷮3﷯ We need to find error in calculating the volume i.e ∆V ∆V = ﷐𝑑𝑣﷮𝑑𝑟﷯ ∆r = ﷐𝑑﷐﷐4﷮3﷯ 𝜋﷐𝑟﷮3﷯﷯﷮𝑑𝑟﷯ ∆r ∆V = ﷐4﷮3﷯𝜋﷐𝑑﷐𝑟﷮3﷯﷮𝑑𝑟﷯ ∆r ∆V = ﷐4﷮3﷯𝜋 ﷐3﷐𝑟﷮2﷯ ﷯﷐0.03 ﷯ = 4𝜋﷐𝑟﷮2﷯(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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