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 solve the problem, we need to determine the increase in length (ΔL) of the thread of liquid in a capillary tube when the temperature is increased by ΔT. The coefficients of volume expansion (γ) for the liquid and linear expansion (α) for the tube material are given.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial length of the liquid thread in the tube be \( L_0 \).
- The initial volume of the liquid can be expressed as \( V_0 = A_0 L_0 \), where \( A_0 \) is the cross-sectional area of the tube.

2. **Volume Expansion of the Liquid**:
- When the temperature is increased by \( \Delta T \), the volume of the liquid expands according to the formula:
\[ V = V_0 (1 + \gamma \Delta T) \]
- Here, \( \gamma \) is the coefficient of volume expansion of the liquid.

3. **Linear Expansion of the Tube**:
- The cross-sectional area of the tube also changes due to thermal expansion. The new area \( A \) can be expressed as:
\[ A = A_0 (1 + 2\alpha \Delta T) \]
- Here, \( \alpha \) is the coefficient of linear expansion of the tube material.

4. **New Length of the Liquid Thread**:
- The new length \( L \) of the liquid thread can be expressed as:
\[ L = \frac{V}{A} = \frac{V_0 (1 + \gamma \Delta T)}{A_0 (1 + 2\alpha \Delta T)} \]

5. **Substituting Initial Values**:
- Substitute \( V_0 = A_0 L_0 \) into the equation:
\[ L = \frac{A_0 L_0 (1 + \gamma \Delta T)}{A_0 (1 + 2\alpha \Delta T)} = \frac{L_0 (1 + \gamma \Delta T)}{1 + 2\alpha \Delta T} \]

6. **Using Binomial Expansion**:
- Since \( \alpha \Delta T \) is small, we can use the binomial expansion:
\[ \frac{1}{1 + 2\alpha \Delta T} \approx 1 - 2\alpha \Delta T \]
- Therefore, we can rewrite \( L \) as:
\[ L \approx L_0 (1 + \gamma \Delta T)(1 - 2\alpha \Delta T) \]

7. **Expanding the Expression**:
- Expanding the product:
\[ L \approx L_0 \left(1 + \gamma \Delta T - 2\alpha \Delta T - 2\alpha \gamma \Delta T^2\right) \]
- For small \( \Delta T \), we can neglect the \( \Delta T^2 \) term:
\[ L \approx L_0 (1 + (\gamma - 2\alpha) \Delta T) \]

8. **Finding the Change in Length (ΔL)**:
- The change in length \( \Delta L \) is given by:
\[ \Delta L = L - L_0 = L_0 (\gamma - 2\alpha) \Delta T \]

9. **Final Result**:
- Thus, the increase in length of the thread of liquid is:
\[ \Delta L = L_0 (\gamma - 2\alpha) \Delta T \]