A copper wire is held at the two ends by rigid supports. At `30^@C,` the wire is just taut with negligible tension, Find the speed of transverse wave in the wave at `10^@C.` Given : Coefficient of linear expansion is `1.7 xx 10^(-5)l^@C` Young's modulus `=1.3 xx 10^(11)"N/m"^2` Density `=9 xx 10^(3)" kg m"^(-3).`

Answer

To find the speed of a transverse wave in a copper wire when the temperature changes from \(30^\circ C\) to \(10^\circ C\), we will follow these steps:

### Step 1: Understand the relationship between temperature change and tension in the wire
The tension \(F\) in the wire due to a change in temperature can be expressed as:
\[ F = Y \cdot A \cdot \alpha \cdot \Delta \theta \]
where:
- \(Y\) is the Young's modulus,
- \(A\) is the cross-sectional area of the wire,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta \theta\) is the change in temperature.

### Step 2: Calculate the change in temperature
Given:
- Initial temperature \(\theta_1 = 10^\circ C\)
- Final temperature \(\theta_2 = 30^\circ C\)

The change in temperature is:
\[ \Delta \theta = \theta_2 - \theta_1 = 30^\circ C - 10^\circ C = 20^\circ C \]

### Step 3: Write down the known values
- Coefficient of linear expansion \(\alpha = 1.7 \times 10^{-5} \, \text{per} \, ^\circ C\)
- Young's modulus \(Y = 1.3 \times 10^{11} \, \text{N/m}^2\)
- Density \(\rho = 9 \times 10^3 \, \text{kg/m}^3\)

### Step 4: Calculate the tension \(F\)
Assuming the cross-sectional area \(A\) cancels out later, we can express \(F\) as:
\[ F = Y \cdot A \cdot \alpha \cdot \Delta \theta \]
Substituting the known values:
\[ F = (1.3 \times 10^{11}) \cdot A \cdot (1.7 \times 10^{-5}) \cdot (20) \]
\[ F = (1.3 \times 10^{11}) \cdot A \cdot (3.4 \times 10^{-4}) \]
\[ F = 4.42 \times 10^{7} \cdot A \, \text{N} \]

### Step 5: Calculate the mass per unit length \(\mu\)
The mass per unit length \(\mu\) is given by:
\[ \mu = \rho \cdot A \]
Substituting the known density:
\[ \mu = (9 \times 10^3) \cdot A \, \text{kg/m} \]

### Step 6: Write the formula for the speed of the wave
The speed \(v\) of the transverse wave is given by:
\[ v = \sqrt{\frac{F}{\mu}} \]
Substituting \(F\) and \(\mu\):
\[ v = \sqrt{\frac{4.42 \times 10^{7} \cdot A}{9 \times 10^3 \cdot A}} \]
The area \(A\) cancels out:
\[ v = \sqrt{\frac{4.42 \times 10^{7}}{9 \times 10^3}} \]
\[ v = \sqrt{4.91 \times 10^{3}} \approx 70 \, \text{m/s} \]

### Final Answer
The speed of the transverse wave in the copper wire at \(10^\circ C\) is approximately \(70 \, \text{m/s}\).

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