Question

If radius of the sphere is measure with error of 2% then what would be the error %age in volume of the sphere

Answer 1

$\textbf{Your answer is --}$

Given,

% error in radius of sphere is 2%

Now,

Volume of sphere = 4/3 πr^3

4/3π is constant .

Let, error in volume of sphere be ∆V

Now,

∆V/V × 100 = (3×∆r/r)×100

=> ∆V/V/× 100 = (3×2)% [ given % error in r = 2% ]

=> ∆V/V×100 = 6%

Hence, % error in volume of sphere is 6%

$\textbf{ HOPE IT HELPS YOU }$