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. (a) `50 Hz`
B. (b) `100 Hz`
C. ( c ) `200 Hz`
D. (d) `25 Hz`

Explanation

To solve the problem, we need to find the frequency \( n \) of the alternating current that causes the metal wire to vibrate in resonance. Here are the steps to arrive at the solution:

### Step 1: Understand the Given Data
- Linear mass density \( \mu = 9.8 \, \text{g/m} = 9.8 \times 10^{-3} \, \text{kg/m} \) (since \( 1 \, \text{g} = 10^{-3} \, \text{kg} \))
- Tension \( T = 10 \, \text{kg-wt} = 10 \times 9.8 \, \text{N} = 98 \, \text{N} \) (using \( g = 9.8 \, \text{m/s}^2 \))
- Length of the wire \( L = 1 \, \text{m} \)

### Step 2: Use the Formula for Fundamental Frequency
The fundamental frequency \( n \) of a vibrating string fixed at both ends is given by the formula:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

### Step 3: Substitute the Values into the Formula
Substituting the values into the formula:
\[ n = \frac{1}{2 \times 1} \sqrt{\frac{98}{9.8 \times 10^{-3}}} \]

### Step 4: Simplify the Expression
Calculate \( \frac{98}{9.8 \times 10^{-3}} \):
\[ \frac{98}{9.8 \times 10^{-3}} = 10000 \]
Now, substituting back:
\[ n = \frac{1}{2} \sqrt{10000} \]
\[ n = \frac{1}{2} \times 100 = 50 \, \text{Hz} \]

### Step 5: Conclusion
The frequency \( n \) of the alternating source is \( 50 \, \text{Hz} \).

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

---