In observing thereal expansion of a liquid a certain part of glass vessel is to be filled with a liquid of volume expansion `gamma=18xx10^(-5)` in order to exclude the effect of the change in remaining volume of the vessel during heating. What fraction of the vessel should be filled wiht the liquid? The coefficient of linear expansion of glass is `=9.0xx10^(-6)`.

Answer

To solve the problem of determining what fraction of the glass vessel should be filled with a liquid to exclude the effect of the change in the remaining volume of the vessel during heating, we can follow these steps:

### Step 1: Understand the relationship between linear and volume expansion
The coefficient of volume expansion (γ) of a material is related to the coefficient of linear expansion (α) by the formula:
\[ \gamma = 3\alpha \]
For glass, we have:
\[ \alpha_g = 9 \times 10^{-6} \, \text{°C}^{-1} \]
Thus, the coefficient of volume expansion for glass is:
\[ \gamma_g = 3\alpha_g = 3 \times (9 \times 10^{-6}) = 27 \times 10^{-6} \, \text{°C}^{-1} \]

### Step 2: Set up the equation for volume changes
Let \( V_g \) be the initial volume of the glass vessel and \( V_l \) be the initial volume of the liquid. The change in volume of the glass vessel when heated by a temperature change \( \Delta T \) is given by:
\[ \Delta V_g = V_g \cdot \gamma_g \cdot \Delta T \]
The change in volume of the liquid is:
\[ \Delta V_l = V_l \cdot \gamma_l \cdot \Delta T \]
where \( \gamma_l = 18 \times 10^{-5} \, \text{°C}^{-1} \) is the coefficient of volume expansion of the liquid.

### Step 3: Set the volume changes equal to each other
To exclude the effect of the change in volume of the vessel, we set the changes in volume equal:
\[ V_g \cdot \gamma_g \cdot \Delta T = V_l \cdot \gamma_l \cdot \Delta T \]
The \( \Delta T \) cancels out, leading to:
\[ V_g \cdot \gamma_g = V_l \cdot \gamma_l \]

### Step 4: Solve for the fraction of the vessel filled
Rearranging the equation gives:
\[ \frac{V_l}{V_g} = \frac{\gamma_g}{\gamma_l} \]
Substituting the values:
\[ \frac{V_l}{V_g} = \frac{27 \times 10^{-6}}{18 \times 10^{-5}} = \frac{27}{180} = \frac{3}{20} \]

### Conclusion
Thus, the fraction of the vessel that should be filled with the liquid is:
\[ \frac{V_l}{V_g} = \frac{3}{20} \]