The time rate of change of the radius of a sphere is 1/2π. When it’s radius is 5cm, what will be the rate of change of the surface of the sphere with time?

The correct option is "20 sq cm".

Let us understand how to solve this problem step-by-step.

First, let the radius of the sphere at any time $t$ be represented by $r$ and its surface area be represented by $S$.

According to the problem statement, we are given:
The time rate of change of the radius of the sphere is:

$\frac{dr}{dt}=\frac{1}{2\pi }\text{cm/s}$

We need to find the rate of change of the surface area of the sphere with respect to time ($\frac{dS}{dt}$) when the radius is:

$r=5\text{cm}$

The formula for the surface area ($S$) of a sphere in terms of its radius ($r$) is:

$S=4\pi r^2$

To find the rate of change of the surface area with respect to time $t$, we differentiate both sides of the equation with respect to $t$ using the chain rule:

$\frac{dS}{dt}=\frac{d}{dt}\left(4\pi r^2\right)$

$\frac{dS}{dt}=4\pi ·2r·\frac{dr}{dt}$

Simplifying the expression, we get:

$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

Now, substitute the given values ($r=5$ and $\frac{dr}{dt}=\frac{1}{2\pi }$) into the differentiated equation:

$\frac{dS}{dt}=8\pi \left(5\right)\left(\frac{1}{2\pi }\right)$

The $\pi$ in the numerator and denominator cancel out:

$\frac{dS}{dt}=40\pi ·\frac{1}{2\pi }$

$\frac{dS}{dt}=20{\text{cm}}^2\text{/s}$

Thus, the rate of change of the surface area of the sphere with time is 20 sq cm.