Steel wire of length 'L' at `40^@C` is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from `40^@C to 30^@C` to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is `10^-5//^@C`, Young's modulus of steel is `10^11 N//m^2` and radius of the wire is 1mm. Assume that `L gt gt` diameter of the wire. Then the value of 'm' in kg is nearly

Explanation

To solve the problem step by step, we will use the concepts of Young's modulus, thermal expansion, and the relationship between stress, strain, and force.

### Step 1: Understand the Problem
We have a steel wire of length \( L \) at \( 40^\circ C \) that is cooled to \( 30^\circ C \). We need to find the mass \( m \) that is hung from the wire to bring it back to its original length \( L \).

### Step 2: Identify Given Values
- Coefficient of linear thermal expansion \( \alpha = 10^{-5} \, ^\circ C^{-1} \)
- Young's modulus \( Y = 10^{11} \, N/m^2 \)
- Radius of the wire \( r = 1 \, mm = 1 \times 10^{-3} \, m \)
- Change in temperature \( \Delta T = 40^\circ C - 30^\circ C = 10^\circ C \)

### Step 3: Calculate the Change in Length Due to Temperature
The change in length \( \Delta L \) due to thermal expansion can be calculated using the formula:
\[ \Delta L = L \alpha \Delta T \]
Substituting the values:
\[ \Delta L = L \cdot (10^{-5}) \cdot (10) = L \cdot 10^{-4} \]

### Step 4: Relate Stress and Strain
Young's modulus is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress \( = \frac{F}{A} = \frac{mg}{A} \)
- Strain \( = \frac{\Delta L}{L} \)

Thus, we can write:
\[ Y = \frac{mg/A}{\Delta L/L} \]
Rearranging gives:
\[ Y = \frac{mgL}{A \Delta L} \]

### Step 5: Substitute \( \Delta L \)
Substituting \( \Delta L = L \cdot 10^{-4} \) into the equation:
\[ Y = \frac{mgL}{A (L \cdot 10^{-4})} \]
This simplifies to:
\[ Y = \frac{mg}{A \cdot 10^{-4}} \]

### Step 6: Solve for Mass \( m \)
Rearranging for \( m \):
\[ m = \frac{Y \cdot A \cdot 10^{-4}}{g} \]

### Step 7: Calculate the Area \( A \)
The area \( A \) of the wire can be calculated as:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, m^2 \]

### Step 8: Substitute Values
Now substituting \( Y \), \( A \), and \( g \):
\[ m = \frac{(10^{11}) \cdot (\pi \times 10^{-6}) \cdot (10^{-4})}{10} \]
This simplifies to:
\[ m = \frac{10^{11} \cdot \pi \cdot 10^{-10}}{10} = \frac{\pi \cdot 10^{1}}{10} = \pi \]

### Step 9: Approximate the Value of \( \pi \)
Using \( \pi \approx 3.14 \):
\[ m \approx 3.14 \, kg \]

### Conclusion
Thus, the mass \( m \) that needs to be hung from the wire is nearly \( 3 \, kg \).