Error Analysis: Practice Problem & Solution
The error in measurement of radius of a sphere is 0.1% then error in its volume is:
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 \(\Delta r/r times 100% = 0.1%\). So, Percentage error in \(V = 3 \times 0.1% = 0.3% \).
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