Two wires having same length and material are stretched by same force. Their diameters are in the ratio 1:3. The ratio of strain energy per unit volume for these two wires (smaller to larger diameter) when stretched is

Options

A. `3 :1`
B. `9 :1`
C. `27 :1`
D. `81 :1`

Answer

To solve the problem, we need to find the ratio of strain energy per unit volume for two wires of the same length and material, stretched by the same force, with diameters in the ratio of 1:3.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the diameters of the two wires be \( d_1 \) and \( d_2 \) such that \( d_1 : d_2 = 1 : 3 \).
- Therefore, we can express \( d_1 = d \) and \( d_2 = 3d \).
- The radius will be \( r_1 = \frac{d}{2} \) and \( r_2 = \frac{3d}{2} \).

2. **Cross-Sectional Area Calculation**:
- The cross-sectional area \( A \) of a wire is given by the formula \( A = \pi r^2 \).
- For wire 1:
\[ A_1 = \pi r_1^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]
- For wire 2:
\[ A_2 = \pi r_2^2 = \pi \left(\frac{3d}{2}\right)^2 = \frac{9\pi d^2}{4} \]

3. **Stress Calculation**:
- Stress \( \sigma \) is defined as force per unit area:
\[ \sigma_1 = \frac{F}{A_1} = \frac{F}{\frac{\pi d^2}{4}} = \frac{4F}{\pi d^2} \]
\[ \sigma_2 = \frac{F}{A_2} = \frac{F}{\frac{9\pi d^2}{4}} = \frac{4F}{9\pi d^2} \]

4. **Strain Calculation**:
- Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{\epsilon} \Rightarrow \epsilon = \frac{\sigma}{Y} \]
- Therefore, the strains for the two wires are:
\[ \epsilon_1 = \frac{\sigma_1}{Y} = \frac{4F}{\pi d^2 Y} \]
\[ \epsilon_2 = \frac{\sigma_2}{Y} = \frac{4F}{9\pi d^2 Y} \]

5. **Strain Energy per Unit Volume Calculation**:
- The strain energy per unit volume \( u \) is given by:
\[ u = \frac{1}{2} \sigma \epsilon \]
- For wire 1:
\[ u_1 = \frac{1}{2} \sigma_1 \epsilon_1 = \frac{1}{2} \left(\frac{4F}{\pi d^2}\right) \left(\frac{4F}{\pi d^2 Y}\right) = \frac{8F^2}{\pi^2 d^4 Y} \]
- For wire 2:
\[ u_2 = \frac{1}{2} \sigma_2 \epsilon_2 = \frac{1}{2} \left(\frac{4F}{9\pi d^2}\right) \left(\frac{4F}{9\pi d^2 Y}\right) = \frac{8F^2}{81\pi^2 d^4 Y} \]

6. **Finding the Ratio of Strain Energy per Unit Volume**:
- The ratio \( \frac{u_1}{u_2} \) is:
\[ \frac{u_1}{u_2} = \frac{\frac{8F^2}{\pi^2 d^4 Y}}{\frac{8F^2}{81\pi^2 d^4 Y}} = \frac{81}{1} \]

7. **Final Answer**:
- The ratio of strain energy per unit volume for the two wires (smaller to larger diameter) is:
\[ \frac{u_1}{u_2} = 81 : 1 \]