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
Answer & explanation
Correct answer: option 3
$\frac{d v}{d t}=3 m^3 / m m$
$\frac{d A}{d t}=5 m^2 / m m$
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