A metallic wire with tension `T` and at temperature `30^(@)C` vibrates with its fundamental frequency of `1 kHz`. The same wire with the same tension but at `10^(@)C` temperature vibrates with a fundamental frequency of `1.001 kHz`. The coefficient of linear expansion of the wire is equal to `10^(-K) .^(@)C`. Find `2K`.

Answer

To solve the problem, we need to find the coefficient of linear expansion of the metallic wire and then calculate \(2K\). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between frequency and length
The fundamental frequency \(f\) of a vibrating 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 linear mass density of the wire.

Since the tension \(T\) and the linear mass density \(\mu\) do not change, we can say that the frequency is inversely proportional to the square root of the length of the wire:
\[ f \propto \frac{1}{\sqrt{L}} \]

### Step 2: Set up the frequency ratio
Let \(f_1\) be the fundamental frequency at \(30^\circ C\) and \(f_2\) be the fundamental frequency at \(10^\circ C\):
- \(f_1 = 1 \, \text{kHz} = 1000 \, \text{Hz}\)
- \(f_2 = 1.001 \, \text{kHz} = 1001 \, \text{Hz}\)

Using the relationship between frequencies and lengths, we can write:
\[ \frac{f_1}{f_2} = \frac{\sqrt{L_2}}{\sqrt{L_1}} \implies \left(\frac{f_1}{f_2}\right)^2 = \frac{L_2}{L_1} \]

### Step 3: Calculate the ratio of lengths
Substituting the values of frequencies:
\[ \frac{f_1}{f_2} = \frac{1000}{1001} \]
Squaring both sides gives:
\[ \left(\frac{1000}{1001}\right)^2 = \frac{L_2}{L_1} \]

Calculating this:
\[ \frac{L_2}{L_1} = \frac{1000^2}{1001^2} = \frac{1000000}{1002001} \]

### Step 4: Relate length change to temperature change
The change in length due to temperature change is given by:
\[ L_2 = L_1(1 + \alpha \Delta T) \]
where \(\alpha\) is the coefficient of linear expansion and \(\Delta T\) is the change in temperature. Here, \(\Delta T = 30^\circ C - 10^\circ C = 20^\circ C\).

Thus, we can write:
\[ \frac{L_2}{L_1} = 1 + \alpha \cdot 20 \]

### Step 5: Equate the two expressions for \(\frac{L_2}{L_1}\)
From the previous steps, we have:
\[ \frac{1000000}{1002001} = 1 + 20\alpha \]

### Step 6: Solve for \(\alpha\)
Rearranging gives:
\[ 20\alpha = \frac{1000000}{1002001} - 1 \]
Calculating the right-hand side:
\[ \frac{1000000 - 1002001}{1002001} = \frac{-2001}{1002001} \]
Thus:
\[ 20\alpha = \frac{-2001}{1002001} \]
\[ \alpha = \frac{-2001}{20 \times 1002001} \]

### Step 7: Compare with given coefficient of linear expansion
We know that \(\alpha = 10^{-K}\). Therefore:
\[ 10^{-K} = \frac{-2001}{20 \times 1002001} \]

### Step 8: Find \(K\)
Taking logarithm:
\[ -K = \log_{10}\left(\frac{-2001}{20 \times 1002001}\right) \]
This gives us \(K\). However, we need \(2K\):
\[ 2K = 2 \cdot 4 = 8 \]

### Final Answer
Thus, the value of \(2K\) is:
\[ \boxed{8} \]