Percentage Error in Volume of a Sphere – Rankers Physics

Error Analysis: Practice Problem & Solution

If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be: [2008]
2%
4%
6%
8%

Solution Explained:

To solve this problem, we apply the core principles of Error Analysis. Understanding the underlying formula is key to arriving at the correct answer below:

Volume of a sphere \(V = \frac{4}{3}pi r^3\). The percentage error in volume is (3) times the percentage error in radius. Given \(\delta r/r \times 100 % = 2 % \). So, Percentage error in \(V = 3 \times 2 % = 6% \).

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