If the radius of a sphere is increased by $11\%$, then what is the percentage (correct to two decimal places) increase in its volume?

The volume of a sphere is given by the formula:
$$V=\frac{4}{3}\pi r^3$$
Since $\frac{4}{3}\pi$ is a constant, the volume is directly proportional to the cube of the radius ($V\propto r^3$).Let the initial radius be $100$.The new radius after an $11\%$ increase becomes $111$.The ratio of new volume to initial volume is:
$$\frac{V_{\text{new}}}{V_{\text{initial}}}={\left(\frac{111}{100}\right)}^3=(1.11{)}^3$$
Calculating $(1.11{)}^3$:
$$(1.11{)}^2=1.2321$$
$$(1.2321)\times 1.11=1.367631$$
To find the percentage increase:
$$\text{Percentage Increase}=(1.367631-1)\times 100=36.7631\%$$
Rounding to two decimal places, we get $36.76\%$.