Two bars `A` and `B` of circular cross section, same volume and made of the same material, are subjected to tension. If the diameter of `A` is half that of `B` and if the force applied to both the rod is the same and it is in the elastic limit, the ratio of extension of `A` to that of `B` will be

Options

A. `16`
B. `8`
C. `4`
D. `2`

Explanation

To solve the problem, we need to find the ratio of the extensions of two bars `A` and `B` made of the same material and subjected to the same force. The diameter of bar `A` is half that of bar `B`.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the diameter of bar `B` be \( d \).
- Then, the diameter of bar `A` will be \( \frac{d}{2} \).
- Both bars have the same volume and are made of the same material.

2. **Volume of the Bars**:
- The volume \( V \) of a cylindrical bar is given by:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the bar.
- The cross-sectional area \( A \) for a circular cross-section is:
\[ A = \frac{\pi d^2}{4} \]

3. **Setting Up the Volume Equation**:
- For bar `A`:
\[ V_A = A_A \cdot L_A = \frac{\pi \left(\frac{d}{2}\right)^2}{4} \cdot L_A = \frac{\pi d^2}{16} \cdot L_A \]
- For bar `B`:
\[ V_B = A_B \cdot L_B = \frac{\pi d^2}{4} \cdot L_B \]
- Since \( V_A = V_B \):
\[ \frac{\pi d^2}{16} \cdot L_A = \frac{\pi d^2}{4} \cdot L_B \]
- Simplifying, we get:
\[ \frac{L_A}{L_B} = 4 \quad \text{(1)} \]

4. **Using Young's Modulus**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
- Rearranging gives:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
- For bar `A`:
\[ \Delta L_A = \frac{F \cdot L_A}{A_A \cdot Y} \]
- For bar `B`:
\[ \Delta L_B = \frac{F \cdot L_B}{A_B \cdot Y} \]

5. **Finding the Areas**:
- The areas are:
\[ A_A = \frac{\pi \left(\frac{d}{2}\right)^2}{4} = \frac{\pi d^2}{16} \]
\[ A_B = \frac{\pi d^2}{4} \]

6. **Substituting Areas into the Extensions**:
- Substitute the areas into the extensions:
\[ \Delta L_A = \frac{F \cdot L_A}{\frac{\pi d^2}{16} \cdot Y} = \frac{16F \cdot L_A}{\pi d^2 \cdot Y} \]
\[ \Delta L_B = \frac{F \cdot L_B}{\frac{\pi d^2}{4} \cdot Y} = \frac{4F \cdot L_B}{\pi d^2 \cdot Y} \]

7. **Finding the Ratio of Extensions**:
- Now, the ratio of extensions:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{16F \cdot L_A}{4F \cdot L_B} = \frac{16L_A}{4L_B} = 4 \cdot \frac{L_A}{L_B} \]
- From equation (1), we know \( \frac{L_A}{L_B} = 4 \):
\[ \frac{\Delta L_A}{\Delta L_B} = 4 \cdot 4 = 16 \]

### Final Answer:
The ratio of the extension of bar `A` to that of bar `B` is:
\[ \frac{\Delta L_A}{\Delta L_B} = 16 \]