At what rate will the lateral surface area of the cylinder increase if the radius is increasing at the rate of 2 cm/s when the radius is 5 cm and height is 10 cm?

The correct option is 40π cm/s.

Let us break down the solution step-by-step to understand why this is the correct answer.

Step 1: Understand the formula for the lateral surface area of a cylinder
The lateral surface area ($A$) of a cylinder with radius $r$ and height $h$ is given by the formula:
$$A=2\pi rh$$

Here, the height $h$ of the cylinder is constant at $10\text{cm}$, while the radius $r$ is changing with respect to time $t$.

Step 2: Identify the given values
From the problem description, we have the following values:
• Height of the cylinder, $h=10\text{cm}$ (constant)
• Radius at the given instant, $r=5\text{cm}$
• Rate of increase of the radius, $\frac{dr}{dt}=2\text{cm/s}$

Step 3: Differentiate the lateral surface area with respect to time
To find the rate at which the lateral surface area is increasing ($\frac{dA}{dt}$), we differentiate the surface area formula with respect to time $t$:
$$\frac{dA}{dt}=\frac{d}{dt}\left(2\pi h·r\right)$$
Since $2\pi h$ is a constant, we can pull it out of the derivative:
$$\frac{dA}{dt}=2\pi h·\frac{dr}{dt}$$

Step 4: Substitute the given values into the differentiated equation
Now, substitute $h=10\text{cm}$ and $\frac{dr}{dt}=2\text{cm/s}$ into our rate equation:
$$\frac{dA}{dt}=2\pi ·10·2$$
Multiplying the numbers gives:
$$\frac{dA}{dt}=40\pi \text{cm}/\text{s}$$

Therefore, the lateral surface area of the cylinder increases at a rate of 40π cm/s.