A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be _____ cm.

Answer

To solve the problem, we need to understand how elongation in a wire is affected by its length and diameter when subjected to the same force. We will use the concept of Young's modulus and the relationship between stress, strain, and elongation.

### Step-by-step Solution:

1. **Understanding the Initial Conditions**:
- Let the original length of the wire be \( L \) and the original diameter be \( d \).
- The elongation of the wire under force \( F \) is given as \( \Delta L = 0.04 \, \text{m} \).

2. **Formula for Elongation**:
- The elongation \( \Delta L \) of a wire can be expressed as:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
- Where:
- \( A \) is the cross-sectional area of the wire.
- \( Y \) is Young's modulus of the material.

3. **Cross-sectional Area**:
- The cross-sectional area \( A \) of the wire is given by:
\[ A = \frac{\pi d^2}{4} \]

4. **Substituting Area in Elongation Formula**:
- Substituting \( A \) into the elongation formula gives:
\[ \Delta L = \frac{F \cdot L}{\left(\frac{\pi d^2}{4}\right) \cdot Y} = \frac{4F \cdot L}{\pi d^2 \cdot Y} \]

5. **Relationship Between Elongation, Length, and Diameter**:
- From the above equation, we can see that:
\[ \Delta L \propto \frac{L}{d^2} \]
- This means elongation is directly proportional to the length and inversely proportional to the square of the diameter.

6. **New Conditions**:
- Now, if the length is doubled (\( L' = 2L \)) and the diameter is also doubled (\( d' = 2d \)), we need to find the new elongation \( \Delta L' \).

7. **Calculating New Elongation**:
- The new elongation can be expressed as:
\[ \Delta L' \propto \frac{L'}{(d')^2} = \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2} \]

8. **Finding the Ratio of New Elongation to Original Elongation**:
- Since \( \Delta L \propto \frac{L}{d^2} \), we can relate the new elongation to the original elongation:
\[ \Delta L' = \frac{1}{2} \Delta L \]
- Substituting the original elongation:
\[ \Delta L' = \frac{1}{2} \cdot 0.04 \, \text{m} = 0.02 \, \text{m} \]

9. **Converting to Centimeters**:
- Finally, converting meters to centimeters:
\[ \Delta L' = 0.02 \, \text{m} = 2 \, \text{cm} \]

### Final Answer:
The elongation, if its length and diameter are doubled and subjected to the same force, will be **2 cm**.