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 \( B \) when subjected to the same force, given that the diameter of \( A \) is half that of \( B \) and both bars have the same volume and are made of the same material.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Let the diameter of bar \( B \) be \( d \).
- Then, the diameter of bar \( A \) is \( \frac{d}{2} \).
- Both bars have the same volume \( V \) and are made of the same material, which means they have the same Young's modulus \( Y \).

2. **Calculate the Cross-Sectional Areas:**
- The cross-sectional area \( A \) of a circular bar is given by the formula:
\[ A = \frac{\pi d^2}{4} \]
- For bar \( A \):
\[ A_A = \frac{\pi \left(\frac{d}{2}\right)^2}{4} = \frac{\pi \frac{d^2}{4}}{4} = \frac{\pi d^2}{16} \]
- For bar \( B \):
\[ A_B = \frac{\pi d^2}{4} \]

3. **Using Young's Modulus:**
- The Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives us the extension \( \Delta L \):
\[ \Delta L = \frac{F L}{A Y} \]

4. **Express Length in Terms of Volume:**
- Since both bars have the same volume \( V \):
\[ V = A \cdot L \implies L = \frac{V}{A} \]
- Substituting this into the extension formula:
\[ \Delta L = \frac{F \cdot \frac{V}{A}}{A Y} = \frac{F V}{A^2 Y} \]

5. **Calculate Extensions for Both Bars:**
- For bar \( A \):
\[ \Delta L_A = \frac{F V}{A_A^2 Y} \]
- For bar \( B \):
\[ \Delta L_B = \frac{F V}{A_B^2 Y} \]

6. **Find the Ratio of Extensions:**
- The ratio of the extensions \( \frac{\Delta L_A}{\Delta L_B} \) is:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{A_B^2}{A_A^2} \]
- Substituting the areas:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{\left(\frac{\pi d^2}{4}\right)^2}{\left(\frac{\pi d^2}{16}\right)^2} = \frac{\left(\frac{d^2}{4}\right)^2}{\left(\frac{d^2}{16}\right)^2} = \frac{\frac{d^4}{16}}{\frac{d^4}{256}} = \frac{256}{16} = 16 \]

### Final Result:
Thus, the ratio of the extension of bar \( A \) to that of bar \( B \) is:
\[ \frac{\Delta L_A}{\Delta L_B} = 16 \]