The surface area of a spherical balloon is increasing at the rate of $2 \text{ cm}^2/\text{s}$. At what rate is the volume of the balloon is increasing when the radius of the balloon is $8 \text{ cm}$?

$8\text{ cm}^3/\text{s}$

The correct answer is Option (2) → $8\text{ cm}^3/\text{s}$ ##

Let $r$ be the radius, $A$ be the surface area and $V$ be the volume of the spherical balloon at time $t$ seconds.

Then, $A = 4\pi r^2$

$∴\frac{dA}{dt} = 8\pi r \frac{dr}{dt} \quad \text{(Differentiating both sides w.r.t. } 't'\text{.)}$

Given, $\frac{dA}{dt} = 2 \text{ cm}^2/\text{s}$

$∴2 = 8\pi r \frac{dr}{dt}$

$⇒\frac{dr}{dt} = \frac{2}{8\pi r} = \frac{1}{4\pi r} \dots (i)$

Now, volume of spherical balloon $V = \frac{4}{3}\pi r^3$

$∴\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} \quad \text{(Differentiating both sides w.r.t. } 't'\text{.)}$

$\text{Or, } \frac{dV}{dt} = 4\pi r^2 \times \frac{1}{4\pi r} \quad [\text{from eq. (i)}]$

$\text{Or, } \frac{dV}{dt} = r$

$\text{Or, } \frac{dV}{dt} = 8 \text{ cm}^3/\text{s} \quad [\text{Given } r = 8 \text{ cm}]$

Therefore, the volume of the balloon is increasing at the rate of $8 \text{ cm}^3/\text{s}$.