Young's modulus of rubber is `10^(4) N//m^(2)` and area of cross section is `2 cm^(-2)`. If force of `2 xx 10^(5) dyn` is applied along its length, then its initial l becomes

Options

A. `3l`
B. `4l`
C. `2l`
D. none of these

Explanation

To solve the problem, we need to find the final length of the rubber after a force is applied. We will use the formula for elongation based on Young's modulus.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Young's modulus of rubber, \( Y = 10^4 \, \text{N/m}^2 \)
- Area of cross-section, \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \) (conversion from cm² to m²)
- Force applied, \( F = 2 \times 10^5 \, \text{dyn} = 2 \times 10^5 \times 10^{-5} \, \text{N} = 2 \times 10^0 \, \text{N} = 2 \, \text{N} \) (conversion from dyn to N)

2. **Use the Formula for Elongation:**
The elongation \( \Delta L \) can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Here, \( L \) is the original length which we do not know yet.

3. **Substituting Values:**
Substitute the known values into the elongation formula:
\[ \Delta L = \frac{(2 \, \text{N}) \cdot L}{(2 \times 10^{-4} \, \text{m}^2) \cdot (10^4 \, \text{N/m}^2)} \]

4. **Simplifying the Expression:**
\[ \Delta L = \frac{2L}{2 \times 10^{-4} \times 10^4} = \frac{2L}{2} = L \]
This means the elongation \( \Delta L \) is equal to the original length \( L \).

5. **Calculate the Final Length:**
The final length \( L_f \) is given by:
\[ L_f = L + \Delta L = L + L = 2L \]

### Conclusion:
The final length of the rubber after the force is applied is \( 2L \).