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 find the speed of transverse waves in a copper wire at a temperature of \(30^\circ C\), we will follow these steps:

### Step 1: Determine the change in temperature
The initial temperature of the wire is \(50^\circ C\) and we want to find the speed at \(30^\circ C\).

\[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 50^\circ C - 30^\circ C = 20^\circ C \]

### Step 2: Calculate the tension in the wire
The tension \(T\) in the wire can be calculated using the formula derived from Young's modulus:

\[ T = Y \cdot \alpha \cdot \Delta T \cdot A \]

However, since we are looking for the speed of the wave, we need to express tension in terms of Young's modulus, coefficient of linear expansion, and density. We can simplify this to:

\[ T = Y \cdot \alpha \cdot \Delta T \cdot A \]

### Step 3: Calculate the mass per unit length (\(\mu\))
The mass per unit length \(\mu\) can be expressed as:

\[ \mu = \rho \cdot A \]

Where \(\rho\) is the density of the wire.

### Step 4: Substitute into the wave speed formula
The speed of transverse waves \(v\) in a stretched string or wire is given by:

\[ v = \sqrt{\frac{T}{\mu}} \]

Substituting the expressions for \(T\) and \(\mu\):

\[ v = \sqrt{\frac{Y \cdot \alpha \cdot \Delta T \cdot A}{\rho \cdot A}} = \sqrt{\frac{Y \cdot \alpha \cdot \Delta T}{\rho}} \]

### Step 5: Plug in the values
Now we can substitute the known values into the equation:

- \(Y = 1.2 \times 10^{11} \, \text{N/m}^2\)
- \(\alpha = 1.6 \times 10^{-5} \, \text{°C}^{-1}\)
- \(\Delta T = 20 \, \text{°C}\)
- \(\rho = 9.2 \times 10^3 \, \text{kg/m}^3\)

Substituting these values into the equation:

\[ v = \sqrt{\frac{(1.2 \times 10^{11}) \cdot (1.6 \times 10^{-5}) \cdot (20)}{9.2 \times 10^3}} \]

### Step 6: Calculate the value
Calculating the numerator:

\[ 1.2 \times 10^{11} \cdot 1.6 \times 10^{-5} \cdot 20 = 3.84 \times 10^{7} \]

Now, calculating the speed:

\[ v = \sqrt{\frac{3.84 \times 10^{7}}{9.2 \times 10^3}} = \sqrt{4164.35} \approx 64.5 \, \text{m/s} \]

### Final Answer
The speed of transverse waves in the copper wire at \(30^\circ C\) is approximately \(64.5 \, \text{m/s}\).

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