The speed of a sound in an aluminiurn bar is 5.1 km/s If its density is 2700 `kg//m^2`. what is its Young's Modulus ?

Answer

To find the Young's Modulus of the aluminum bar given the speed of sound and its density, we can follow these steps:

### Step-by-step Solution:

1. **Convert the speed of sound to standard units:**
The speed of sound in the aluminum bar is given as 5.1 km/s. We need to convert this to meters per second (m/s).
\[ V = 5.1 \, \text{km/s} = 5.1 \times 10^3 \, \text{m/s} \]

2. **Identify the density:**
The density (\(\rho\)) of aluminum is given as 2700 kg/m³. This is already in standard units.

3. **Use the formula relating speed of sound, Young's Modulus, and density:**
The speed of sound in a material is given by the formula:
\[ V = \sqrt{\frac{Y}{\rho}} \]
where \(Y\) is Young's Modulus and \(\rho\) is the density.

4. **Square both sides of the equation:**
To eliminate the square root, we square both sides:
\[ V^2 = \frac{Y}{\rho} \]

5. **Rearrange the equation to solve for Young's Modulus:**
Multiply both sides by \(\rho\) to isolate \(Y\):
\[ Y = V^2 \cdot \rho \]

6. **Substitute the values:**
Now, substitute the values of \(V\) and \(\rho\):
\[ Y = (5.1 \times 10^3)^2 \cdot 2700 \]

7. **Calculate \(V^2\):**
First, calculate \(V^2\):
\[ V^2 = (5.1 \times 10^3)^2 = 26.01 \times 10^6 \, \text{m}^2/\text{s}^2 \]

8. **Calculate Young's Modulus:**
Now multiply \(V^2\) by \(\rho\):
\[ Y = 26.01 \times 10^6 \cdot 2700 \]
\[ Y = 70.227 \times 10^9 \, \text{N/m}^2 \]

9. **Express in scientific notation:**
Finally, we can express the result in scientific notation:
\[ Y \approx 7.0 \times 10^{10} \, \text{N/m}^2 \]

### Final Answer:
The Young's Modulus of the aluminum bar is approximately \(7.0 \times 10^{10} \, \text{N/m}^2\).

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