The speed of longitudinal wave is ten times the speed of transverse waves in a tight brass wire. If young's modulus of the wire is Y , then strain in the wire is

Options

A. 100
B. `1/(100)`
C. `1/(10)`
D. 10

Answer

To solve the problem, we need to find the strain in the brass wire given that the speed of longitudinal waves is ten times that of transverse waves. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( S_L \) be the speed of longitudinal waves.
- Let \( S_T \) be the speed of transverse waves.
- According to the problem, we have the relationship:
\[ S_L = 10 S_T \]

2. **Formulas for Wave Speeds**:
- The speed of longitudinal waves in a material is given by:
\[ S_L = \sqrt{\frac{E}{\rho}} \]
where \( E \) is Young's modulus and \( \rho \) is the density of the material.
- The speed of transverse waves is given by:
\[ S_T = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the linear mass density.

3. **Relate Tension and Young's Modulus**:
- The tension \( T \) can be expressed in terms of Young's modulus and strain:
\[ T = E \cdot \text{strain} = E \cdot \frac{\Delta L}{L} \]
where \( \Delta L \) is the change in length and \( L \) is the original length.

4. **Substituting Tension in Transverse Wave Speed**:
- Substitute \( T \) into the formula for \( S_T \):
\[ S_T = \sqrt{\frac{E \cdot \frac{\Delta L}{L}}{\mu}} \]

5. **Relate Linear Mass Density to Density**:
- The linear mass density \( \mu \) can be expressed as:
\[ \mu = \rho \cdot A \]
where \( A \) is the cross-sectional area of the wire.

6. **Substituting Linear Mass Density**:
- Substitute \( \mu \) into the equation for \( S_T \):
\[ S_T = \sqrt{\frac{E \cdot \frac{\Delta L}{L}}{\rho \cdot A}} \]

7. **Setting Up the Ratio**:
- Now we have:
\[ \frac{S_T}{S_L} = \frac{\sqrt{\frac{E \cdot \frac{\Delta L}{L}}{\rho \cdot A}}}{\sqrt{\frac{E}{\rho}}} \]
- This simplifies to:
\[ \frac{S_T}{S_L} = \sqrt{\frac{\Delta L}{L \cdot A}} \]

8. **Using the Given Relationship**:
- From the earlier relationship \( S_L = 10 S_T \), we can write:
\[ \frac{S_T}{S_L} = \frac{1}{10} \]
- Thus:
\[ \sqrt{\frac{\Delta L}{L \cdot A}} = \frac{1}{10} \]

9. **Squaring Both Sides**:
- Squaring both sides gives:
\[ \frac{\Delta L}{L \cdot A} = \frac{1}{100} \]

10. **Finding Strain**:
- Since strain is defined as \( \frac{\Delta L}{L} \), we can express it as:
\[ \text{strain} = \frac{\Delta L}{L} = \frac{1}{100} \]

### Final Answer:
The strain in the wire is:
\[ \text{strain} = \frac{1}{100} \]