A thread of liquid is in a uniform capillary tube of length L. As measured by a ruler. The temperature of the tube and thread of liquid is raised by `DeltaT`. If `gamma` be the coefficient of volume expansion of the liquid and `alpha` be the coefficient of linear expansion of the material of the tube, then the increase `DeltaL` in the length of the thread, again measured by the ruler will be

Options

A. `DeltaL=L(gamma-alpha)DeltaT`
B. `DeltaL=L(gamma-2alpha)DeltaT`
C. `DeltaL=L(gamma-3alpha)DeltaT`
D. `DeltaL=LgammaDeltaT`

Explanation

To find the increase in length \(\Delta L\) of the liquid thread in a capillary tube when the temperature is raised by \(\Delta T\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial length of the liquid thread be \(L\).
- Let the initial volume of the liquid be \(V_0\).
- The area of cross-section of the capillary tube is \(A_0\).

2. **Understand Volume Expansion**:
- The coefficient of volume expansion of the liquid is denoted by \(\gamma\).
- When the temperature increases by \(\Delta T\), the change in volume \(\Delta V\) of the liquid can be expressed as:
\[ \Delta V = V_0 \cdot \gamma \cdot \Delta T \]

3. **Understand Linear Expansion of the Tube**:
- The coefficient of linear expansion of the tube material is denoted by \(\alpha\).
- The change in length \(\Delta L_t\) of the tube due to temperature change can be expressed as:
\[ \Delta L_t = L \cdot \alpha \cdot \Delta T \]

4. **Calculate Change in Area of the Tube**:
- The area of cross-section of the tube will also change due to thermal expansion. The change in area \(\Delta A\) can be approximated using the formula for area expansion:
\[ \Delta A = A_0 \cdot 2\alpha \cdot \Delta T \]

5. **Relate Changes in Length and Volume**:
- The length of the liquid thread can be expressed in terms of its volume and the area of the tube:
\[ L + \Delta L = \frac{V_0 + \Delta V}{A_0 + \Delta A} \]
- Rearranging gives:
\[ \Delta L = \frac{\Delta V}{A_0} - \frac{V_0 \cdot \Delta A}{A_0^2} \]

6. **Substituting for \(\Delta V\) and \(\Delta A\)**:
- Substitute the expressions for \(\Delta V\) and \(\Delta A\):
\[ \Delta L = \frac{V_0 \cdot \gamma \cdot \Delta T}{A_0} - \frac{V_0 \cdot (2\alpha \cdot \Delta T)}{A_0} \]

7. **Simplifying the Expression**:
- Factor out \(V_0\) and \(\Delta T\):
\[ \Delta L = V_0 \cdot \Delta T \left(\frac{\gamma - 2\alpha}{A_0}\right) \]

8. **Final Expression**:
- The increase in length of the liquid thread can be expressed as:
\[ \Delta L = \left(\gamma - 2\alpha\right) \cdot \Delta T \cdot L \]
- Here, \(L\) is the original length of the liquid thread.

### Final Answer:
\[ \Delta L = (\gamma - 2\alpha) \cdot \Delta T \cdot L \]