The lengths of two wires of same material are in the ratio 1:2, their tensions are in the ratio 1:2 and their diameters are in the ratio 1:3. the ratio of the notes they emits when sounded together by the same source is

Options

A. `sqrt(2)`
B. `sqrt3`
C. `2sqrt(3)`
D. `3sqrt(2)`

Explanation

To solve the problem, we need to find the ratio of the frequencies of two wires made of the same material, given the ratios of their lengths, tensions, and diameters.

### Step-by-Step Solution:

1. **Understanding the Frequency Formula**:
The frequency \( f \) of a wire is given by the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the velocity of the wave in the wire and \( L \) is the length of the wire.

2. **Velocity of the Wave**:
The velocity \( V \) can be expressed in terms of tension \( T \) and linear mass density \( \mu \):
\[ V = \sqrt{\frac{T}{\mu}} \]
Therefore, the frequency can be rewritten as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

3. **Linear Mass Density \( \mu \)**:
The linear mass density \( \mu \) is defined as:
\[ \mu = \frac{m}{L} = \text{Area} \times \text{Density} \]
For a wire with diameter \( d \):
\[ \text{Area} = \frac{\pi d^2}{4} \]
Thus, we can write:
\[ \mu = \frac{\pi d^2}{4} \cdot \rho \]
where \( \rho \) is the density of the material.

4. **Finding the Frequencies**:
For the first wire (Wire 1):
\[ f_1 \propto \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu_1}} \]
For the second wire (Wire 2):
\[ f_2 \propto \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu_2}} \]

5. **Setting Up the Ratios**:
Given:
- Lengths: \( L_1 : L_2 = 1 : 2 \) (i.e., \( L_2 = 2L_1 \))
- Tensions: \( T_1 : T_2 = 1 : 2 \) (i.e., \( T_2 = 2T_1 \))
- Diameters: \( d_1 : d_2 = 1 : 3 \) (i.e., \( d_2 = 3d_1 \))

The linear mass densities can be expressed as:
\[ \mu_1 = \frac{\pi d_1^2}{4} \cdot \rho \quad \text{and} \quad \mu_2 = \frac{\pi d_2^2}{4} \cdot \rho = \frac{\pi (3d_1)^2}{4} \cdot \rho = \frac{9\pi d_1^2}{4} \cdot \rho \]

6. **Calculating the Frequency Ratio**:
The ratio of frequencies \( \frac{f_1}{f_2} \) can be expressed as:
\[ \frac{f_1}{f_2} = \frac{\sqrt{T_1/\mu_1}}{\sqrt{T_2/\mu_2}} \cdot \frac{L_2}{L_1} \]
Substituting the ratios:
\[ \frac{f_1}{f_2} = \frac{\sqrt{T_1}}{\sqrt{2T_1}} \cdot \frac{L_2}{L_1} \cdot \frac{\sqrt{\mu_2}}{\sqrt{\mu_1}} = \frac{1}{\sqrt{2}} \cdot \frac{2L_1}{L_1} \cdot \frac{3}{1} \]
Simplifying gives:
\[ \frac{f_1}{f_2} = \frac{3}{\sqrt{2}} \]

7. **Final Result**:
Therefore, the ratio of the frequencies is:
\[ \frac{f_1}{f_2} = 3\sqrt{2} \]