The radius of a soap bubble is increasing at the rate of `0.2 cms^(-1)` then the rate of increases of its surface area when radius 4 cm is

Options

A. `7.3pi cm^(2)s^(-1)`
B. `7.4pi cm^(2)s^(-1)`
C. `6.4pi cm^(2)s^(-1)`
D. `8.6 pi cm^(2) s^(-1)`

Answer

C

Explanation

Given `r=4, (dr)/(dt)=0.2 ms^(-1)`
`:'` Surface area of soap but bubble `s=pir^(2)`
`implies (ds)/(dt)=8pi r (dr)/(dt)`
`=8pixx(0.2)xx4`
`=6.4pi cm^(2)s^(-1)`