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 :

6 %

Volume of sphere = \(\frac{4}{3}\)\(\pi\)R³

\(\frac{\Delta V}{V} = \frac{3 \Delta R}{R} \)

Percentage error:

(\(\frac{\Delta V}{V}×100) = 3(\frac{\Delta R}{R} ×100)\)

                                          =\(3 ×2%\)

                                          = 6%