Question

A spherical balloon is being inflated at the rate of 35 cc/min. The rate of increase in the surface area (in cm²/min.) of the balloon when its diameter is 14 cm, is ______.

पर्याय

• 10

• `sqrt(10)`

• 100

• `10sqrt(10)`

Answer 1

A spherical balloon is being inflated at the rate of 35 cc/min. The rate of increase in the surface area (in cm²/min.) of the balloon when its diameter is 14 cm, is 10.

Explanation:

Volume of sphere V = `4/3 πr^3`

`(dV)/(dt) = 4/3 . π . 3r^2 . (dr)/(dt)`

35 = `4πr^2. (dr)/(dt)` or `(dr)/(dt) = 35/(4πr^2)`  ...(i)

Surface area of sphere = S = 4πr²

`(dS)/(dt) = 4π xx 2r xx (dr)/(dt) = 8πr. (dr)/(dt)`

`\implies (dS)/(dt) = 70/r`  ...(By using (i))

Now, diameter = 14 cm

`\implies` r = 7

∴ `(dS)/(dt)` = 10