A wire having a linear density of `0.05 g//cm`is stretched between two rigid supports with a tension of `450N`. It is observed that the wire resonates at a frequency of `420 Hz`. The next higher frequency at which the same wire resonates is `490 Hz`. Find the length of the wire.

Options

A. 314 cm
B. 254 cm
C. 214 cm
D. 354 cm

Answer

To find the length of the wire, we can follow these steps:

### Step 1: Calculate the Fundamental Frequency
The fundamental frequency \( f_0 \) can be calculated as the difference between the two resonant frequencies observed:
\[ f_0 = f_2 - f_1 = 490 \, \text{Hz} - 420 \, \text{Hz} = 70 \, \text{Hz} \]

### Step 2: Calculate the Linear Density in SI Units
The linear density \( \mu \) is given as \( 0.05 \, \text{g/cm} \). We need to convert this to kg/m:
\[ \mu = 0.05 \, \text{g/cm} = 0.05 \times 10^{-3} \, \text{kg/m} = 0.00005 \, \text{kg/m} \]

### Step 3: Calculate the Wave Speed
The wave speed \( V \) in the wire can be calculated using the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire. Substituting the values:
\[ V = \sqrt{\frac{450 \, \text{N}}{0.00005 \, \text{kg/m}}} \]
Calculating this gives:
\[ V = \sqrt{9000000} = 3000 \, \text{m/s} \]

### Step 4: Relate Wave Speed to Frequency and Length
The fundamental frequency is related to the wave speed and the length of the wire \( L \) by the formula:
\[ f_0 = \frac{V}{2L} \]
Rearranging to solve for \( L \):
\[ L = \frac{V}{2f_0} \]
Substituting the values we calculated:
\[ L = \frac{3000 \, \text{m/s}}{2 \times 70 \, \text{Hz}} = \frac{3000}{140} \approx 21.43 \, \text{m} \]

### Step 5: Convert Length to Centimeters
To convert the length from meters to centimeters:
\[ L = 21.43 \, \text{m} \times 100 \, \text{cm/m} = 2143 \, \text{cm} \]

### Final Answer
The length of the wire is approximately \( 2143 \, \text{cm} \).

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