Two wires of the same material and length but diameters in the ratio 1 : 2 are stretched by the same force. The potential energy per unit volume for the two wire when stretched will be in the ratio

Options

A. `16 : 1`
B. `4 : 1`
C. `2 : 1`
D. `1 : 1`

Explanation

To solve the problem of finding the ratio of potential energy per unit volume for two wires of the same material and length but with diameters in the ratio of 1:2 when stretched by the same force, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let the diameters of the two wires be \(d_1\) and \(d_2\) such that \(d_1:d_2 = 1:2\).
- This implies \(d_2 = 2d_1\).
- Both wires have the same length \(L\) and are made of the same material.

2. **Calculating the Cross-Sectional Area**:
- The cross-sectional area \(A\) of a wire is given by the formula:
\[ A = \frac{\pi d^2}{4} \]
- For wire 1:
\[ A_1 = \frac{\pi d_1^2}{4} \]
- For wire 2:
\[ A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (2d_1)^2}{4} = \frac{\pi \cdot 4d_1^2}{4} = \pi d_1^2 \]

3. **Calculating Stress**:
- Stress (\(\sigma\)) is defined as the force (\(F\)) applied per unit area:
\[ \sigma = \frac{F}{A} \]
- For wire 1:
\[ \sigma_1 = \frac{F}{A_1} = \frac{F}{\frac{\pi d_1^2}{4}} = \frac{4F}{\pi d_1^2} \]
- For wire 2:
\[ \sigma_2 = \frac{F}{A_2} = \frac{F}{\pi d_1^2} = \frac{F}{\pi d_1^2} \]

4. **Calculating Strain**:
- Young's modulus (\(Y\)) relates stress and strain (\(\epsilon\)):
\[ Y = \frac{\sigma}{\epsilon} \Rightarrow \epsilon = \frac{\sigma}{Y} \]
- For wire 1:
\[ \epsilon_1 = \frac{\sigma_1}{Y} = \frac{4F}{\pi d_1^2 Y} \]
- For wire 2:
\[ \epsilon_2 = \frac{\sigma_2}{Y} = \frac{F}{\pi d_1^2 Y} \]

5. **Calculating Potential Energy per Unit Volume**:
- The potential energy per unit volume (\(u\)) is given by:
\[ u = \frac{1}{2} \text{stress} \times \text{strain} \]
- For wire 1:
\[ u_1 = \frac{1}{2} \sigma_1 \epsilon_1 = \frac{1}{2} \left(\frac{4F}{\pi d_1^2}\right) \left(\frac{4F}{\pi d_1^2 Y}\right) = \frac{8F^2}{\pi^2 d_1^4 Y} \]
- For wire 2:
\[ u_2 = \frac{1}{2} \sigma_2 \epsilon_2 = \frac{1}{2} \left(\frac{F}{\pi d_1^2}\right) \left(\frac{F}{\pi d_1^2 Y}\right) = \frac{F^2}{2\pi^2 d_1^4 Y} \]

6. **Finding the Ratio of Potential Energies**:
- Now, we find the ratio \(u_1 : u_2\):
\[ \frac{u_1}{u_2} = \frac{\frac{8F^2}{\pi^2 d_1^4 Y}}{\frac{F^2}{2\pi^2 d_1^4 Y}} = \frac{8}{\frac{1}{2}} = 16 \]
- Therefore, the ratio of potential energy per unit volume for the two wires is:
\[ u_1 : u_2 = 16 : 1 \]

### Final Answer:
The potential energy per unit volume for the two wires when stretched will be in the ratio \(16:1\).