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Question

If the percentage error in measuring the radius of a sphere is 2%, what is the percentage error in its calculated volume (assuming small errors)?

This question was previously asked in
UPTET 2026 Paper 2 Social Studies Question Paper (3-Jul-2026) (Shift 1)
The correct answer is

6%

The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\), so the percentage error in volume relates to the percentage error in radius through the power rule for error propagation.

Since \(V \propto r^3\), the relative error is \(\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}\).

Substituting the given percentage error in radius, \(\frac{\Delta V}{V} \times 100 = 3 \times 2\% = 6\%\).

Hence, the percentage error in the calculated volume is 6%.

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