A wire is stretched by 0.01 m by a certain force F. Another wire of the same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be

Options

A. `0.005m`
B. `0.01m`
C. `0.02m`
D. `0.002m`

Explanation

To solve the problem, we need to analyze the elongation of the two wires based on their dimensions and the force applied. Let's break it down step by step.

### Step 1: Understand the relationship between stress, strain, and Young's modulus.
The formula for elongation (ΔL) of a wire is given by:
\[ \Delta L = \frac{F L}{A Y} \]
where:
- \( F \) = force applied,
- \( L \) = original length of the wire,
- \( A \) = cross-sectional area of the wire,
- \( Y \) = Young's modulus of the material.

### Step 2: Calculate the area of the original wire.
For a wire with diameter \( D_1 \), the cross-sectional area \( A_1 \) is:
\[ A_1 = \frac{\pi D_1^2}{4} \]

### Step 3: Calculate the elongation of the original wire.
Given that the original wire is stretched by 0.01 m under force \( F \), we have:
\[ \Delta L_1 = 0.01 \, \text{m} \]

### Step 4: Analyze the second wire.
The second wire has:
- Length \( L_2 = 2L_1 \)
- Diameter \( D_2 = 2D_1 \)

### Step 5: Calculate the area of the second wire.
The cross-sectional area \( A_2 \) of the second wire is:
\[ A_2 = \frac{\pi D_2^2}{4} = \frac{\pi (2D_1)^2}{4} = \frac{\pi (4D_1^2)}{4} = \pi D_1^2 \]

### Step 6: Substitute the values into the elongation formula for the second wire.
Using the elongation formula for the second wire:
\[ \Delta L_2 = \frac{F L_2}{A_2 Y} \]
Substituting \( L_2 = 2L_1 \) and \( A_2 = \pi D_1^2 \):
\[ \Delta L_2 = \frac{F (2L_1)}{\pi D_1^2 Y} \]

### Step 7: Relate the elongation of the two wires.
From the elongation of the first wire:
\[ \Delta L_1 = \frac{F L_1}{A_1 Y} = \frac{F L_1}{\frac{\pi D_1^2}{4} Y} = \frac{4 F L_1}{\pi D_1^2 Y} \]
Now, we can express the ratio of elongations:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{4 F L_1}{\pi D_1^2 Y}}{\frac{F (2L_1)}{\pi D_1^2 Y}} = \frac{4}{2} = 2 \]
Thus, we have:
\[ \Delta L_2 = \frac{\Delta L_1}{2} = \frac{0.01}{2} = 0.005 \, \text{m} \]

### Final Answer:
The elongation of the second wire is \( \Delta L_2 = 0.005 \, \text{m} \).