Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

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

Options:

$8 \pi ~\mathrm{cm}^2 / \mathrm{sec}$

$12 \pi ~\mathrm{cm}^2 / \mathrm{sec}$

$160 \pi ~\mathrm{cm}^2 / \mathrm{sec}$

$200 \mathrm{~cm}^2 / \mathrm{sec}$

Correct Answer:

$160 \pi ~\mathrm{cm}^2 / \mathrm{sec}$

Explanation:

Let r be the radius and S be the surface area of the sphere at any time t. Then,

$S=4 \pi r^2$

$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$

$\Rightarrow \left(\frac{d S}{d t}\right)_{r=200}=8 \pi \times 200 \times 0.1=160 \pi \mathrm{cm}^2 / \mathrm{sec}$