A spherical balloon is being inflated so that its volume increase uniformly at the rate of 40 cm3 /minute. The rate of increase in its surface area when the radius is 8 cm, is

# A spherical balloon is being inflated so that its volume increase uniformly at the rate of 40 cm3 /minute. The rate of increase in its surface area when the radius is 8 cm, is

1. A

2. B

3. C

4. D

none of these

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### Solution:

Let V be the volume, S be the surface area and r be the radius of the balloon. Then,

and $S=4\pi {r}^{2}$

and

When and

$\frac{dV}{dt}=4{\pi }^{2}\frac{dr}{dt}⇒40=4\pi ×{8}^{2}\frac{dr}{dt}⇒\frac{dr}{dt}=\frac{5}{32\pi }$

putting , $r=8,\frac{dr}{dt}=\frac{5}{32\pi }$ in $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

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