Additional Mathematics – Calculus | e-Consult

For a sphere, \(V=\frac{4}{3}\pi r^{3}\).

Differentiate: \(\displaystyle dV = 4\pi r^{2}\,dr\).

At \(r=10\) cm and \(dr=0.2\) cm:

\(\displaystyle dV = 4\pi (10)^{2}(0.2)=4\pi \times100 \times0.2 = 80\pi\) cm³.

Numerically, \(80\pi \approx 251.33\) cm³.

Hence the volume increases by approximately \(\boxed{2.5\times10^{2}\text{ cm}^{3}}\) (to two‑significant figures).