Two wires are fixed in a sanometer. Their tension are in the ratio `8:1` The lengths are in the ratio `36:35` The diameter are in the ratio `4:1` Densities of the materials are in the ratio `1:2` if the lower frequency in the setting is `360Hz`. The beat frequency when the two wires are sounded together is

Explanation

To solve the problem step by step, we will analyze the given data and apply the relevant formulas.

### Step 1: Write down the given data
- Tension ratio: \( T_1 : T_2 = 8 : 1 \)
- Length ratio: \( L_1 : L_2 = 36 : 35 \)
- Diameter ratio: \( D_1 : D_2 = 4 : 1 \)
- Density ratio: \( \rho_1 : \rho_2 = 1 : 2 \)
- Lower frequency: \( f_1 = 360 \, \text{Hz} \)

### Step 2: Calculate the ratios of the relevant parameters
- Since the diameter ratio is \( 4 : 1 \), the radius ratio \( R_1 : R_2 = 2 : 1 \).
- The cross-sectional area ratio \( A_1 : A_2 = R_1^2 : R_2^2 = 4 : 1 \).

### Step 3: Express the mass per unit length (μ) in terms of density and area
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \rho \cdot A \]
Thus, the ratios of mass per unit length for the two wires can be expressed as:
\[ \mu_1 : \mu_2 = \rho_1 \cdot A_1 : \rho_2 \cdot A_2 \]

### Step 4: Substitute the ratios into the frequency formula
The frequency of a wire in a sonometer is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
The ratio of frequencies \( \frac{f_1}{f_2} \) can be expressed as:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} \cdot \sqrt{\frac{T_1}{T_2}} \cdot \sqrt{\frac{\rho_2}{\rho_1}} \cdot \sqrt{\frac{A_2}{A_1}} \]

### Step 5: Substitute the known ratios
Now substituting the known ratios:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} \cdot \sqrt{\frac{T_1}{T_2}} \cdot \sqrt{\frac{\rho_2}{\rho_1}} \cdot \sqrt{\frac{A_2}{A_1}} \]
Substituting the values:
- \( \frac{L_2}{L_1} = \frac{35}{36} \)
- \( \frac{T_1}{T_2} = 8 \)
- \( \frac{\rho_2}{\rho_1} = 2 \)
- \( \frac{A_2}{A_1} = \frac{1}{4} \)

Thus,
\[ \frac{f_1}{f_2} = \frac{35}{36} \cdot \sqrt{8} \cdot \sqrt{2} \cdot \sqrt{\frac{1}{4}} \]

### Step 6: Simplify the expression
Calculating the square roots:
\[ \sqrt{8} = 2\sqrt{2}, \quad \sqrt{2} = \sqrt{2}, \quad \sqrt{\frac{1}{4}} = \frac{1}{2} \]
So,
\[ \frac{f_1}{f_2} = \frac{35}{36} \cdot 2\sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = \frac{35}{36} \cdot 2 \]
Thus,
\[ \frac{f_1}{f_2} = \frac{35 \cdot 2}{36} = \frac{70}{36} = \frac{35}{18} \]

### Step 7: Calculate \( f_2 \)
Given \( f_1 = 360 \, \text{Hz} \):
\[ \frac{360}{f_2} = \frac{35}{18} \]
Cross-multiplying gives:
\[ 360 \cdot 18 = 35 \cdot f_2 \]
\[ 6480 = 35 f_2 \implies f_2 = \frac{6480}{35} = 185.14 \, \text{Hz} \]

### Step 8: Calculate the beat frequency
The beat frequency \( f_b \) is given by:
\[ f_b = |f_1 - f_2| = |360 - 185.14| = 174.86 \, \text{Hz} \]

Thus, the beat frequency when the two wires are sounded together is approximately **174.86 Hz**.