Water has a unique property: it reaches its maximum density at 4°C. This means that as water is heated or cooled from 4°C, it expands. Here’s why:

1. If the water is heated above 4°C: It begins to expand, decreasing in density. Since the beaker is already full, this expansion causes water to overflow.

2. If the water is cooled below 4°C: Water also expands when cooled below 4°C due to the formation of an open, hexagonal structure in its molecules (especially as it approaches freezing). This expansion also leads to overflow.

Therefore, the beaker will overflow if the temperature changes from 4°C in either direction: heating above 4°C or cooling below 4°C.

The coefficient of apparent expansion of mercury in a vessel is given by:

\[
\gamma_{\text{apparent}} = \gamma_{\text{mercury}} - \alpha_{\text{vessel}}
\]

For the glass vessel:
\[
153 \times 10^{-6} = \gamma_{\text{mercury}} - \alpha_{\text{glass}}
\]

For the steel vessel:
\[
144 \times 10^{-6} = \gamma_{\text{mercury}} - \alpha_{\text{steel}}
\]

Subtracting these two equations, we get:

\[
153 \times 10^{-6} - 144 \times 10^{-6} = \alpha_{\text{steel}} - \alpha_{\text{glass}}
\]

Substitute \(\alpha_{\text{steel}} = 12 \times 10^{-6}/^\circ \text{C}\):

\[
9 \times 10^{-6} = 12 \times 10^{-6} - \alpha_{\text{glass}}
\]

Solving for \(\alpha_{\text{glass}}\):
\[
\alpha_{\text{glass}} = 3 \times 10^{-6}/^\circ \text{C}
\]

Let the length of the other rod be \( L \).

Since the difference in their lengths remains the same at all temperatures, the expansion of each rod must be identical.

For the first rod:
\[
\Delta L_1 = 40 \times \alpha_1 \times \Delta T
\]

For the second rod:
\[
\Delta L_2 = L \times \alpha_2 \times \Delta T
\]

Since \(\Delta L_1 = \Delta L_2\):
\[
40 \times \alpha_1 = L \times \alpha_2
\]

Substitute \(\alpha_1 = 6 \times 10^{-6}/^\circ \text{C}\) and \(\alpha_2 = 4 \times 10^{-6}/^\circ \text{C}\):
\[
40 \times 6 \times 10^{-6} = L \times 4 \times 10^{-6}
\]

Dividing both sides by \(4 \times 10^{-6}\):
\[
L = \frac{40 \times 6}{4} = 60 \, \text{cm}
\]

So, the length of the other rod is \( 60 \, \text{cm} \).

For the two rods to expand by the same length, their expansions \(\Delta l_1\) and \(\Delta l_2\) should be equal. The linear expansion for each rod can be written as:

\[
\Delta l_1 = l_1 \alpha_a t \quad \text{and} \quad \Delta l_2 = l_2 \alpha_s t
\]

Since \(\Delta l_1 = \Delta l_2\), we get:

\[
l_1 \alpha_a t = l_2 \alpha_s t
\]

Dividing both sides by \(t\) (assuming \(t \neq 0\)):

\[
l_1 \alpha_a = l_2 \alpha_s
\]

Now, to find the ratio \(\frac{l_1}{l_1 + l_2}\), divide both sides by \(\alpha_a + \alpha_s\):

\[
\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}
\]

So, the required ratio is:

\[
\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}
\]

For a liquid in a U-tube with different temperatures in each arm, the expansion of the liquid in each arm is affected by the temperature difference.

Let:
- \( l_1 \) and \( l_2 \) be the heights of the liquid columns at temperatures \( t_1 \) and \( t_2 \), respectively.
- \( \beta \) be the coefficient of volume expansion of the liquid.

Since the pressure at the same horizontal level in both arms must be equal, we have:
\[
l_1 (1 + \beta t_1) = l_2 (1 + \beta t_2)
\]

Rearranging, we get:
\[
l_1 + l_1 \beta t_1 = l_2 + l_2 \beta t_2
\]

Solving for \( \beta \):
\[
\beta = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}
\]

Thus, the coefficient of volume expansion of the liquid is:
\[
\beta = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}
\]