Length of an elastic string is x then tension 5N is applied and its length will be y when tension 7 N. What will be its original length ?

Options

A. `(2x+y)/(2)`
B. `(2y-yx)/(2)`
C. `(7x-5y)/(2)`
D. `(7y+5x)/(2)`

Explanation

To solve the problem step by step, we will use the concepts of elasticity, specifically Young's modulus, and the relationship between force, extension, and original length.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Length of the elastic string under tension of 5 N is \( x \).
- Length of the elastic string under tension of 7 N is \( y \).
- We need to find the original length \( L_1 \) of the string.

2. **Understand the Concept of Young's Modulus:**
- Young's modulus \( Y \) is defined as the ratio of stress to strain.
- Stress \( \sigma = \frac{F}{A} \) (Force per unit area).
- Strain \( \epsilon = \frac{\Delta L}{L_0} \) (Change in length/original length).

3. **Set Up the Equations:**
- For the first tension (5 N), we can express the extension \( \Delta L_1 \) as:
\[ \Delta L_1 = x - L_1 \]
- For the second tension (7 N), the extension \( \Delta L_2 \) can be expressed as:
\[ \Delta L_2 = y - L_1 \]

4. **Relate the Forces and Extensions:**
- Since Young's modulus is constant for the same material, we can write:
\[ \frac{F_1}{\Delta L_1} = \frac{F_2}{\Delta L_2} \]
- Substituting the forces and extensions:
\[ \frac{5}{x - L_1} = \frac{7}{y - L_1} \]

5. **Cross Multiply to Solve for \( L_1 \):**
- Cross multiplying gives:
\[ 5(y - L_1) = 7(x - L_1) \]
- Expanding both sides:
\[ 5y - 5L_1 = 7x - 7L_1 \]

6. **Rearranging the Equation:**
- Bringing all terms involving \( L_1 \) to one side:
\[ 7L_1 - 5L_1 = 7x - 5y \]
- Simplifying gives:
\[ 2L_1 = 7x - 5y \]

7. **Solve for the Original Length \( L_1 \):**
- Dividing both sides by 2:
\[ L_1 = \frac{7x - 5y}{2} \]

### Final Answer:
The original length of the elastic string is:
\[ L_1 = \frac{7x - 5y}{2} \]