Two wires of same material of radii 2r and r are welded together end to end The combination is used as a sonometer wire and is kept under tension T. The welded point lies midway between the bridges. What wil be the ratio of the number of loops formed in the wires, such that the joint is node when the stationary waves are set up in the wire?

Answer

To solve the problem of finding the ratio of the number of loops formed in two welded wires of different radii, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have two wires of the same material with radii \(2r\) and \(r\).
- The wires are welded together end to end and are under tension \(T\).
- The welded point is located midway between the bridges.

2. **Define the Variables**:
- Let \(n_1\) be the number of loops formed in the wire with radius \(2r\).
- Let \(n_2\) be the number of loops formed in the wire with radius \(r\).

3. **Understand the Relationship of Frequency**:
- For a sonometer wire, the frequency \(F\) is given by the formula:
\[ F = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
- Here, \(L\) is the length of the wire, \(T\) is the tension, and \(\mu\) is the mass per unit length.

4. **Calculate Mass per Unit Length**:
- The mass per unit length \(\mu\) is given by:
\[ \mu = \pi r^2 \rho \]
- For the first wire (radius \(2r\)):
\[ \mu_1 = \pi (2r)^2 \rho = 4\pi r^2 \rho \]
- For the second wire (radius \(r\)):
\[ \mu_2 = \pi r^2 \rho \]

5. **Set Up the Frequency Equations**:
- Since the welded wire is continuous, the frequencies of both sections must be equal:
\[ F_1 = F_2 \]
- Therefore, we can write:
\[ \frac{n_1}{2L} \sqrt{\frac{T}{\mu_1}} = \frac{n_2}{2L} \sqrt{\frac{T}{\mu_2}} \]

6. **Simplify the Equation**:
- Cancel \(2L\) and \(T\) from both sides:
\[ n_1 \sqrt{\frac{1}{\mu_1}} = n_2 \sqrt{\frac{1}{\mu_2}} \]
- Rearranging gives:
\[ \frac{n_1}{n_2} = \sqrt{\frac{\mu_1}{\mu_2}} \]

7. **Substitute the Values of \(\mu_1\) and \(\mu_2\)**:
- Substitute \(\mu_1\) and \(\mu_2\):
\[ \frac{n_1}{n_2} = \sqrt{\frac{4\pi r^2 \rho}{\pi r^2 \rho}} = \sqrt{4} = 2 \]

8. **Final Ratio**:
- Thus, the ratio of the number of loops formed in the wires is:
\[ \frac{n_1}{n_2} = 2:1 \]

### Conclusion:
The ratio of the number of loops formed in the wires is \(2:1\).