A spherical balloon is being inflated so that its volume increases at…

The correct answer is Option (3) → $\frac{1}{4\pi} \text{ cm/s}$ ##

The volume $V$ of a sphere with radius $r$ is given by:

$V = \frac{4}{3}\pi r^3$

To find the rate at which the radius $r$ is changing with respect to time, differentiate both sides with respect to $t$:

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

We are given $\frac{dV}{dt} = 100 \text{ cm}^3/\text{s}$ and $r = 10 \text{ cm}$.

$100 = 4\pi(10)^2 \frac{dr}{dt}$

$100 = 4\pi \cdot 100 \cdot \frac{dr}{dt}$

$100 = 400\pi \frac{dr}{dt}$

$\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \text{ cm/s}$

The rate at which the radius of the balloon is increasing when the radius is $10 \text{ cm}$ is:

$\frac{1}{4\pi} \approx 0.0796 \text{ cm/s}$