A liquid (coefficient of cubical expansion `gamma_(1)`) is contained in a glass vessel of volume `V_(g)` (coefficient of cubical expansion `gamma_(g)`) at a temperature. The volume of liquid at this temperature is `V_(l)`. Now the system is heated and it is found that at all temperatures, the volume of vessel, unoccupied by liquid remains always same, then

Options

A. `(V_(g))/(V_(l))=(gamma_(l))/(gamma_(g))`
B. `(V_(g))/(V_(l))=(gamma_(g))/(gamma_(l))`
C. `V_(g)-V_(l)=gamma_(g)-gamma_(l)`
D. `V_(g)+V_(l)=gamma_(g)+gamma_(l)`

Answer

To solve the problem, we need to analyze the relationship between the volume of the liquid and the volume of the glass vessel when both are subjected to temperature changes.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a liquid with a coefficient of cubical expansion \( \gamma_L \) contained in a glass vessel with a coefficient of cubical expansion \( \gamma_g \).
- The initial volume of the liquid is \( V_L \) and the volume of the glass vessel is \( V_g \).
- Upon heating, the volume of the vessel that is unoccupied by the liquid remains constant.

2. **Volume Expansion Formulas**:
- The change in volume of the glass vessel when heated can be expressed as:
\[ \Delta V_g = \gamma_g \cdot V_g \cdot \Delta T \]
- The change in volume of the liquid when heated can be expressed as:
\[ \Delta V_L = \gamma_L \cdot V_L \cdot \Delta T \]

3. **Setting Up the Equation**:
- Since the volume of the vessel that is unoccupied by the liquid remains constant, the volume expansion of the glass must equal the volume expansion of the liquid:
\[ \Delta V_g = \Delta V_L \]

4. **Equating the Volume Expansions**:
- From the expressions for \( \Delta V_g \) and \( \Delta V_L \), we can set them equal to each other:
\[ \gamma_g \cdot V_g \cdot \Delta T = \gamma_L \cdot V_L \cdot \Delta T \]

5. **Cancelling Common Terms**:
- Since \( \Delta T \) is common on both sides, we can cancel it out (assuming \( \Delta T \neq 0 \)):
\[ \gamma_g \cdot V_g = \gamma_L \cdot V_L \]

6. **Rearranging the Equation**:
- We can rearrange the equation to find the relationship between \( V_g \) and \( V_L \):
\[ \frac{V_g}{V_L} = \frac{\gamma_L}{\gamma_g} \]

### Final Result:
Thus, the relationship between the volumes of the glass vessel and the liquid is given by:
\[ \frac{V_g}{V_L} = \frac{\gamma_L}{\gamma_g} \]