A fluid having a thermal coefficient of volume expansion `gamma` is filled in a cylindrical vessel up to a height `h_0`. The coefficient of linear thermal expansion of the material of the vessel is `alpha` . If the fluid is heated with the vessel , then find the level of liquid when temperature increases by `Deltatheta`.

Options

A. `(h_0(1+gammaDeltatheta))/((1+2alphaDeltatheta))`
B. `(h_0(1+gammaDeltatheta))/((1+3alphaDeltatheta))`
C. `(h_0(1+gammaDeltatheta))/((1+alphaDeltatheta))`
D. `h_0 (1+ gammaDeltatheta)`

Answer

To solve the problem, we will analyze the changes in volume of the fluid and the dimensions of the cylindrical vessel when the temperature increases by \( \Delta \theta \).

### Step-by-Step Solution:

1. **Initial Conditions**:
- The initial height of the fluid in the vessel is \( h_0 \).
- The area of the base of the cylindrical vessel is \( A_0 \).
- The volume of the fluid initially is given by:
\[ V_0 = A_0 h_0 \]

2. **Change in Volume of the Fluid**:
- The coefficient of volume expansion of the fluid is \( \gamma \).
- When the temperature increases by \( \Delta \theta \), the change in volume \( \Delta V \) of the fluid can be expressed as:
\[ \Delta V = V_0 \cdot \gamma \Delta \theta = A_0 h_0 \cdot \gamma \Delta \theta \]
- Therefore, the new volume \( V \) of the fluid after heating is:
\[ V = V_0 + \Delta V = A_0 h_0 + A_0 h_0 \cdot \gamma \Delta \theta = A_0 h_0 (1 + \gamma \Delta \theta) \]

3. **Change in Dimensions of the Vessel**:
- The coefficient of linear thermal expansion of the vessel material is \( \alpha \).
- The area of the base of the vessel changes due to thermal expansion:
\[ A = A_0 (1 + 2\alpha \Delta \theta) \]
- The height of the vessel after heating is denoted as \( h \). Thus, the new volume can also be expressed as:
\[ V = A \cdot h = A_0 (1 + 2\alpha \Delta \theta) \cdot h \]

4. **Equating the Two Expressions for Volume**:
- We have two expressions for the volume of the fluid:
\[ A_0 h_0 (1 + \gamma \Delta \theta) = A_0 (1 + 2\alpha \Delta \theta) \cdot h \]
- We can cancel \( A_0 \) from both sides (assuming \( A_0 \neq 0 \)):
\[ h_0 (1 + \gamma \Delta \theta) = (1 + 2\alpha \Delta \theta) \cdot h \]

5. **Solving for \( h \)**:
- Rearranging the equation to solve for \( h \):
\[ h = \frac{h_0 (1 + \gamma \Delta \theta)}{1 + 2\alpha \Delta \theta} \]

### Final Answer:
The level of the liquid when the temperature increases by \( \Delta \theta \) is:
\[ h = \frac{h_0 (1 + \gamma \Delta \theta)}{1 + 2\alpha \Delta \theta} \]