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

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

$V = \frac{4}{3} π r^{3}$ and $S = 4 π r^{2}$

$\Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ and $\frac{d S}{d t} = 8 π r \frac{d r}{d t}$

When $r = 8 cm$ and $\frac{d V}{d t} = 40 {cm}^{3} / minute$

$\frac{d V}{d t} = 4 π^{2} \frac{d r}{d t} \Rightarrow 40 = 4 π \times 8^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{5}{32 π}$

putting , $r = 8, \frac{d r}{d t} = \frac{5}{32 π}$ in $\frac{d S}{d t} = 8 π r \frac{d r}{d t}$

$\frac{d S}{d t} = 8 π \times 8 \times \frac{5}{32 π} = 10 {cm}^{2} / minute$

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