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, we will follow these steps:

### Step 1: Understand the problem
We need to find the mass 'm' that is hung from the steel wire such that when the wire is cooled from 40°C to 30°C, it regains its original length 'L'.

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

### Step 3: Calculate the change in length due to temperature
The change in length \( \Delta L \) due to temperature change can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]

### Step 4: Calculate the cross-sectional area of the wire
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Relate stress and strain using Young's modulus
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Where:
- Stress = \( \frac{F}{A} = \frac{mg}{A} \)
- Strain = \( \frac{\Delta L}{L} \)

### Step 6: Substitute the expressions into Young's modulus equation
Substituting the expressions for stress and strain into the equation for Young's modulus gives:
\[ Y = \frac{mg}{A} \cdot \frac{L}{\Delta L} \]

Rearranging this equation to find \( m \):
\[ m = \frac{Y \cdot A \cdot \Delta L}{g \cdot L} \]

### Step 7: Substitute known values into the equation
We know:
- \( \Delta L = L \cdot \alpha \cdot \Delta T \)
Substituting this into the equation for \( m \):
\[ m = \frac{Y \cdot A \cdot (L \cdot \alpha \cdot \Delta T)}{g \cdot L} \]
The \( L \) cancels out:
\[ m = \frac{Y \cdot A \cdot \alpha \cdot \Delta T}{g} \]

### Step 8: Calculate the mass 'm'
Substituting the known values:
- \( Y = 10^{11} \, \text{N/m}^2 \)
- \( A = \pi \times 10^{-6} \, \text{m}^2 \)
- \( \alpha = 10^{-5} \, \text{°C}^{-1} \)
- \( \Delta T = 10 \, \text{°C} \)
- \( g = 10 \, \text{m/s}^2 \)

\[ m = \frac{10^{11} \cdot \pi \times 10^{-6} \cdot 10^{-5} \cdot 10}{10} \]

This simplifies to:
\[ m = \pi \times 10^{11} \cdot 10^{-6} \cdot 10^{-5} = \pi \times 10^{0} = \pi \approx 3 \, \text{kg} \]

### Final Answer
The value of 'm' is approximately 3 kg.

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