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

4%

6%

8%

2%

Solution

Volume space of space straight a space sphere space equals space 4 over 3 straight pi space left parenthesis radius right parenthesis cubed or space straight V space equals space 4 over 3 πR cubed Taking space logarithium space on space both space sides comma space we space have space log space straight V space equals space log space 4 over 3 space πR cubed Taking space logarithium space on space both space sides comma space we space have log space straight V space equals space log space 4 over 3 straight pi space plus 3 space log space straight R Differentiating comma space we space get comma fraction numerator increment straight V over denominator straight V end fraction space equals space 0 space plus fraction numerator 3 increment straight R over denominator straight R end fraction Accordingly comma space fraction numerator increment straight R over denominator straight R end fraction space equals 2 percent sign Thus comma space fraction numerator increment straight V over denominator straight V end fraction space equals space 3 space straight x space 2 percent sign space equals space 6 percent sign

Some More Questions From Units and Measurement Chapter

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