Riemann surface
matlab

Describe briefly three different instruments that can be used for the accurate measurement of temperature and state roughly the temperature range in which they are useful and one important advantage of each instrument. Include at least one instrument that is capable of measuring temperatures down to $1 \mathrm{~K}$.
(Wisconsin)

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Solution:

Magnetic thermometer: Its principle is Curie’s law $\chi=C / T$, where $\chi$ is the susceptibility of the paramagnetic substance used, $T$ is its absolute temperature and $C$ is a constant. Its advantage is that it can measure temperatures below $1 \mathrm{~K}$.

Optical pyrometer: It is based on the principle that we can find the temperature of a hot body by measuring the energy radiated from it, using the formula of radiation. While taking measurements, it does not come into direct contact with the measured body. Therefore, it is usually used to measure the temperatures of celestial bodies.

Vapor pressure thermometer: It is a kind of thermometer used to measure low temperatures. Its principle is as follows. There exists a definite relation between the saturation vapor pressure of a chemically pure material and its boiling point. If this relation is known, we can determine temperature by measuring vapor pressure. It can measure temperatures greater than $14 \mathrm{~K}$, and is the thermometer usually used to measure low temperatures.

A bimetallic strip of total thickness $x$ is straight at temperature $T$. What is the radins of curvature of the strip, $R$, when it is heated to temperature $T+\Delta T$ ? The coefficients of linear expansion of the two metals are $\alpha_1$ and $\alpha_2$, respectively, with $\alpha_2>\alpha_1$. You may assume that each metal has thickness $x / 2$, and you may assume that $x \& R$.
(Wisconsin)

Solution:
We assume that the initial length is $l_0$. After heating, the lengths of the mid-lines of the two metallic strips are respectively
\begin{aligned} & l_1=l_0\left(1+\alpha_1 \Delta T\right), \ & l_2=l_0\left(1+\alpha_2 \Delta T\right) . \end{aligned}

Assuming that the radius of curvature is $R$, the subtending angle of the strip is $\theta$, and the change of thickness is negligible, we have
$$\begin{gathered} l_2=\left(R+\frac{x}{4}\right) \theta, \quad l_1=\left(R-\frac{x}{4}\right) \theta \ l_2-l_1=\frac{x}{2} \theta=\frac{x}{2} \frac{l_1+l_2}{2 R}=\frac{x l_0}{4 R}\left[2+\left(\alpha_1+\alpha_2\right) \Delta T\right] . \end{gathered}$$
From (1) and (2) we obtain
$$l_2-l_1=l_0 \Delta T\left(\alpha_2-\alpha_1\right)$$
(3) and (4) then give
$$R=\frac{x}{4} \frac{\left[2+\left(\alpha_1+\alpha_2\right) \Delta T\right]}{\left(\alpha_2-\alpha_1\right) \Delta T}$$

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