A metal wire of linear mass density of `9.8g//m` is stretched with a tension of `10 kg-wt` between two rigid support `1meter` 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. 50 Hz
B. 100 Hz
C. 200 Hz
D. 25 Hz

Answer

To solve the problem, we need to find the frequency \( n \) of the alternating current that causes the wire to vibrate in resonance. We will use the formula for the resonant frequency of a vibrating string:

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

where:
- \( n \) 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 \( \mu = 9.8 \, \text{g/m} = 9.8 \times 10^{-3} \, \text{kg/m} \) (convert grams to kilograms)
- Tension \( T = 10 \, \text{kg-wt} = 10 \times 9.8 \, \text{N} = 98 \, \text{N} \) (convert kg-wt to Newtons using \( g \approx 9.8 \, \text{m/s}^2 \))
- Length of the wire \( L = 1 \, \text{m} \)

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

\[ n = \frac{1}{2 \times 1} \sqrt{\frac{98}{9.8 \times 10^{-3}}} \]

### Step 3: Simplify the expression
Calculating the term inside the square root:

\[ \frac{98}{9.8 \times 10^{-3}} = \frac{98}{0.0098} = 10000 \]

Now, taking the square root:

\[ \sqrt{10000} = 100 \]

### Step 4: Calculate the frequency
Now substituting back into the frequency formula:

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

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

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