Two wires of different densities but same area of cross-section are soldered together at one end and are stretched to a tension `T`. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of density of the first wire to that of the second wire.

Explanation

To solve the problem, we need to find the ratio of the densities of the two wires given that the velocity of a transverse wave in the first wire is double that in the second wire. Here’s how to approach the problem step by step:

### Step 1: Understand the relationship between wave velocity, tension, and density.
The velocity \( V \) of a transverse wave in a wire is given by the formula:
\[ V = \sqrt{\frac{T}{\rho}} \]
where:
- \( V \) is the wave velocity,
- \( T \) is the tension in the wire,
- \( \rho \) is the density of the wire.

### Step 2: Set up the equations for both wires.
Let:
- \( V_1 \) be the velocity of the wave in the first wire,
- \( V_2 \) be the velocity of the wave in the second wire.

According to the problem, we have:
\[ V_1 = 2V_2 \]

### Step 3: Write the expressions for the velocities of both wires.
Using the wave velocity formula for both wires, we can write:
\[ V_1 = \sqrt{\frac{T}{\rho_1}} \quad \text{(for the first wire)} \]
\[ V_2 = \sqrt{\frac{T}{\rho_2}} \quad \text{(for the second wire)} \]

### Step 4: Substitute \( V_1 \) in terms of \( V_2 \).
Since \( V_1 = 2V_2 \), we can substitute this into the equation:
\[ 2V_2 = \sqrt{\frac{T}{\rho_1}} \]

### Step 5: Square both sides of the equation.
Squaring both sides gives:
\[ (2V_2)^2 = \frac{T}{\rho_1} \]
\[ 4V_2^2 = \frac{T}{\rho_1} \]

### Step 6: Write the equation for \( V_2 \).
From the second wire, we have:
\[ V_2^2 = \frac{T}{\rho_2} \]

### Step 7: Substitute \( V_2^2 \) into the equation from Step 5.
Substituting \( V_2^2 \) into the equation gives:
\[ 4 \left( \frac{T}{\rho_2} \right) = \frac{T}{\rho_1} \]

### Step 8: Cancel \( T \) from both sides.
Assuming \( T \neq 0 \), we can cancel \( T \):
\[ 4 \cdot \frac{1}{\rho_2} = \frac{1}{\rho_1} \]

### Step 9: Rearrange to find the ratio of densities.
Rearranging the equation gives:
\[ \frac{\rho_1}{\rho_2} = \frac{1}{4} \]

### Conclusion:
Thus, the ratio of the density of the first wire to that of the second wire is:
\[ \frac{\rho_1}{\rho_2} = \frac{1}{4} \]