A wire of length L and radius r is clamped at one end. On stretching the other end of the wire with a force F, the increase in its length is 1. If another wire of same material but of length 2L and radius 2r is stretched with a force 2F, the increase in its leagth will be

Options

A. `(l)/(4)`
B. `(l)/(2)`
C. l
D. 2l

Explanation

To solve the problem, we will use the concept of Young's modulus and the relationship between stress, strain, and the properties of the wires.

### Step-by-Step Solution:

1. **Identify Given Information for Wire 1:**
- Length \( L_1 = L \)
- Radius \( r_1 = r \)
- Force \( F_1 = F \)
- Increase in length \( \Delta L_1 = 1 \)

2. **Calculate the Area for Wire 1:**
\[ A_1 = \pi r_1^2 = \pi r^2 \]

3. **Calculate Young's Modulus for Wire 1:**
Using the formula for Young's modulus \( Y \):
\[ Y_1 = \frac{F_1}{A_1} \cdot \frac{L_1}{\Delta L_1} \]
Substituting the values:
\[ Y_1 = \frac{F}{\pi r^2} \cdot \frac{L}{1} = \frac{FL}{\pi r^2} \]

4. **Identify Given Information for Wire 2:**
- Length \( L_2 = 2L \)
- Radius \( r_2 = 2r \)
- Force \( F_2 = 2F \)

5. **Calculate the Area for Wire 2:**
\[ A_2 = \pi r_2^2 = \pi (2r)^2 = 4\pi r^2 \]

6. **Calculate Young's Modulus for Wire 2:**
Using the same formula for Young's modulus \( Y \):
\[ Y_2 = \frac{F_2}{A_2} \cdot \frac{L_2}{\Delta L_2} \]
Substituting the values:
\[ Y_2 = \frac{2F}{4\pi r^2} \cdot \frac{2L}{\Delta L_2} \]

7. **Set Young's Modulus of Wire 1 Equal to Wire 2:**
Since both wires are made of the same material, \( Y_1 = Y_2 \):
\[ \frac{FL}{\pi r^2} = \frac{2F}{4\pi r^2} \cdot \frac{2L}{\Delta L_2} \]

8. **Simplify the Equation:**
Cancel out common terms:
\[ \frac{L}{1} = \frac{2L}{2\Delta L_2} \]
This simplifies to:
\[ 1 = \frac{2}{2\Delta L_2} \]

9. **Solve for \( \Delta L_2 \):**
Rearranging gives:
\[ 2\Delta L_2 = 2 \implies \Delta L_2 = 1 \]

### Final Answer:
The increase in length of the second wire \( \Delta L_2 \) is **1**.