The percentage error in the calculated volume of a sphere, if there...

The volume of a sphere in terms of its diameter $D$ is given by $V = \dfrac{\pi}{6}D^3$

Taking the natural logarithm on both sides, we get $\ln V = \ln\left(\dfrac{\pi}{6}\right) + 3 \ln D$

Differentiating for small errors, we obtain $\dfrac{\Delta V}{V} = 3 \dfrac{\Delta D}{D}$

Multiplying by $100$ to find the percentage error, $\dfrac{\Delta V}{V} \times 100 = 3 \left(\dfrac{\Delta D}{D} \times 100\right)$

Given that the percentage error in the diameter is $2\%$, we have $\dfrac{\Delta D}{D} \times 100 = 2$

Therefore, the percentage error in the volume is $3 \times 2 = 6\%$

Answer: $6$