Errors in measurementMedium
Question
The percentage error in the calculated volume of a sphere, if there is $2\%$ error in its diameter measurement, is $\_\_\_\_$ .
Options
A.1
B.2
C.6
D.8
Solution
Sol. [Key Concept: For a sphere $V = \frac{4}{3}\pi\ R^{3}$, so $\frac{\Delta\ V}{V} = \frac{3\Delta\ R}{R} = 3 \times 2\% = 6\%$.]
$$\ v = \frac{4}{3}\pi R^{3}$$
$$\begin{matrix} & \frac{\Delta v}{v} = \frac{3\Delta R}{R} \\ & \ = 6\% \end{matrix}$$
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