The radius of a sphere is changing at the rate of $0.1 \mathrm{~cm} / \mathrm{sec}$. The rate of change of its surface area when the radius is $200 \mathrm{~cm}$ is
Answer & explanation
Correct answer: option 3
Let r be the radius and S be the surface area of the sphere at any time t. Then,
$S=4 \pi r^2$
$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$
$\Rightarrow \left(\frac{d S}{d t}\right)_{r=200}=8 \pi \times 200 \times 0.1=160 \pi \mathrm{cm}^2 / \mathrm{sec}$