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 have a steel wire of length \( L \) at \( 40^\circ C \). When the wire is cooled down to \( 30^\circ C \), it regains its original length \( L \) by hanging a mass \( m \) from its end. We need to find the value of \( m \).

### Step 2: Use the formula for linear thermal expansion
The change in length due to temperature change can be expressed using the linear thermal expansion formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
Where:
- \( \Delta L \) is the change in length,
- \( \alpha \) is the coefficient of linear thermal expansion,
- \( \Delta T \) is the change in temperature.

In this case, \( \Delta T = 40^\circ C - 30^\circ C = 10^\circ C \).

### Step 3: Calculate the change in length
Substituting the values into the formula:
\[ \Delta L = L \cdot (10^{-5} \, /^\circ C) \cdot (10^\circ C) = L \cdot 10^{-4} \]

### Step 4: Relate change in length to strain
The strain \( \epsilon \) in the wire is given by:
\[ \epsilon = \frac{\Delta L}{L} = \frac{L \cdot 10^{-4}}{L} = 10^{-4} \]

### Step 5: Use Young's modulus
Young's modulus \( Y \) relates stress and strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Stress can be expressed as force per unit area:
\[ \text{Stress} = \frac{F}{A} \]
Where \( A \) is the cross-sectional area of the wire. The area \( A \) for a circular wire is given by:
\[ A = \pi r^2 \]
Given the radius \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \):
\[ A = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 6: Substitute into Young's modulus equation
From the Young's modulus equation:
\[ Y = \frac{F}{A \cdot \epsilon} \]
Rearranging gives:
\[ F = Y \cdot A \cdot \epsilon \]

### Step 7: Substitute known values
Substituting the known values:
- \( Y = 10^{11} \, \text{N/m}^2 \)
- \( A = \pi \times 10^{-6} \, \text{m}^2 \)
- \( \epsilon = 10^{-4} \)

So,
\[ F = 10^{11} \cdot (\pi \times 10^{-6}) \cdot (10^{-4}) = \pi \times 10^{1} \, \text{N} \]

### Step 8: Relate force to mass
The force \( F \) is also equal to the weight of the mass \( m \):
\[ F = m \cdot g \]
Where \( g \approx 10 \, \text{m/s}^2 \). Therefore,
\[ m = \frac{F}{g} = \frac{\pi \times 10^{1}}{10} = \pi \, \text{kg} \]

### Step 9: Calculate the approximate value of \( m \)
Using \( \pi \approx 3.14 \):
\[ m \approx 3.14 \, \text{kg} \]

### Final Answer
The value of \( m \) is approximately \( 3.14 \, \text{kg} \).

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