If the error in the measurement of the radius of a sphere is 2%2\%, then the error in the determination of its volume will be:

Model Answer & Options

Source: Extra Practice

4%

2%

8%

6%

Explanation

The volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3. The percentage error in VV is ΔVV×100=3(Δrr×100)\frac{\Delta V}{V} \times 100 = 3 \left( \frac{\Delta r}{r} \times 100 \right). Given the percentage error in rr is 2%2\%, the percentage error in V=3×2%=6%V = 3 \times 2\% = 6\%. Constants like 4/34/3 and π\pi do not contribute to the error.

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