A spherical balloon expands at a rate proportional to its surface area. Initially its radius is 2 cm and 5 minutes later it increases upto 7 cm, then surface area of spherical balloon after 12 minutes will be

Let $V$ be the volume and $S$ be the surface area of the spherical balloon.

Given that the balloon expands at a rate proportional to its surface area:

$\dfrac{dV}{dt} \propto S$

$\dfrac{dV}{dt} = kS$

Using $V = \dfrac{4}{3}\pi r^3$ and $S = 4\pi r^2$, differentiating $V$ with respect to $t$ gives:

$\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} = S \dfrac{dr}{dt}$

Equating the two expressions for $\dfrac{dV}{dt}$:

$S \dfrac{dr}{dt} = kS \Rightarrow \dfrac{dr}{dt} = k$

Integrating both sides with respect to $t$:

$r = kt + C$

Initially, at $t = 0$, $r = 2$ cm:

$2 = k(0) + C \Rightarrow C = 2$

At $t = 5$ minutes, $r = 7$ cm:

$7 = 5k + 2 \Rightarrow 5k = 5 \Rightarrow k = 1$

Thus, the radius at any time $t$ is given by:

$r = t + 2$

After $12$ minutes ($t = 12$):

$r = 12 + 2 = 14$ cm

The surface area of the balloon at $t = 12$ minutes is:

$S = 4\pi r^2 = 4 \times \dfrac{22}{7} \times (14)^2$

$S = 4 \times \dfrac{22}{7} \times 196 = 4 \times 22 \times 28 = 2464$ sq.cm.

Answer: $2464$ sq.cm.