A U-shaped glass tube is $1 \mathrm{~m}$ long at $0^{\circ}$ and has a cross-sectional area of $1 \mathrm{~cm}^{2}$. If it is placed vertically and mercury is poured into it, at what height must the mercury stand so that the center of gravity of the mercury column is at a constant distance from the upper end of the tube at any temperature. The cubic expansion coefficient of the glass is $1: 38700$ and that of mercury is $1: 5550$.
Let the height of the mercury in the tube at $0^{\circ}$ be $x$ meters, then at $t$ degrees it would be $x(1+\alpha t)$, where $\alpha=1: 5550$, while at the same time the cross-sectional area of the tube would not increase from $1 \mathrm{~cm}^{2}$ to $1+\frac{2}{3} k t$, where $\frac{2}{3} k$ is the surface expansion coefficient of the glass. Due to this expansion, the height of the mercury

$$
x \frac{1+\alpha t}{1+\frac{2}{3} k t}-\text {, }
$$

decreases. This latter value can also be approximately written in the following form:

$$
x\left[1+\left(\alpha-\frac{2}{3} k\right) t\right]
$$

As the length of the glass tube increases from 1 meter to $\left(1+\frac{k}{3} t\right)$ meters ( $\frac{k}{3}$ is the linear expansion coefficient of the glass), the distance of the center of gravity of the mercury column from the upper end of the glass tube is

$$
\left(1+\frac{k}{3} t\right)-\frac{x}{2}\left[1+\left(\alpha-\frac{2}{3} k\right) t\right]
$$

for this value to be independent of $t$, the coefficient of $t$

$$
\frac{k}{3}-\frac{x}{2}\left(\alpha-\frac{2}{3} k\right)
$$

must be equal to zero, from which the sought $x$

$$
x=\frac{\frac{2}{3} k}{\alpha-\frac{2}{3} k}=0.106 \mathrm{~m}
$$