Accuracy; precision of instruments and errors in measurements

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Question

If the error in the measurement of the volume of sphere is 6 %, then the error (in percent) in the measurement of its surface area will be

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Solution

$\begin{array}{l} \frac{ΔV}{V} = 3 \frac{Δr}{r} or 6 % = 3 \frac{Δr}{r} or \frac{Δr}{r} = 2 % \\ Now surface area s = 4 π r^{2} or \log s = \log 4 π + 2 \log r \\ ∴ \frac{Δs}{s} = 2 \frac{Δr}{r} = 2 \times 2 % = 4 % \end{array}$

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