The inner tube of a Pyrex glass mercury thermometer has a diameter of $0.120 \mathrm{mm} .\( The bulb at the bottom of the thermometer contains \)0.200 \mathrm{cm}^{3}$ of mercury. How far will the thread of mercury move for a change of \(1.00^{\circ} \mathrm{C} ?\) Remember to take into account the expansion of the glass. (W) tutorial thermometer)

To find the initial volume of the mercury thread inside the thermometer, we will first determine the radius of the inner tube. The diameter of the inner tube is given as \(0.120\mathrm{mm}\), which means the radius is half of the diameter, or \(0.060\mathrm{mm}\). To work with SI units, we need to convert the radius to meters: \(0.060\mathrm{mm} \times \frac{1 \mathrm{m}}{1000 \mathrm{mm}} = 6.0 \times 10^{-5}\mathrm{m}\) Now, we can use the given volume of mercury to find the initial height of the mercury thread. The volume of the mercury is given as \(0.200\mathrm{cm}^3\). To convert this to SI units, we multiply by \(\frac{1 \mathrm{m}^3}{1000^3 \mathrm{cm}^3} = 2.0\times 10^{-7} \mathrm{m}^3\). Using the volume of a cylinder formula, we can find the initial height of the mercury thread (\(h\)): \(h = \frac{V}{\pi r^2} = \frac{2.0\times 10^{-7}\mathrm{m}^3}{\pi (6.0\times 10^{-5}\mathrm{m})^2} = 0.0709\mathrm{m}\)

We will calculate the volume expansion coefficients for both mercury and the Pyrex glass using the coefficient of volume expansion formula: \(\beta = 3 \alpha\) The linear expansion coefficients for mercury and Pyrex glass are approximately \(\alpha_\text{Hg} = 1.81 \times 10^{-4}\, \mathrm{K^{-1}}\) and \(\alpha_\text{Pyrex} = 3.25 \times 10^{-6}\, \mathrm{K^{-1}}\), respectively. Using these values, we can find the volume expansion coefficients: \(\beta_\text{Hg} = 3 \alpha_\text{Hg} = 3 (1.81 \times 10^{-4}\, \mathrm{K^{-1}}) = 5.43 \times 10^{-4}\, \mathrm{K^{-1}}\) Similarly, for Pyrex glass: \(\beta_\text{Pyrex} = 3 \alpha_\text{Pyrex} = 3 (3.25 \times 10^{-6}\, \mathrm{K^{-1}}) = 9.75 \times 10^{-6}\, \mathrm{K^{-1}}\)

To determine the volume of mercury and glass after expansion, we use the volume after expansion formula: \(V' = V (1 + \beta \Delta T)\) For mercury, with \(\Delta T = 1.00^{\circ}\mathrm{C}\): \(V'_\text{Hg} = V_\text{Hg} (1 + \beta_\text{Hg} \Delta T) = 2.0\times 10^{-7}\mathrm{m}^3 (1 + 5.43\times 10^{-4}\,\mathrm{K^{-1}}\times 1.00\,\mathrm{K}) = 2.01086\times 10^{-7}\,\mathrm{m}^3\) For Pyrex glass: \(V'_\text{Pyrex} = V_\text{Pyrex} (1 + \beta_\text{Pyrex} \Delta T) = 2.0\times 10^{-7}\mathrm{m}^3 (1 + 9.75\times 10^{-6}\,\mathrm{K^{-1}}\times 1.00\,\mathrm{K}) = 2.000195\times 10^{-7}\,\mathrm{m}^3\)

Using the volume after expansion of both the mercury and the Pyrex glass, we can find the new height of the mercury thread (\(h'\)): \(h'_\text{Hg} = \frac{V'_\text{Hg}}{\pi r^2} = \frac{2.01086\times 10^{-7}\mathrm{m}^3}{\pi (6.0\times 10^{-5}\mathrm{m})^2} = 0.0714\mathrm{m}\) And for the Pyrex glass: \(h'_\text{Pyrex} = \frac{V'_\text{Pyrex}}{\pi r^2} = \frac{2.000195\times 10^{-7}\mathrm{m}^3}{\pi (6.0\times 10^{-5}\mathrm{m})^2} = 0.0710\mathrm{m}\)

Finally, we can find the difference in height between the two heights obtained in the previous step: \(\Delta h = h'_\text{Hg} - h'_\text{Pyrex} = 0.0714\,\mathrm{m} - 0.0710\,\mathrm{m} = 0.0004\,\mathrm{m}\) So, the mercury thread will move \(0.0004\,\mathrm{m}\) or \(0.4\,\mathrm{mm}\) for a change of \(1.00^{\circ}\mathrm{C}\).