A copper sphere is being heated. Its volume as a result of increasing at the rate of `1mm^(3)` every second. Find the rate at which its surface area is increasing when its radius 10 cm.

Options

A. `0.2mm^(2)//s`
B. `0.02mm^(2)//s`
C. `0.02cm^(2)//s`
D. none

Explanation

To solve the problem, we need to find the rate at which the surface area of a copper sphere is increasing when its radius is 10 cm, given that its volume is increasing at a rate of 1 mm³ per second.

### Step-by-Step Solution:

1. **Understand the given information:**
- The volume of the sphere is increasing at a rate of \( \frac{dV}{dt} = 1 \, \text{mm}^3/\text{s} \).
- We need to find the rate of change of the surface area \( \frac{dS}{dt} \) when the radius \( r = 10 \, \text{cm} \).

2. **Formulas for Volume and Surface Area:**
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- The surface area \( S \) of a sphere is given by:
\[ S = 4 \pi r^2 \]

3. **Differentiate the Volume with respect to time:**
- Differentiating \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

4. **Rearranging to find \( \frac{dr}{dt} \):**
- From the equation above, we can express \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{1}{4 \pi r^2} \frac{dV}{dt} \]

5. **Substituting the known values:**
- We know \( \frac{dV}{dt} = 1 \, \text{mm}^3/\text{s} \) and \( r = 10 \, \text{cm} = 100 \, \text{mm} \):
\[ \frac{dr}{dt} = \frac{1}{4 \pi (100)^2} = \frac{1}{4 \pi \cdot 10000} = \frac{1}{40000 \pi} \, \text{mm/s} \]

6. **Differentiate the Surface Area with respect to time:**
- Differentiating \( S \) with respect to \( t \):
\[ \frac{dS}{dt} = 8 \pi r \frac{dr}{dt} \]

7. **Substituting \( r \) and \( \frac{dr}{dt} \) into the Surface Area equation:**
- Substitute \( r = 100 \, \text{mm} \) and \( \frac{dr}{dt} = \frac{1}{40000 \pi} \):
\[ \frac{dS}{dt} = 8 \pi (100) \left(\frac{1}{40000 \pi}\right) \]
- Simplifying this:
\[ \frac{dS}{dt} = 8 \cdot 100 \cdot \frac{1}{40000} = \frac{800}{40000} = \frac{1}{50} \, \text{mm}^2/\text{s} \]

8. **Convert to a more convenient unit:**
- \( \frac{1}{50} \, \text{mm}^2/\text{s} = 0.02 \, \text{mm}^2/\text{s} \)

### Final Answer:
The rate at which the surface area is increasing when the radius is 10 cm is:
\[ \frac{dS}{dt} = 0.02 \, \text{mm}^2/\text{s} \]