Percentage Error in Calculated Volume of Sphere

To find the percentage error in the calculated volume of a sphere given the percentage error in its diameter measurement, we can express the volume $V$ in terms of its diameter $D$.

The formula for the volume $V$ of a sphere in terms of its radius $r$ is: $V = \frac{4}{3}\pi r^3$

Since the diameter $D$ is related to the radius $r$ by $r = \frac{D}{2}$, substituting this into the volume equation gives: $V = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{\pi}{6}D^3$

To relate the fractional change (or relative error) in volume to the fractional change in diameter, we take the natural logarithm on both sides: $\ln V = \ln\left(\frac{\pi}{6}\right) + 3 \ln D$

Differentiating both sides yields: $\frac{dV}{V} = 3 \frac{dD}{D}$

For small relative errors, this can be written in terms of finite errors as: $\frac{\Delta V}{V} = 3 \frac{\Delta D}{D}$

Multiplying by $100$ gives the percentage error relationship: $\left(\frac{\Delta V}{V} \times 100\right)\% = 3 \times \left(\frac{\Delta D}{D} \times 100\right)\%$

Given that the percentage error in diameter measurement is $2\%$: $\frac{\Delta D}{D} \times 100\% = 2\%$

Substituting this value into the equation: $\text{Percentage Error in Volume} = 3 \times 2\% = 6\%$

Thus, the percentage error in the calculated volume of the sphere is $6\%$.

Correct Option: C ($6$)