The volume of a sphere is given by
`V=4/3 piR^(3)`
where `R` is the radius of the radius of the sphere. Find the change in volume of the sphere as the radius is increased from `10.0 cm` to `10.1 cm`. Assume that the rate does not appreciably change between `R=10.0 cm` to `R=10.1 cm`

Options

A. `10 pi cm^(3)`
B. `20 pi cm^(3)`
C. `30 pi cm^(3)`
D. `40 pi cm^(3)`

Answer

D

Explanation

(a) `V=4/3 pi R^(3)`
or, `(dV)/(dR) =4/3 pi d/(dr) (R)^(3) =4/3 pi . 3R^(2) =4piR^(2)`.
(b) At R=10 cm, the rate of change of volume with the radius is
`(dV)/(dR)=4piR^(3) =4pi(100 cm^(2))`
`=400 pi cm^(2)`
The change in volume as the raduis change from 10.0 cm to 10.1 cm is
`DeltaV=(dV)/(dR) DeltaR`
`=(400 pi cm^(2)) (0.1 cm)`
`=40 pi cm^(2)`.