Two wire of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by same force. The ratio of work done in stretching them is

Options

A. `5 : 2`
B. `2 : 5`
C. `1 : 3`
D. `3 : 1`

Explanation

To solve the problem, we need to find the ratio of work done in stretching two wires of the same material and diameter but different lengths. Let's denote the lengths of the two wires as \( L_1 \) and \( L_2 \), where \( L_1 : L_2 = 2 : 5 \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Length ratio: \( L_1 : L_2 = 2 : 5 \)
- Both wires are made of the same material and have the same diameter.
- They are stretched by the same force.

2. **Understand the Formula for Work Done:**
The work done \( W \) in stretching a wire can be expressed as:
\[ W = \frac{1}{2} \cdot \frac{Y \cdot A \cdot (\Delta L)^2}{L} \]
where:
- \( Y \) is Young's modulus (constant for the same material),
- \( A \) is the cross-sectional area (constant for the same diameter),
- \( \Delta L \) is the extension in length,
- \( L \) is the original length of the wire.

3. **Establish the Ratio of Work Done:**
Since \( Y \) and \( A \) are constants, we can express the ratio of work done in stretching the two wires as:
\[ \frac{W_1}{W_2} = \frac{\Delta L_1^2 / L_1}{\Delta L_2^2 / L_2} \]
This simplifies to:
\[ \frac{W_1}{W_2} = \frac{\Delta L_1^2 \cdot L_2}{\Delta L_2^2 \cdot L_1} \]

4. **Relate the Extensions \( \Delta L_1 \) and \( \Delta L_2 \):**
From the definition of Young's modulus:
\[ Y = \frac{F}{A \cdot \frac{\Delta L}{L}} \]
Since the force \( F \) is the same for both wires, we can write:
\[ \frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2} \]
Rearranging gives:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{L_1}{L_2} \]
Substituting the length ratio \( L_1 : L_2 = 2 : 5 \):
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{2}{5} \]

5. **Substitute Back into the Work Done Ratio:**
Now we can substitute \( \Delta L_1 = \frac{2}{5} \Delta L_2 \) into the work done ratio:
\[ \frac{W_1}{W_2} = \frac{\left(\frac{2}{5} \Delta L_2\right)^2 \cdot L_2}{\Delta L_2^2 \cdot L_1} \]
Simplifying gives:
\[ \frac{W_1}{W_2} = \frac{\frac{4}{25} \Delta L_2^2 \cdot L_2}{\Delta L_2^2 \cdot L_1} = \frac{4L_2}{25L_1} \]

6. **Substituting the Lengths:**
Since \( L_1 = 2k \) and \( L_2 = 5k \) for some constant \( k \):
\[ \frac{W_1}{W_2} = \frac{4 \cdot 5k}{25 \cdot 2k} = \frac{20}{50} = \frac{2}{5} \]

### Final Answer:
The ratio of work done in stretching the two wires is:
\[ \frac{W_1}{W_2} = \frac{2}{5} \]