A spherical balloon is pumped at the constant rate of 3 m³/min. The rate of increase of it’s surface area at certain instant is found to be 5 m²/min. At this instant, it’s radius is equal to

$\frac{6}{5} m$

$\frac{d v}{d t}=3 m^3 / m m$

$\frac{d A}{d t}=5 m^2 / m m$

We know $v=\frac{4}{3} \pi r^3$ and $A=4 \pi r^2$

$\Rightarrow \frac{d v}{d t}=4 \pi r^2 \frac{d r}{d t}$  and  $\frac{d A}{d t}=8 \pi r \frac{d r}{d t}$

⇒ r = 6/5 m