Question

If the error in the measurement of the surface area of a sphere is 1.2%, then the error in the determination of the volume of the sphere is

• A. 2.4%
• B. 1.8%
• C. 1.2%
• D. 0.6%

Hint:

For related measurements, use relative errors via differentiation or logarithmic differentiation to propagate errors. Percentage error is relative error times 100.

Answer 1

The Correct Option is B

For a sphere with radius \( r \), the surface area is \( S = 4\pi r^2 \) and the volume is \( V = \frac{4}{3}\pi r^3 \). Given the percentage error in surface area is 1.2%, i.e., \( \frac{\delta S}{S} \times 100 = 1.2 \), find the percentage error in volume, \( \frac{\delta V}{V} \times 100 \). Use logarithmic differentiation or relative errors: \[ S = 4\pi r^2 \implies \ln S = \ln (4\pi) + 2 \ln r \] \[ \frac{dS}{S} = 2 \frac{dr}{r} \implies \frac{\delta S}{S} \approx 2 \frac{\delta r}{r} \] \[ V = \frac{4}{3}\pi r^3 \implies \ln V = \ln \left( \frac{4}{3}\pi \right) + 3 \ln r \] \[ \frac{dV}{V} = 3 \frac{dr}{r} \implies \frac{\delta V}{V} \approx 3 \frac{\delta r}{r} \] Relate the errors: \[ \frac{\delta V}{V} = \frac{3 \frac{\delta r}{r}}{2 \frac{\delta r}{r}} \cdot \frac{\delta S}{S} = \frac{3}{2} \cdot \frac{\delta S}{S} \] \[ \frac{\delta V}{V} \times 100 = \frac{3}{2} \cdot 1.2 = 1.8% \] Option (2) is correct. Options (1), (3), and (4) do not match.