JEE MainPhysicsEasyError analysis

Error in the measurement of radius of a sphere is1.What is the
percentage error in the calculated value of its volume?

Show answer and solution

V = \frac{4}{3}\pi r^{3}

Take natural log on both side

ln(V) = ln\left( \frac{4}{3}\pi \right) + 3ln(r)

Differentiate both side

\frac{\Delta V}{V} = 3\frac{\Delta r}{r}

\%\frac{\Delta V}{V} = 3 \times 1 = 3\%

Source: Grass-G / P-05 Physics paper 10 October.docx

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