Two uniform strings `A and B` made of steel are made to vibrate under the same tension. If the first overtone of `A` is equal to the second overtone of `B` and if the radius of `A` is twice that of `B`, the ratio of the lengths of the strings is

Options

A. `2 : 1`
B. `3 : 2`
C. `3 : 4`
D. `1 : 3`

Explanation

To solve the problem, we need to find the ratio of the lengths of two strings A and B, given that the first overtone of string A is equal to the second overtone of string B, and the radius of string A is twice that of string B.

### Step-by-Step Solution:

1. **Understanding Overtone Frequencies**:
- The first overtone of string A corresponds to the frequency \( n_1 \) and is given by \( n_1 = \frac{2}{L_1} \sqrt{\frac{T}{m_1}} \).
- The second overtone of string B corresponds to the frequency \( n_2 \) and is given by \( n_2 = \frac{3}{L_2} \sqrt{\frac{T}{m_2}} \).
- According to the problem, we have \( 2n_1 = 3n_2 \).

2. **Setting Up the Equation**:
- From the relationship \( 2n_1 = 3n_2 \), we can substitute the expressions for \( n_1 \) and \( n_2 \):
\[ 2 \left( \frac{2}{L_1} \sqrt{\frac{T}{m_1}} \right) = 3 \left( \frac{3}{L_2} \sqrt{\frac{T}{m_2}} \right) \]
- Simplifying this gives:
\[ \frac{4}{L_1} \sqrt{\frac{T}{m_1}} = \frac{9}{L_2} \sqrt{\frac{T}{m_2}} \]

3. **Canceling Tension**:
- Since both strings are under the same tension \( T \), we can cancel \( \sqrt{T} \) from both sides:
\[ \frac{4}{L_1 \sqrt{m_1}} = \frac{9}{L_2 \sqrt{m_2}} \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ \frac{L_1}{L_2} = \frac{4 \sqrt{m_2}}{9 \sqrt{m_1}} \]

5. **Finding Mass per Unit Length**:
- The mass per unit length \( m \) of a string can be expressed as:
\[ m = A \cdot \rho \]
- Where \( A \) is the cross-sectional area and \( \rho \) is the density. For circular strings, \( A = \pi r^2 \).
- Thus, we have:
\[ m_1 = \pi r_1^2 \rho \quad \text{and} \quad m_2 = \pi r_2^2 \rho \]

6. **Substituting the Areas**:
- Substituting these into our length ratio gives:
\[ \frac{L_1}{L_2} = \frac{4 \sqrt{\pi r_2^2 \rho}}{9 \sqrt{\pi r_1^2 \rho}} = \frac{4 r_2}{9 r_1} \]

7. **Using the Radius Relationship**:
- Given that \( r_1 = 2r_2 \):
\[ \frac{L_1}{L_2} = \frac{4 r_2}{9 (2 r_2)} = \frac{4}{18} = \frac{2}{9} \]

8. **Final Ratio of Lengths**:
- Therefore, the ratio of the lengths of the strings is:
\[ \frac{L_1}{L_2} = \frac{2}{9} \]

### Final Answer:
The ratio of the lengths of strings A and B is \( \frac{2}{9} \).