The ratio of frequencies of two wires having same length and same tension and made of the same material is 2:3.If the diameter of one wire be 0.09 cm , then determine the diameter of the other.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between frequency and mass per unit length (μ)
The frequency (F) of a wire is given by the formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.

Since the length \( L \) and tension \( T \) are the same for both wires, the frequency is inversely proportional to the square root of the mass per unit length:
\[ \frac{F_1}{F_2} = \sqrt{\frac{\mu_2}{\mu_1}} \]

### Step 2: Set up the ratio of frequencies
Given that the ratio of frequencies \( F_1 : F_2 = 2 : 3 \), we can write:
\[ \frac{F_1}{F_2} = \frac{2}{3} \]

### Step 3: Relate the mass per unit length to the diameter
From the relationship established, we have:
\[ \frac{2}{3} = \sqrt{\frac{\mu_2}{\mu_1}} \]
Squaring both sides gives:
\[ \left(\frac{2}{3}\right)^2 = \frac{\mu_2}{\mu_1} \]
\[ \frac{4}{9} = \frac{\mu_2}{\mu_1} \]

### Step 4: Express mass per unit length in terms of diameter
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \rho \cdot A \]
where \( \rho \) is the density and \( A \) is the cross-sectional area. For a circular wire, the area \( A \) is given by:
\[ A = \frac{\pi}{4} d^2 \]
Thus,
\[ \mu = \rho \cdot \frac{\pi}{4} d^2 \]

### Step 5: Set up the ratio of mass per unit lengths
For the two wires, we have:
\[ \mu_1 = \rho \cdot \frac{\pi}{4} d_1^2 \]
\[ \mu_2 = \rho \cdot \frac{\pi}{4} d_2^2 \]
Now substituting these into the ratio:
\[ \frac{\mu_2}{\mu_1} = \frac{\frac{\pi}{4} d_2^2}{\frac{\pi}{4} d_1^2} = \frac{d_2^2}{d_1^2} \]

### Step 6: Substitute the ratio of mass per unit lengths
From the earlier step, we have:
\[ \frac{4}{9} = \frac{d_2^2}{d_1^2} \]

### Step 7: Substitute the known diameter
We know \( d_1 = 0.09 \, \text{cm} \):
\[ \frac{4}{9} = \frac{d_2^2}{(0.09)^2} \]

### Step 8: Solve for \( d_2^2 \)
Cross-multiplying gives:
\[ 4 \cdot (0.09)^2 = 9 \cdot d_2^2 \]
\[ d_2^2 = \frac{4 \cdot (0.09)^2}{9} \]

### Step 9: Calculate \( d_2 \)
Calculating \( (0.09)^2 \):
\[ (0.09)^2 = 0.0081 \]
Thus,
\[ d_2^2 = \frac{4 \cdot 0.0081}{9} = \frac{0.0324}{9} = 0.0036 \]
Now taking the square root:
\[ d_2 = \sqrt{0.0036} = 0.06 \, \text{cm} \]

### Final Answer
The diameter of the other wire \( d_2 \) is \( 0.06 \, \text{cm} \).

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