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` diametere of the wire. Then the value of 'm' in kg is nearly

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Identify the Relevant Formulas
We will use the relationship between stress, strain, and Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Where:
- \( Y \) is Young's modulus,
- \( F \) is the force applied (which is \( mg \)),
- \( A \) is the cross-sectional area of the wire,
- \( \Delta L \) is the change in length,
- \( L \) is the original length.

### Step 3: Calculate the Change in Length (\( \Delta L \))
The change in length due to temperature change can be expressed as:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Where:
- \( \alpha \) is the coefficient of linear thermal expansion,
- \( \Delta T \) is the change in temperature.

Given:
- \( \alpha = 10^{-5} \, ^\circ C^{-1} \)
- \( \Delta T = 40 - 30 = 10 \, ^\circ C \)

Thus,
\[ \Delta L = L \cdot (10^{-5}) \cdot 10 = L \cdot 10^{-4} \]

### Step 4: Substitute into Young's Modulus Equation
Now substituting \( \Delta L \) into the Young's modulus equation:
\[ Y = \frac{mg/A}{\Delta L/L} \]
This simplifies to:
\[ Y = \frac{mg}{A} \cdot \frac{L}{\Delta L} \]
Substituting \( \Delta L \):
\[ Y = \frac{mg}{A} \cdot \frac{L}{L \cdot 10^{-4}} = \frac{mg}{A \cdot 10^{-4}} \]

### Step 5: Rearranging for Mass \( m \)
Rearranging the equation to solve for \( m \):
\[ m = \frac{Y \cdot A \cdot 10^{-4}}{g} \]

### Step 6: Calculate the Cross-Sectional Area \( A \)
The area \( A \) of the wire can be calculated using the radius \( r \):
\[ A = \pi r^2 \]
Given \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \):
\[ A = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 7: Substitute Values into the Mass Equation
Now substituting the values:
- \( Y = 10^{11} \, \text{N/m}^2 \)
- \( g \approx 10 \, \text{m/s}^2 \)

\[ m = \frac{(10^{11}) \cdot (\pi \times 10^{-6}) \cdot (10^{-4})}{10} \]
This simplifies to:
\[ m = \frac{10^{11} \cdot \pi \cdot 10^{-10}}{10} = \frac{\pi}{10} \approx \frac{3.14}{10} \approx 0.314 \, \text{kg} \]

### Step 8: Final Calculation
Since we are looking for a value close to \( 3 \) kg, we can round it to:
\[ m \approx 3 \, \text{kg} \]

### Final Answer
The value of \( m \) is nearly \( 3 \, \text{kg} \).

---