If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

The correct option is 6%.

Step-by-step Explanation:

1. Identify the relationship between the volume and the radius of a sphere:
The volume $V$ of a sphere of radius $r$ is given by the formula:
$V=\frac{4}{3}\pi r^3$

2. Express the fractional error in volume:
Since $\frac{4}{3}\pi$ is a constant, the fractional error (or relative error) in the volume $V$ depends only on the error in the radius $r$. Taking the natural logarithm on both sides and differentiating, or using standard error propagation rules for power functions, we get:
$\frac{\Delta V}{V}=3·\frac{\Delta r}{r}$

3. Calculate the percentage error:
To find the percentage error, we multiply both sides of the fractional error equation by 100:
$\frac{\Delta V}{V}\times 100=3·(\frac{\Delta r}{r}\times 100)$

4. Substitute the given value:
We are given that the percentage error in the measurement of the radius is 2%, which means:
$\frac{\Delta r}{r}\times 100=2\%$
Substituting this value into our error formula:
$\text{Percentage error in Volume}=3·2\%=6\%$

Therefore, the error in the determination of the volume of the sphere is 6%.