Answer 1

Let r be the radius and V be the volume of the sphere at any time t. Then,

`V = 4/3pir^3`

`=> (dV)/(dt) = 4pir^2 (dr)/(dt)`

`=> (dr)/(dt) = 1/(4pir^2) (dV)/(dT)`

`=> (dr)/(dt) = 8/(4pi(12)^2)`       [∵ r = 12 cm and `(dV)/(dt) = 8 cm^3"/sec"`]

`=> (dr)/(dt) =  1/(72pi) "cm/sec"`

Now, let S be the surface area of the sphere at any time t. Then,

S = 4πr²

`=> (dS)/(dt) = 8pir (dr)/(dt)`

`=> (dS)/(dt) = 8pi(12)xx 1/(72pi)`   [∵  r = 12 cm and `(dr)/(dt) = 1/(72pi) "cm/sec"]`

`=> (dS)/(dt) = 4/3 cm^2"/sec"`

Hence, the surface area of the sphere is increasing at the rate of `4/3 cm^2"/sec"`