If error in radius is 3%, what is error in volume of sphere ?

The correct option is 9%.

To find the percentage error in the volume of a sphere, we start with the formula for the volume $V$ of a sphere in terms of its radius $r$:

$V=\frac{4}{3}\pi r^3$

Taking the natural logarithm on both sides of the equation:

$\ln \left(V\right)=\ln \left(\frac{4}{3}\pi \right)+3\ln \left(r\right)$

Now, differentiating both sides to find the relation between the relative changes, we get:

$\frac{dV}{V}=0+3\frac{dr}{r}$

To express this in terms of percentage errors, we multiply both sides of the equation by 100:

$\left(\frac{dV}{V}\times 100\right)=3\times \left(\frac{dr}{r}\times 100\right)$

This indicates that the percentage error in the volume is 3 times the percentage error in the radius.

Given that the percentage error in the radius $\left(\frac{dr}{r}\times 100\right)$ is 3%:

$\text{Percentage error in volume}=3\times 3\%=9\%$