Question:
The radius of a sphere is changing at the rate of 0.1 cm/sec. The rate of change of its surface area when the radius is 200 cm is
(a) $8 \pi \mathrm{cm}^{2} / \mathrm{sec}$
(b) $12 \pi \mathrm{cm}^{2} / \mathrm{sec}$
(c) $160 \pi \mathrm{cm}^{2} / \mathrm{sec}$
(d) $200 \mathrm{~cm}^{2} / \mathrm{sec}$
Solution:
(c) $160 \pi \mathrm{cm}^{2} / \mathrm{sec}$
Let $r$ be the radius and $S$ be the surface area of the sphere at any time $t .$ Then,
$S=4 \pi r^{2}$
$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$
$\Rightarrow \frac{d S}{d t}=8 \pi(200)(0.1)$
$\Rightarrow \frac{d S}{d t}=160 \pi \mathrm{cm}^{2} / \mathrm{sec}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.