To find the extension in the string, we will use the relationship defined by Young's modulus and the wave velocity. Let's break down the solution step by step.
### Step 1: Convert Units
First, we need to convert all the given values to SI units.
- Length of the string, \( L = 60 \, \text{cm} = 0.6 \, \text{m} \)
- Mass of the string, \( m = 6 \, \text{g} = 0.006 \, \text{kg} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Velocity of wave, \( v = 90 \, \text{m/s} \)
- Young's modulus, \( Y = 1.6 \times 10^{11} \, \text{N/m}^2 \)
### Step 2: Calculate Mass per Unit Length
Next, we calculate the mass per unit length (\( \mu \)) of the string.
\[
\mu = \frac{m}{L} = \frac{0.006 \, \text{kg}}{0.6 \, \text{m}} = 0.01 \, \text{kg/m}
\]
### Step 3: Calculate Tension in the String
Using the wave velocity formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
We can rearrange this to find the tension \( T \):
\[
T = \mu v^2
\]
Substituting the values:
\[
T = 0.01 \, \text{kg/m} \times (90 \, \text{m/s})^2 = 0.01 \times 8100 = 81 \, \text{N}
\]
### Step 4: Calculate Extension in the String
Using the formula for Young's modulus:
\[
Y = \frac{T}{A} \cdot \frac{L}{\Delta L}
\]
Rearranging to find the extension \( \Delta L \):
\[
\Delta L = \frac{T \cdot L}{A \cdot Y}
\]
Substituting the values:
\[
\Delta L = \frac{81 \, \text{N} \cdot 0.6 \, \text{m}}{1 \times 10^{-6} \, \text{m}^2 \cdot 1.6 \times 10^{11} \, \text{N/m}^2}
\]
Calculating the numerator:
\[
81 \cdot 0.6 = 48.6 \, \text{N m}
\]
Calculating the denominator:
\[
1 \times 10^{-6} \cdot 1.6 \times 10^{11} = 1.6 \times 10^{5} \, \text{N}
\]
Now substituting back:
\[
\Delta L = \frac{48.6}{1.6 \times 10^{5}} = 3.0375 \times 10^{-4} \, \text{m} = 0.00030375 \, \text{m}
\]
Converting to millimeters:
\[
\Delta L = 0.30375 \, \text{mm} \approx 0.3 \, \text{mm}
\]
### Final Answer
The extension in the string is approximately \( \Delta L = 0.3 \, \text{mm} \).
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