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 ??
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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 %
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