1.6 A hockey puck, whose diameter is approximately 3 inches, is to be used to determine the value of \(\pi\) to three significant figures by carefully measuring its diameter and its circumference. For this calculation to be done properly, the measurements must be made to the nearest _____________. a) hundredth of a \(\mathrm{mm}\) c) \(\mathrm{mm}\) e) in b) tenth of a \(\mathrm{mm}\) d) \(\mathrm{cm}\)

By calculating \(\pi\), we use the formula \(\pi = \frac{C}{d}\), where \(C\) represents the circumference, and \(d\) represents the diameter of a circle (in this case, the hockey puck). We want to calculate the value of \(\pi\) to three significant figures to determine the accuracy we need for the measurements.

Since we want the value of \(\pi\) to be accurate to three significant figures, we can use the approximation \(\pi \approx 3.14\). In our calculations, we want errors to be smaller than half of the last significant figure, i.e., smaller than 0.005.

The problem states that the diameter of the hockey puck is approximately 3 inches. We will use this value to express the circumference. For a circle, circumference \(C = \pi d\). For a 3-inch diameter, the approximate circumference would be \(C = 3.14 \times 3 \approx 9.42\) inches.

To determine the level of accuracy in measurement required to have the value of \(\pi\) accurate to three significant figures, we need to make sure that the potential errors in measurement are smaller than 0.005. Let's assume that our measurements have an error of \(e_d\) for the diameter and \(e_C\) for the circumference. To hold the specified \(\pi\) accuracy, we need: \(\frac{e_C}{e_d} < 0.005\) \(\Rightarrow e_C < 0.005 \cdot e_d\) Given that we know the approximate diameter (3 inches) and circumference (9.42 inches) of the hockey puck, we can calculate the required errors in inches.

With the help of the equation we got in Step 4, we can now find possible values for \(e_d\) and \(e_C\) in different units. Note that there are 2.54 cm (25.4 mm) in one inch. a) hundredth of a mm: \(\Rightarrow e_d = \frac{1}{100} \cdot \frac{1}{10} \,\text{in} \approx 0.000393701 \,\text{in}\) \(\Rightarrow e_C = 0.005 \cdot e_d \approx 0.00001968504 \,\text{in}\) b) tenth of a mm: \(\Rightarrow e_d = \frac{1}{10} \cdot \frac{1}{10} \,\text{in} \approx 0.00393701 \,\text{in}\) \(\Rightarrow e_C = 0.005 \cdot e_d \approx 0.0001968504 \,\text{in}\) c) mm: \(\Rightarrow e_d = \frac{1}{10} \,\text{in} \approx 0.0393701 \,\text{in}\) \(\Rightarrow e_C = 0.005 \cdot e_d \approx 0.0001968504 \,\text{in}\) d) cm: \(\Rightarrow e_d = \frac{1}{2.54} \,\text{in} \approx 0.393701 \,\text{in}\) \(\Rightarrow e_C = 0.005 \cdot e_d \approx 0.001968504 \,\text{in}\) e) in: \(\Rightarrow e_d = 1 \,\text{in}\) \(\Rightarrow e_C = 0.005 \cdot e_d = 0.005 \,\text{in}\)

Now, we compare the required errors found in Step 5 and the options provided. We can observe that: a) hundredth of a mm: 0.000393701 in (diameter) and 0.00001968504 in (circumference), which is less than the allowed error; b) tenth of a mm: 0.00393701 in (diameter) and 0.0001968504 in (circumference), which is less than the allowed error; c) mm: 0.0393701 in (diameter) and 0.001968504 in (circumference), which is higher than the allowed error; d) cm: 0.393701 in (diameter) and 0.001968504 in (circumference), which is higher than the allowed error; e) in: 1 in (diameter) and 0.005 in (circumference), which is higher than the allowed error.

From the analysis in Step 6, the best option would be option b) tenth of a mm. Since it provides a level of accuracy that is less than the allowed error, it would help in determining the value of \(\pi\) to three significant figures.