Given the volume of a sphere \(V = \frac{4}{3}\pi r^3\), differentiate with respect to \(r\):

\(\frac{dV}{dr} = 4\pi r^2\).

Substitute \(r = 10\):

\(\frac{dV}{dr} = 4\pi \times 10^2 = 400\pi\).

We know \(\frac{dV}{dt} = 50\) cm³/s. Use the chain rule:

\(\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}\).

Rearrange to find \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr} = \frac{50}{400\pi} = \frac{1}{8\pi}\).

Thus, the rate of increase of the radius is \(\frac{1}{8\pi}\) or approximately 0.0398 cm/s.