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 analyze the relationship between the fundamental frequency of a vibrating wire, its tension, length, and temperature. Let's break it down step by step.

### Step 1: Understand the relationship between frequency and length
The fundamental frequency \( f \) of a wire fixed at both ends 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.

### Step 2: Analyze the effect of temperature on length
The length of the wire changes with temperature due to thermal expansion. The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = L_0 \alpha \Delta T \]
where:
- \( L_0 \) is the original length,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.

Given that the temperature changes from \( 30^\circ C \) to \( 10^\circ C \), we have:
\[ \Delta T = 10 - 30 = -20^\circ C \]
Thus, the new length \( L' \) can be expressed as:
\[ L' = L_0 (1 + \alpha \Delta T) = L_0 (1 - 20\alpha) \]

### Step 3: Relate the frequencies at different temperatures
Let \( f_1 = 1 \, \text{kHz} \) (at \( 30^\circ C \)) and \( f_2 = 1.001 \, \text{kHz} \) (at \( 10^\circ C \)). We can express the frequencies in terms of their respective lengths:
\[ f_1 = \frac{1}{2L_0} \sqrt{\frac{T}{\mu}} \quad \text{and} \quad f_2 = \frac{1}{2L'} \sqrt{\frac{T}{\mu}} \]

### Step 4: Set up the ratio of frequencies
Taking the ratio of the two frequencies:
\[ \frac{f_2}{f_1} = \frac{L_0}{L'} \]
Substituting \( L' \):
\[ \frac{1.001}{1} = \frac{L_0}{L_0(1 - 20\alpha)} \]
This simplifies to:
\[ 1.001 = \frac{1}{1 - 20\alpha} \]

### Step 5: Solve for \( \alpha \)
Rearranging gives:
\[ 1 - 20\alpha = 0.999 \implies 20\alpha = 0.001 \implies \alpha = \frac{0.001}{20} = 0.00005 \, ^\circ C^{-1} \]

### Step 6: Express \( \alpha \) in terms of \( k \)
From the problem, we know that:
\[ \alpha = 10^{-k} \, ^\circ C^{-1} \]
Setting this equal to our calculated value:
\[ 10^{-k} = 0.00005 = 5 \times 10^{-5} \]
This implies:
\[ -k = -5 \implies k = 5 \]

### Step 7: Find \( 2k \)
Finally, we need to find \( 2k \):
\[ 2k = 2 \times 5 = 10 \]

### Final Answer
Thus, the value of \( 2k \) is:
\[ \boxed{10} \]