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+gamma Delta theta))/((1+2 alphaDeltatheta))`
B. `(h_(0)(1+gamma Delta theta))/((1+3alpha Delta theta))`
C. `(h_(0)(1+gammaDelta theta))/((1+alpha Delta theta))`
D. `h_(0)(1+gamma Delta theta)`

Answer

To solve the problem, we need to determine the new height of the liquid in a cylindrical vessel after both the fluid and the vessel are heated by a temperature change of \(\Delta \theta\).

### Step-by-Step Solution:

1. **Initial Parameters**:
- Let the initial height of the fluid be \( h_0 \).
- The volume expansion coefficient of the fluid is \( \gamma \).
- The linear expansion coefficient of the vessel material is \( \alpha \).

2. **Volume Expansion of the Fluid**:
- The initial volume \( V_0 \) of the fluid can be expressed as:
\[ V_0 = A_0 h_0 \]
where \( A_0 \) is the initial cross-sectional area of the cylindrical vessel.
- When the temperature increases by \( \Delta \theta \), the new volume \( V' \) of the fluid becomes:
\[ V' = V_0 (1 + \gamma \Delta \theta) = A_0 h_0 (1 + \gamma \Delta \theta) \]

3. **Area Expansion of the Vessel**:
- The area of the vessel expands due to the temperature change. The new area \( A' \) can be calculated using the linear expansion coefficient:
\[ A' = A_0 (1 + 2\alpha \Delta \theta) \]
(Note: The area expansion coefficient is twice the linear expansion coefficient).

4. **Finding the New Height**:
- The new height \( h' \) of the fluid can be found using the relationship between volume and area:
\[ V' = A' h' \]
- Substituting the expressions for \( V' \) and \( A' \):
\[ A_0 h_0 (1 + \gamma \Delta \theta) = A_0 (1 + 2\alpha \Delta \theta) h' \]
- Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ h_0 (1 + \gamma \Delta \theta) = (1 + 2\alpha \Delta \theta) h' \]
- Rearranging for \( h' \):
\[ h' = \frac{h_0 (1 + \gamma \Delta \theta)}{1 + 2\alpha \Delta \theta} \]

5. **Final Expression**:
- Thus, the final height of the liquid after the temperature increase is:
\[ h' = h_0 \frac{1 + \gamma \Delta \theta}{1 + 2\alpha \Delta \theta} \]