A string of length L is stretched by `L//20` and speed transverse wave alon it is v. The speed of wave ehen it is stretched by `L//10` will be (assume that Hooke law is applicable)

Options

A. 2v
B. `(v)/(sqrt2)`
C. `sqrt2v`
D. 4v

Answer

To solve the problem, we need to analyze how the speed of a transverse wave in a string changes when the string is stretched by different amounts. We will use the principles of wave motion and Hooke's law.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a string of length \( L \) which is initially stretched by \( \frac{L}{20} \). The speed of the transverse wave on this string is given as \( v \). We need to find the speed of the wave when the string is stretched by \( \frac{L}{10} \).

2. **Using Hooke's Law**:
According to Hooke's law, the force exerted by the string is proportional to the extension. The Young's modulus \( Y \) is defined as:
\[ Y = \frac{F/A}{\Delta L/L} \]
where \( F \) is the force, \( A \) is the cross-sectional area, \( \Delta L \) is the extension, and \( L \) is the original length of the string.

3. **Setting Up the Proportions**:
- For the first case (extension \( \Delta L_1 = \frac{L}{20} \)):
\[ Y = \frac{F_1/A}{\frac{L/20}{L}} = \frac{F_1}{A} \cdot 20 \]
- For the second case (extension \( \Delta L_2 = \frac{L}{10} \)):
\[ Y = \frac{F_2/A}{\frac{L/10}{L}} = \frac{F_2}{A} \cdot 10 \]

4. **Relating Forces**:
Since \( Y \) is constant for the same material, we can equate the two expressions:
\[ \frac{F_1}{A} \cdot 20 = \frac{F_2}{A} \cdot 10 \]
Simplifying this gives:
\[ F_1 \cdot 20 = F_2 \cdot 10 \implies F_2 = 2F_1 \]

5. **Finding the Relationship of Pressure**:
The pressure \( P \) in the string can be defined as:
\[ P = \frac{F}{A} \]
Thus,
- For the first case:
\[ P_1 = \frac{F_1}{A} \]
- For the second case:
\[ P_2 = \frac{F_2}{A} = \frac{2F_1}{A} = 2P_1 \]

6. **Relating Wave Speed to Pressure**:
The speed of a transverse wave \( v \) in a medium is given by:
\[ v = \sqrt{\frac{P}{\rho}} \]
where \( \rho \) is the density of the string. Since the density remains constant, we can relate the speeds:
\[ \frac{v_1}{v_2} = \sqrt{\frac{P_1}{P_2}} \]
Substituting \( P_2 = 2P_1 \):
\[ \frac{v_1}{v_2} = \sqrt{\frac{P_1}{2P_1}} = \sqrt{\frac{1}{2}} \implies v_2 = \sqrt{2} v_1 \]

7. **Final Result**:
Since \( v_1 = v \), we have:
\[ v_2 = \sqrt{2} v \]

### Conclusion:
The speed of the wave when the string is stretched by \( \frac{L}{10} \) will be \( \sqrt{2} v \).