A particular force (F) applied on a wire increases its length by `2 xx 10^(-3)` m. To increases the wire's length by `4 xx 10^(-3)` m, the applied force will be

Answer

To solve the problem, we need to determine the force required to increase the length of a wire by \(4 \times 10^{-3}\) m, given that a certain force \(F\) increases its length by \(2 \times 10^{-3}\) m.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus \(Y\) for a material is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
where:
- \(F\) is the applied force,
- \(L\) is the original length of the wire,
- \(A\) is the cross-sectional area of the wire,
- \(\Delta L\) is the change in length.

2. **Identifying Constants**:
Since we are using the same wire, the material properties (Young's modulus, original length, and cross-sectional area) remain constant. Therefore, we can say that the force applied is directly proportional to the change in length:
\[ F \propto \Delta L \]

3. **Setting Up the Proportionality**:
Let \(F_1\) be the force that increases the length by \(2 \times 10^{-3}\) m, and let \(F_2\) be the force required to increase the length by \(4 \times 10^{-3}\) m. We can express this relationship as:
\[ \frac{F_1}{F_2} = \frac{\Delta L_1}{\Delta L_2} \]

4. **Substituting Known Values**:
Here, \(\Delta L_1 = 2 \times 10^{-3}\) m and \(\Delta L_2 = 4 \times 10^{-3}\) m. Thus, we have:
\[ \frac{F}{F_2} = \frac{2 \times 10^{-3}}{4 \times 10^{-3}} \]

5. **Simplifying the Ratio**:
The ratio simplifies to:
\[ \frac{F}{F_2} = \frac{2}{4} = \frac{1}{2} \]
This implies:
\[ F_2 = 2F \]

6. **Conclusion**:
Therefore, to increase the length of the wire by \(4 \times 10^{-3}\) m, the applied force \(F_2\) will be twice the original force \(F\):
\[ F_2 = 2F \]

### Final Answer:
To increase the wire's length by \(4 \times 10^{-3}\) m, the applied force will be \(2F\).