The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If in

Question

The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is\

Professor PC Thomas Classes · Answer

(a) : Let $A$ be the surface area of balloon.\ According to question, we have\ $ \frac{d A}{d t}=k $\ Integrating both sides, we get $$ A=k t+C $$ $\Rightarrow \quad 4 \pi r^{2}=k t+C \ldots(\mathrm{i})$, where $r$ is the radius of bailoon.\ Now, at $t=0, r=3$ units\ $	herefore \quad 36 \pi=C \quad$...(ii)\ (Using (i))\ Also, at $t=5 \mathrm{sec}, r=7$ units\ $	herefore \quad 196 \pi=5 k+36 \pi \quad$ (Using (i) & (ii))\ $\Rightarrow \quad k=32 \pi$\ So, from (i), we have $$ 4 \pi r^{2}=32 \pi t+36 \pi $$ $\Rightarrow \quad r^{2}=8 t+9$\ $	herefore \quad$ At $t=9$ secs, we have $$ r=\sqrt{8 	imes 9+9}=9 	ext { units } $$