The speed of longitudinal wave is 100 times, then the speed of transverse wave in a brass wire. What is the stress in wire ? TheYoung's modulus of brass Is `1.0 xx 10^(11)N//m^(2)`

Answer

To solve the problem, we need to find the stress in the brass wire given that the speed of longitudinal waves is 100 times the speed of transverse waves, and we know the Young's modulus of brass.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( v_t \) be the speed of transverse wave.
- Let \( v_l \) be the speed of longitudinal wave.
- Given that \( v_l = 100 \times v_t \).

2. **Use the Formula for Wave Speeds**:
- The speed of a transverse wave in a wire is given by:
\[ v_t = \sqrt{\frac{T}{m}} \]
where \( T \) is the tension in the wire and \( m \) is the mass per unit length of the wire.
- The mass per unit length can be expressed as:
\[ m = \rho \cdot A \]
where \( \rho \) is the density of the material and \( A \) is the cross-sectional area of the wire. For a circular cross-section, \( A = \pi r^2 \).

3. **Express the Speed of Longitudinal Wave**:
- The speed of a longitudinal wave is given by:
\[ v_l = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is the Young's modulus.

4. **Set Up the Relationship**:
- From the problem, we know:
\[ v_l = 100 \times v_t \]
- Substituting the expressions for \( v_l \) and \( v_t \):
\[ \sqrt{\frac{Y}{\rho}} = 100 \times \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]

5. **Square Both Sides**:
- Squaring both sides gives:
\[ \frac{Y}{\rho} = 10000 \times \frac{T}{\rho \cdot \pi r^2} \]

6. **Cancel \( \rho \)**:
- Since \( \rho \) is on both sides, we can cancel it out:
\[ Y = 10000 \times \frac{T}{\pi r^2} \]

7. **Relate Stress to Young's Modulus**:
- Stress \( \sigma \) is defined as:
\[ \sigma = \frac{T}{A} = \frac{T}{\pi r^2} \]
- Rearranging gives:
\[ T = \sigma \cdot \pi r^2 \]

8. **Substitute Back**:
- Substitute \( T \) back into the equation for \( Y \):
\[ Y = 10000 \times \frac{\sigma \cdot \pi r^2}{\pi r^2} \]
- This simplifies to:
\[ Y = 10000 \sigma \]

9. **Solve for Stress**:
- Rearranging gives:
\[ \sigma = \frac{Y}{10000} \]
- Substitute the given value of Young's modulus:
\[ \sigma = \frac{1.0 \times 10^{11}}{10000} = 1.0 \times 10^{7} \, \text{N/m}^2 \]

### Final Answer:
The stress in the wire is \( \sigma = 1.0 \times 10^{7} \, \text{N/m}^2 \).

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