Question

If the volume of a sphere is increased by 25%, what is the percentage increase in the surface area, correct to the nearest whole?

Answer 1

Percentage incease in surface area is `16.04%`

Explanation:

If radius of sphere is `r`, volume `V` is given by `4/3pir^3` and surface area `S` is `4pir^2`

Hence `Vpropr^3` and `Spropr^2` i.e. `rpropS^(1/2)`

and hence `VpropS^(3/2)` or `SpropV^(2/3)`

as volume rises by `25%`, `V->1.25V`

hence `S->Sxx(1.25)^(2/3)=1.1604S`

i.e. percentage incease in surface area is `16.04%`