If the surface area of a sphere of radius r is increasing uniformly at the rate `8(cm)^2/s`, then the rate of change of its volume is:

Options

A. constant
B. proportional to `sqrt(R)`
C. proportional to `r^(2)`
D. proportional to `r`

Answer

D

Explanation

Surface area of the sphere `S=4pir^(2)`
`implies(dS)/(dt)=8pir.(dr)/(dt)implies8pir.(dr)/(dt)=8 [ :' (dS)/(dt)=8cm^(2)s^(-1)]`
`implies(dr)/(dt)=1/(pir)`
Now `V=(4pir^(3))/3implies(dV)/(dt)=4pir^(2) . (dr)/(dt)=4pir^(2). 1/(pir)=4r`
`:.(dV)/(dt)propr`.