Question

Example 83. The error in the measurement of radius of a
sphere is 2%. What would be the error in the volume of the
sphere?
Solution. Given :
3 x 100 = 3 x 2 = 6%.
here why didn't we take /add 4/3π ??​

Answer 1

Explanation:

Percentage error in radius is given as 2% i.e.  rΔr×100=2 %

Volume of sphere   V=34πr3

Percentage error in volume   VΔV×100=3×rΔr×100=3×2=6 %