The bar shown in the figure is made of a single piece of material. It is fixed at one end. It consists of two segments of equal lengh `L//2` each but different cross sectional area `A` and `2A`. Find the ratio of total elongation in the bar to the elongation produced in thicker segment under the action of an axial force `F`. Consider the shape of the joint to remain circular. (Given :`Y` is Young's modulus.

Explanation

To solve the problem, we need to find the ratio of the total elongation in the bar to the elongation produced in the thicker segment under the action of an axial force \( F \). The bar consists of two segments of equal length \( \frac{L}{2} \) but different cross-sectional areas \( A \) (thinner segment) and \( 2A \) (thicker segment).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The bar is fixed at one end and has two segments of equal length \( \frac{L}{2} \).
- The first segment has a cross-sectional area \( A \) and the second segment has a cross-sectional area \( 2A \).

2. **Using the Formula for Elongation**:
- The elongation \( \Delta L \) in a segment can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
- Where \( F \) is the axial force, \( L \) is the length of the segment, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus.

3. **Calculating Elongation in Each Segment**:
- For the first segment (cross-sectional area \( A \)):
- Length of the segment = \( \frac{L}{2} \)
- Elongation \( \Delta L_1 \):
\[ \Delta L_1 = \frac{F \cdot \frac{L}{2}}{A \cdot Y} = \frac{F \cdot L}{2A \cdot Y} \]

- For the second segment (cross-sectional area \( 2A \)):
- Length of the segment = \( \frac{L}{2} \)
- Elongation \( \Delta L_2 \):
\[ \Delta L_2 = \frac{F \cdot \frac{L}{2}}{2A \cdot Y} = \frac{F \cdot L}{4A \cdot Y} \]

4. **Total Elongation in the Bar**:
- The total elongation \( \Delta L_{total} \) in the bar is the sum of the elongations in both segments:
\[ \Delta L_{total} = \Delta L_1 + \Delta L_2 = \frac{F \cdot L}{2A \cdot Y} + \frac{F \cdot L}{4A \cdot Y} \]
- To add these fractions, we need a common denominator:
\[ \Delta L_{total} = \frac{2F \cdot L}{4A \cdot Y} + \frac{F \cdot L}{4A \cdot Y} = \frac{3F \cdot L}{4A \cdot Y} \]

5. **Finding the Ratio of Total Elongation to Elongation in Thicker Segment**:
- We need to find the ratio \( \frac{\Delta L_{total}}{\Delta L_2} \):
\[ \frac{\Delta L_{total}}{\Delta L_2} = \frac{\frac{3F \cdot L}{4A \cdot Y}}{\frac{F \cdot L}{4A \cdot Y}} = \frac{3F \cdot L}{4A \cdot Y} \cdot \frac{4A \cdot Y}{F \cdot L} = 3 \]

### Final Answer:
The ratio of the total elongation in the bar to the elongation produced in the thicker segment is \( 3 \).