A spherical balloon is inflated such that its radius is increasing at the rate of `1/11 cm//sec`.At what rate would the volume be increasing at the instant when the radius is 7 cm ? `[pi=22/7]`

Explanation

Let r be the radius of the spherical balloon.
Then, volume of the balloon, `V=4/3pir^3`
`therefore`Rate of change(increase)of volume of the balloon, `(dV)/dt=4pir^2(dr)/dt`
When r = 7cm and `(dr)/dt=1/11 cm//sec`(given), then rate of change(increase)of volume of the balloon, `(dV)/dt=4pi. 7^2. 1/11=4. 22/7. 7^2. 1/11=56 cm^3//sec`.