A liquid having a coefficient of volume expansion `gamma` is filled in a cylindrical vessel made out of glass having coefficient of lienar expansion `alpha = (gamma)/( 5)` . At room temperature , the level of liquid in the vessel is`l_(0)` and when the temperature is increased by `Delta T`, the level of liquid in the vessel becomes `l~~ l_(0)= ( 1+ n alpha Delta T)`. What is the value of n ? [Given`, alpha Delta T lt lt 1`]

Answer

To find the value of \( n \) in the equation \( l = l_0(1 + n \alpha \Delta T) \), we need to analyze the expansion of both the liquid and the cylindrical vessel due to the temperature increase \( \Delta T \).

### Step-by-Step Solution:

1. **Understanding the Coefficients of Expansion**:
- The coefficient of volume expansion of the liquid is given as \( \gamma \).
- The coefficient of linear expansion of the glass vessel is given as \( \alpha = \frac{\gamma}{5} \).

2. **Initial Setup**:
- At room temperature, the height of the liquid in the vessel is \( l_0 \).
- When the temperature increases by \( \Delta T \), the new height of the liquid is \( l \).

3. **Volume Expansion of the Liquid**:
- The volume of the liquid initially is:
\[ V_0 = A l_0 \]
where \( A \) is the cross-sectional area of the vessel.
- The change in volume of the liquid when the temperature increases by \( \Delta T \) is given by:
\[ \Delta V = V_0 \gamma \Delta T = A l_0 \gamma \Delta T \]
- The new volume of the liquid is:
\[ V = V_0 + \Delta V = A l_0 + A l_0 \gamma \Delta T = A l_0 (1 + \gamma \Delta T) \]

4. **Expansion of the Vessel**:
- The linear expansion of the vessel causes its height to increase. The change in height \( \Delta h \) of the vessel due to the temperature increase is:
\[ \Delta h = \alpha \Delta T \cdot l_0 \]
- Therefore, the new height of the vessel is:
\[ l_{vessel} = l_0 + \Delta h = l_0 + \alpha \Delta T \cdot l_0 = l_0(1 + \alpha \Delta T) \]

5. **Relating the New Height of the Liquid**:
- The new height of the liquid \( l \) must equal the new height of the vessel, which is:
\[ l = \frac{V}{A} = \frac{A l_0 (1 + \gamma \Delta T)}{A} = l_0 (1 + \gamma \Delta T) \]

6. **Setting the Equations Equal**:
- We have two expressions for the new height of the liquid:
\[ l = l_0(1 + n \alpha \Delta T) \quad \text{and} \quad l = l_0(1 + \gamma \Delta T) \]
- Equating these gives:
\[ l_0(1 + n \alpha \Delta T) = l_0(1 + \gamma \Delta T) \]

7. **Simplifying the Equation**:
- Dividing both sides by \( l_0 \) (assuming \( l_0 \neq 0 \)):
\[ 1 + n \alpha \Delta T = 1 + \gamma \Delta T \]
- Thus, we can equate the coefficients of \( \Delta T \):
\[ n \alpha = \gamma \]

8. **Substituting \( \alpha \)**:
- Recall that \( \alpha = \frac{\gamma}{5} \):
\[ n \left(\frac{\gamma}{5}\right) = \gamma \]
- Solving for \( n \):
\[ n = 5 \]

### Final Answer:
The value of \( n \) is \( 5 \).