To determine the correct measurement and uncertainty of the diameter of the solid sphere, we need to follow these steps:

1. Calculate the mean diameter ($\bar{x}$): $\bar{x} = \frac{\sum x_i}{n}$ where $x_i$ are the individual measurements and $n$ is the number of measurements.

2. Calculate the standard deviation ($s$): $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

3. Determine the standard error ($SE$): $SE = \frac{s}{\sqrt{n}}$

4. Express the measurement with the uncertainty.

First, we find the mean diameter: $\bar{x} = \frac{8.15 + 8.10 + 8.00 + 8.15 + 8.00}{5} = \frac{40.40}{5} = 8.08 \text{ cm}$

Next, we calculate the standard deviation: $s = \sqrt{\frac{(8.15 - 8.08)^2 + (8.10 - 8.08)^2 + (8.00 - 8.08)^2 + (8.15 - 8.08)^2 + (8.00 - 8.08)^2}{4}}$ $s = \sqrt{\frac{0.0049 + 0.0004 + 0.0064 + 0.0049 + 0.0064}{4}}$ $s = \sqrt{\frac{0.023}{4}} = \sqrt{0.00575} \approx 0.076 \text{ cm}$

Then, we determine the standard error: $SE = \frac{s}{\sqrt{5}} = \frac{0.076}{2.236} \approx 0.034 \text{ cm}$

Thus, the correct way to write the diameter with its uncertainty is: $8.08 \pm 0.034 \text{ cm}$

So the correct answer is: $\text{d. } (8.080 \pm 0.034) \text{ cm}$

Do you want any further details or have any questions?

Here are 5 related questions for further practice:

1. How would the mean diameter change if the measurement 8.15 cm was recorded as 8.20 cm?
2. Calculate the mean diameter and uncertainty if a sixth measurement of 8.05 cm was added.
3. What is the significance of using the standard error in reporting measurements?
4. Explain why the standard deviation is divided by the square root of the number of measurements to find the standard error.
5. How would the result change if one of the measurements was significantly different, such as 8.50 cm?

Tip: When calculating the standard deviation, make sure to use the correct formula for the sample standard deviation, especially when working with a small number of measurements.