Two wires of the same material and length but diameter in the ratio 1: 2 are stretched by the same load. The ratio of elastic potential energy per unit volume for the two wires is

Options

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

Explanation

To solve the problem, we need to find the ratio of elastic potential energy per unit volume for two wires made of the same material and length, but with diameters in the ratio of 1:2.

### Step-by-Step Solution:

1. **Understanding Elastic Potential Energy per Unit Volume**:
The elastic potential energy (U) per unit volume (u) can be expressed using the formula:
\[ u = \frac{1}{2} \frac{\text{Stress}^2}{Y} \]
where \( Y \) is the Young's modulus of the material.

2. **Identifying Stress**:
Stress is defined as the force (F) applied per unit area (A):
\[ \text{Stress} = \frac{F}{A} \]
Since both wires are subjected to the same load (F), we can express the stress for each wire in terms of their respective cross-sectional areas.

3. **Calculating the Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the wire. Given that the diameters are in the ratio of 1:2, the radii will be in the ratio of 1:2 as well:
\[ r_1 = r \quad \text{and} \quad r_2 = 2r \]
Therefore, the areas will be:
\[ A_1 = \pi r_1^2 = \pi r^2 \]
\[ A_2 = \pi r_2^2 = \pi (2r)^2 = 4\pi r^2 \]

4. **Calculating Stress for Each Wire**:
Now we can find the stress for each wire:
\[ \text{Stress}_1 = \frac{F}{A_1} = \frac{F}{\pi r^2} \]
\[ \text{Stress}_2 = \frac{F}{A_2} = \frac{F}{4\pi r^2} \]

5. **Finding the Ratio of Stresses**:
The ratio of the stresses can be calculated as:
\[ \frac{\text{Stress}_1}{\text{Stress}_2} = \frac{\frac{F}{\pi r^2}}{\frac{F}{4\pi r^2}} = \frac{4}{1} = 4 \]

6. **Calculating the Ratio of Elastic Potential Energy per Unit Volume**:
Using the formula for elastic potential energy per unit volume:
\[ \frac{u_1}{u_2} = \frac{\frac{1}{2} \text{Stress}_1^2}{\frac{1}{2} \text{Stress}_2^2} = \frac{\text{Stress}_1^2}{\text{Stress}_2^2} \]
Substituting the ratio of stresses:
\[ \frac{u_1}{u_2} = \left(\frac{\text{Stress}_1}{\text{Stress}_2}\right)^2 = 4^2 = 16 \]

### Final Result:
The ratio of elastic potential energy per unit volume for the two wires is:
\[ \frac{u_1}{u_2} = 16 \]