If the error in the measurement of radius of a sphere in `2%` then the error in the determination of volume of the spahere will be

Options

A. `4%`
B. `6%`
C. `8%`
D. `2%`

Answer

b

Explanation

Volume of a sphere `= (4)/(3) pi (radius)^(3)`
or `V = (4)/(3) pi R^(2)`
Taking legarithen on both sides we have
`log V = log" (4)/(3) pi = 3 log R`
Differentiating , we get `(DeltaV)/(V) = 0 + 3(DeltaR)/(R)`
Accodingly, `(DeltaR/(R)) = 2%`
Thus, `(DeltaV)/(V) = 3 xx 2% = 6%`