A metal wire of linear mass density `10`g/m is stretched with a tension of 100 N weight between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is :

Options

A. 25Hz
B. 50Hz
C. 100Hz
D. 200Hz

Explanation

To find the frequency \( n \) of the alternating source for the vibrating metal wire, we can follow these steps:

### Step 1: Identify the given values
- Linear mass density \( \mu = 10 \, \text{g/m} = 10 \times 10^{-3} \, \text{kg/m} \) (since 1 g = \( 10^{-3} \) kg)
- Tension \( T = 100 \, \text{N} \)
- Length of the wire \( L = 1 \, \text{m} \)

### Step 2: Use the formula for the frequency of a vibrating string
For a wire fixed at both ends, the fundamental frequency \( f \) is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the frequency formula:
\[ f = \frac{1}{2 \times 1} \sqrt{\frac{100}{10 \times 10^{-3}}} \]

### Step 4: Simplify the expression
Calculating the term inside the square root:
\[ \frac{100}{10 \times 10^{-3}} = \frac{100}{0.01} = 10000 \]
Now, taking the square root:
\[ \sqrt{10000} = 100 \]
Thus, substituting back into the frequency formula:
\[ f = \frac{1}{2} \times 100 = 50 \, \text{Hz} \]

### Step 5: Conclusion
Since the wire vibrates in resonance with the alternating current, the frequency \( n \) of the alternating source is:
\[ n = 50 \, \text{Hz} \]

### Final Answer
The frequency \( n \) of the alternating source is \( 50 \, \text{Hz} \).

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