Two wires of the same material and length but diameter in the ratic 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 of finding the ratio of elastic potential energy per unit volume for two wires of the same material and length but with diameters in the ratio of 1:2, we can follow these steps:

### Step 1: Understand the relationship between stress and area
The stress (σ) in a wire is defined as the force (F) applied per unit area (A). Mathematically, this is given by:
\[ \sigma = \frac{F}{A} \]
Since both wires are subjected to the same load (F), the stress will depend on the cross-sectional area of each wire.

### Step 2: Calculate the areas of the wires
The area (A) of a circular wire can be calculated using the formula:
\[ A = \pi r^2 \]
Given that the diameters of the wires are in the ratio of 1:2, the radii (r) will also be in the ratio of 1:2. Let’s denote the radius of the first wire as \( r_1 \) and the second wire as \( r_2 \):
- \( r_1 = r \)
- \( r_2 = 2r \)

Now, we can calculate the areas:
- Area of the first wire:
\[ A_1 = \pi r_1^2 = \pi r^2 \]
- Area of the second wire:
\[ A_2 = \pi r_2^2 = \pi (2r)^2 = \pi (4r^2) = 4\pi r^2 \]

### Step 3: Determine the stress in each wire
Using the areas calculated:
- Stress in the first wire:
\[ \sigma_1 = \frac{F}{A_1} = \frac{F}{\pi r^2} \]
- Stress in the second wire:
\[ \sigma_2 = \frac{F}{A_2} = \frac{F}{4\pi r^2} \]

### Step 4: Find the ratio of stresses
Now, we can find the ratio of the stresses:
\[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{F}{\pi r^2}}{\frac{F}{4\pi r^2}} = \frac{4}{1} = 4 \]

### Step 5: Relate the elastic potential energy per unit volume to stress
The elastic potential energy per unit volume (u) is given by:
\[ u = \frac{1}{2} \frac{\sigma^2}{Y} \]
where \( Y \) is Young's modulus. Since both wires are made of the same material, \( Y \) is constant for both wires.

### Step 6: Calculate the ratio of elastic potential energy per unit volume
The ratio of elastic potential energy per unit volume for the two wires can be expressed as:
\[ \frac{u_1}{u_2} = \frac{\frac{1}{2} \sigma_1^2}{\frac{1}{2} \sigma_2^2} = \frac{\sigma_1^2}{\sigma_2^2} \]

Substituting the ratio of stresses:
\[ \frac{u_1}{u_2} = \left(\frac{\sigma_1}{\sigma_2}\right)^2 = (4)^2 = 16 \]

### Final Answer
Thus, the ratio of elastic potential energy per unit volume for the two wires is:
\[ \frac{u_1}{u_2} = 16:1 \]

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