A load of 40 kg is suspended from a wire of radius `10^(-3) m` and density `8xx 10 kgm^(-3) 75%` of the work done to increase the length is converted into heat. Calculate the change in temperature (in kelvin) of the wire . Given that `Y= 7.5 xx 10^(10) Nm^(-2)` and heat capacity of the material of the wire is `490 J kg K^(-1)` respectively. (Take `pi^(2)=9.8 and g =9.8 m//s^(2))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Load in Newtons
The load (force) \( F \) can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m = 40 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)

Calculating:
\[ F = 40 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 392 \, \text{N} \]

### Step 2: 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:
\[ A = \pi r^2 \]
where:
- \( r = 10^{-3} \, \text{m} \)

Calculating:
\[ A = \pi (10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]
Using \( \pi^2 = 9.8 \):
\[ A \approx 9.8 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the Change in Length of the Wire
Using Young's modulus \( Y \):
\[ Y = \frac{F / A}{\Delta L / L} \]
Rearranging gives:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

### Step 4: Calculate the Work Done on the Wire
The work done \( W \) to increase the length of the wire can be calculated using:
\[ W = \frac{1}{2} F \Delta L \]
Substituting \( \Delta L \):
\[ W = \frac{1}{2} F \left( \frac{F \cdot L}{A \cdot Y} \right) = \frac{F^2 \cdot L}{2 A Y} \]

### Step 5: Calculate the Heat Converted
Given that 75% of the work done is converted to heat:
\[ Q = 0.75 W = 0.75 \left( \frac{F^2 \cdot L}{2 A Y} \right) \]

### Step 6: Calculate the Mass of the Wire
The mass \( m_{\text{wire}} \) of the wire can be calculated using:
\[ m_{\text{wire}} = \text{Volume} \times \text{Density} = A \cdot L \cdot \rho \]
where \( \rho = 8 \times 10^3 \, \text{kg/m}^3 \).

### Step 7: Relate Heat to Temperature Change
Using the specific heat capacity \( S \):
\[ Q = m_{\text{wire}} \cdot S \cdot \Delta T \]
Substituting for \( Q \):
\[ 0.75 \left( \frac{F^2 \cdot L}{2 A Y} \right) = (A \cdot L \cdot \rho) \cdot S \cdot \Delta T \]

### Step 8: Solve for the Change in Temperature \( \Delta T \)
Rearranging gives:
\[ \Delta T = \frac{0.75 \left( \frac{F^2}{2 A Y} \right)}{A \cdot \rho \cdot S} \]

### Step 9: Substitute Values
Substituting the values:
- \( F = 392 \, \text{N} \)
- \( A = \pi \times 10^{-6} \)
- \( Y = 7.5 \times 10^{10} \, \text{N/m}^2 \)
- \( \rho = 8 \times 10^3 \, \text{kg/m}^3 \)
- \( S = 490 \, \text{J/(kg K)} \)

Calculating:
\[ \Delta T = \frac{0.75 \left( \frac{392^2}{2 \cdot \pi \times 10^{-6} \cdot 7.5 \times 10^{10}} \right)}{(\pi \times 10^{-6}) \cdot (8 \times 10^3) \cdot 490} \]

### Final Calculation
After substituting and simplifying, we find:
\[ \Delta T \approx 0.02 \, \text{K} \]