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

Options

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

Explanation

To solve the problem, we need to find the ratio of the number of loops formed in two wires of different radii when they are welded together and used as a sonometer wire under tension. The key points to consider are the properties of standing waves, tension, and the relationship between frequency, wavelength, and the physical properties of the wires.

### Step-by-Step Solution

1. **Identify the Properties of the Wires**:
- Let the radius of the first wire (Wire 1) be \( r \) and the radius of the second wire (Wire 2) be \( 2r \).
- The tension \( T \) is the same for both wires since they are welded and under the same tension.

2. **Calculate the Linear Mass Density (\( \mu \))**:
- The linear mass density \( \mu \) is given by:
\[ \mu = \frac{m}{L} = \rho A \]
- For Wire 1 (radius \( r \)):
\[ \mu_1 = \rho \cdot \pi r^2 \]
- For Wire 2 (radius \( 2r \)):
\[ \mu_2 = \rho \cdot \pi (2r)^2 = 4\rho \cdot \pi r^2 \]

3. **Determine the Wave Speeds in Each Wire**:
- The wave speed \( v \) in a wire is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- For Wire 1:
\[ v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]
- For Wire 2:
\[ v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{4\rho \cdot \pi r^2}} = \frac{1}{2} \sqrt{\frac{T}{\rho \cdot \pi r^2}} = \frac{1}{2} v_1 \]

4. **Set Up the Frequency Relations**:
- The frequency \( f \) of the standing wave is related to the number of loops \( N \) and the length \( L \) of the wire:
\[ f = \frac{N v}{2L} \]
- Since the welded point is a node, the lengths of the two wires are equal, and we can denote the common length as \( L \).

5. **Equate Frequencies**:
- The frequencies for both wires must be equal since they are part of the same system:
\[ f_1 = f_2 \]
- Thus:
\[ N_1 \cdot v_1 = N_2 \cdot v_2 \]
- Substituting \( v_2 = \frac{1}{2} v_1 \):
\[ N_1 \cdot v_1 = N_2 \cdot \frac{1}{2} v_1 \]
- Canceling \( v_1 \) from both sides (since \( v_1 \neq 0 \)):
\[ N_1 = \frac{1}{2} N_2 \]

6. **Find the Ratio of the Number of Loops**:
- Rearranging gives:
\[ \frac{N_1}{N_2} = \frac{1}{2} \]

### Final Answer
The ratio of the number of loops formed in the wires is:
\[ \frac{N_1}{N_2} = \frac{1}{2} \]