Answer 1

Let r be the radius, A be the surface area and V be the volume of the spherical ballon at time t seconds.
Then,`A=4pir2`

`dA/dt=4pi(2r((dr)/dt))`

`Given `

`2=8pirdr/dt`

`r dr/dt=2/8pi=1/4pi .............(1)`

`Now , V=4/3 pir^3`

`(dV)/dt=4pi/3xx3r^2 (dr)/dt=(4pir)(r (dr)/dt)`

Given : ` r=6, (d)V/dt=?`

`therefore (dV)/dt=4pi(6)(1/(4pi))=6...........from(1)`

Therefore, the volume of the balloon is increasing at the rate of 6cm³/sec.