Two wires of same diameter of the same material having the length `l` and `2l` If the force `F` is applied on each, the ratio of the work done in two wires will be

Options

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

Explanation

To solve the problem of finding the ratio of the work done in two wires of the same diameter and material but different lengths (one of length `l` and the other of length `2l`), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two wires: Wire 1 with length `l` and Wire 2 with length `2l`.
- Both wires have the same diameter and are made of the same material.

2. **Identifying the Relevant Formula**:
- The work done (or potential energy stored) in a wire when a force is applied can be expressed using the formula for elastic potential energy:
\[ P.E. = \frac{1}{2} F x \]
- Here, `F` is the applied force and `x` is the extension of the wire.

3. **Finding the Extension (x)**:
- The extension `x` in a wire can be calculated using Hooke's law:
\[ x = \frac{F L}{Y A} \]
- Where:
- `F` is the applied force,
- `L` is the original length of the wire,
- `Y` is Young's modulus of the material,
- `A` is the cross-sectional area of the wire.

4. **Calculating the Work Done for Each Wire**:
- For Wire 1 (length `l`):
\[ x_1 = \frac{F l}{Y A} \]
\[ P.E_1 = \frac{1}{2} F x_1 = \frac{1}{2} F \left(\frac{F l}{Y A}\right) = \frac{F^2 l}{2 Y A} \]
- For Wire 2 (length `2l`):
\[ x_2 = \frac{F (2l)}{Y A} = \frac{2F l}{Y A} \]
\[ P.E_2 = \frac{1}{2} F x_2 = \frac{1}{2} F \left(\frac{2F l}{Y A}\right) = \frac{F^2 (2l)}{2 Y A} = \frac{F^2 l}{Y A} \]

5. **Finding the Ratio of Work Done**:
- Now, we can find the ratio of the work done in the two wires:
\[ \text{Ratio} = \frac{P.E_1}{P.E_2} = \frac{\frac{F^2 l}{2 Y A}}{\frac{F^2 l}{Y A}} = \frac{1}{2} \]

### Conclusion:
The ratio of the work done in the two wires is:
\[ \text{Ratio} = 1:2 \]