A metal wire of linear mass density `10^(-1)` 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 solve the problem, we need to find the frequency \( n \) of the vibrating wire. We will use the formula for the frequency of a vibrating string fixed at both ends:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire.

### Step 1: Identify the given values
- Linear mass density \( \mu = 10 \, \text{g/m} = 0.01 \, \text{kg/m} \) (since \( 1 \, \text{g} = 0.001 \, \text{kg} \))
- Tension \( T = 100 \, \text{N} \)
- Length \( L = 1 \, \text{m} \)

### Step 2: Substitute the values into the formula
Now we can substitute the values into the frequency formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{2 \times 1} \sqrt{\frac{100}{0.01}} \]

### Step 3: Simplify the expression
Calculate \( \frac{100}{0.01} \):

\[ \frac{100}{0.01} = 10000 \]

Now take the square root:

\[ \sqrt{10000} = 100 \]

Now substitute this back into the frequency formula:

\[ f = \frac{1}{2} \times 100 = 50 \, \text{Hz} \]

### Final Answer
The frequency \( n \) of the alternating current is:

\[ n = 50 \, \text{Hz} \]

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