A copper wire is held at the two ends by rigid supports . At ` 50^(@)C` the wire is just taut , with negligible tension . If `Y=1.2 xx 10^(11) N//m^(2) , alpha = 1.6 xx 10^(-5)//""^(@)C` and ` rho = 9.2 xx 10^(3) kg//m^(3)` , then the speed of transverse waves in this wire at ` 30^(@)C` is

Options

A. 64. 6 m/s
B. 16.2 m/s
C. 23.2 m/s
D. 32.2 m/s

Answer

To solve the problem of finding the speed of transverse waves in a copper wire at 30°C, we will follow these steps:

### Step 1: Identify the given data
- Young's modulus (Y) = \(1.2 \times 10^{11} \, \text{N/m}^2\)
- Coefficient of linear expansion (\(\alpha\)) = \(1.6 \times 10^{-5} \, \text{°C}^{-1}\)
- Density (\(\rho\)) = \(9.2 \times 10^{3} \, \text{kg/m}^3\)
- Initial temperature = \(50 \, \text{°C}\)
- Final temperature = \(30 \, \text{°C}\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 50 \, \text{°C} - 30 \, \text{°C} = 20 \, \text{°C} \]

### Step 3: Calculate the change in length (\(\Delta L\))
Using the formula for change in length due to temperature change:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Assuming the initial length of the wire is \(L\):
\[ \Delta L = L \cdot (1.6 \times 10^{-5}) \cdot (20) = L \cdot 3.2 \times 10^{-4} \]

### Step 4: Calculate the tension (T) in the wire
The tension in the wire can be calculated using the formula:
\[ T = \frac{\Delta L}{L} \cdot Y \cdot A \]
Where \(A\) is the cross-sectional area of the wire. Substituting \(\Delta L\):
\[ T = (3.2 \times 10^{-4}) \cdot Y \cdot A = (3.2 \times 10^{-4}) \cdot (1.2 \times 10^{11}) \cdot A \]
\[ T = 3.84 \times 10^{7} \cdot A \]

### Step 5: Calculate the mass per unit length (\(\mu\))
The mass per unit length is given by:
\[ \mu = \rho \cdot A \]
Substituting the value of density:
\[ \mu = (9.2 \times 10^{3}) \cdot A \]

### Step 6: Calculate the speed of transverse waves (v)
The speed of transverse waves in the wire is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values of \(T\) and \(\mu\):
\[ v = \sqrt{\frac{3.84 \times 10^{7} \cdot A}{9.2 \times 10^{3} \cdot A}} \]
The area \(A\) cancels out:
\[ v = \sqrt{\frac{3.84 \times 10^{7}}{9.2 \times 10^{3}}} \]
Calculating the value:
\[ v = \sqrt{4165.217} \approx 64.6 \, \text{m/s} \]

### Conclusion
The speed of transverse waves in the copper wire at \(30 \, \text{°C}\) is approximately \(64.6 \, \text{m/s}\).

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