Radius of a sphere is increasing at the rate of $2\text{cm/s}$. The rate at which its volume increasing when its radius is $10\text{cm}$ is ______

Volume of a sphere, $V=\frac{4}{3}\pi r^3$
Given, rate of change of radius, $\frac{dr}{dt}=2\text{cm/s}$ and radius, $r=10\text{cm}$
Rate of change of volume is given by $\frac{dV}{dt}$

By using chain rule,
$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
$=4\pi \times {10}^2\times 2$
$=800\pi {\text{cm}}^3\text{/s}$