Two steel wires of the same radius have their lengths in the ratio of `1:2`. If they are stretched by the same force, then the strains produced in the two wires will be in the ratio of

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1:4`

Explanation

To solve the problem, let's analyze the situation step by step.

### Given:
- Two steel wires of the same radius.
- Lengths of the wires are in the ratio of \(1:2\).
- Both wires are stretched by the same force.

### To Find:
- The ratio of the strains produced in the two wires.

### Step 1: Define the lengths of the wires
Let the lengths of the two wires be:
- \(L_1 = L\) (for the first wire)
- \(L_2 = 2L\) (for the second wire)

### Step 2: Define the radius and area of the wires
Since both wires have the same radius \(r\), the cross-sectional area \(A\) for both wires can be calculated as:
\[ A = \pi r^2 \]

### Step 3: Calculate the stress in each wire
Stress (\(\sigma\)) is defined as the force (\(F\)) applied per unit area (\(A\)):
\[ \sigma = \frac{F}{A} \]
Since both wires are subjected to the same force and have the same area:
\[ \sigma_1 = \sigma_2 = \frac{F}{\pi r^2} \]

### Step 4: Relate stress and strain using Young's Modulus
Young's modulus (\(Y\)) is defined as the ratio of stress to strain (\(\epsilon\)):
\[ Y = \frac{\sigma}{\epsilon} \]
From this, we can express strain as:
\[ \epsilon = \frac{\sigma}{Y} \]

### Step 5: Calculate the strains in both wires
Since Young's modulus is a property of the material and both wires are made of steel, \(Y\) is the same for both wires. Thus, we can write:
- For wire 1:
\[ \epsilon_1 = \frac{\sigma_1}{Y} = \frac{F/\pi r^2}{Y} \]
- For wire 2:
\[ \epsilon_2 = \frac{\sigma_2}{Y} = \frac{F/\pi r^2}{Y} \]

### Step 6: Find the ratio of strains
Since both strains \(\epsilon_1\) and \(\epsilon_2\) are equal:
\[ \frac{\epsilon_1}{\epsilon_2} = \frac{F/\pi r^2}{Y} \div \frac{F/\pi r^2}{Y} = 1 \]

### Conclusion
The ratio of the strains produced in the two wires is:
\[ \epsilon_1 : \epsilon_2 = 1 : 1 \]

### Final Answer
The strains produced in the two wires will be in the ratio of \(1:1\).

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