If there is an error of `+-0.04cm` in themeasurement of the diameter of sphere then the percentage error in its volume, when radius is `10 cm`

Options

A. `+-1.2`
B. `+-1.0`
C. `+-0.8`
D. `+-0.6`

Answer

D

Explanation

Given error in diameter `=+-0.04`
`:.` Error in radius `Delta=+-0.02`
Percentage error in the volume of sphere
`=(DeltaV)/Vxx 100=(Delta(4/3 pir^(3)))/(4/3pir^(3))xx100=(3Deltar)/rxx100`
`=(3xx(+-0.02))/10xx100=+-0.6`