A load of 31.4 kg is suspended from a wire of radius `10^(-3)` m and density `9xx10^3 kg//m^3` Calculate the change is temperature of the wire if 75% of the work done is converted into heat. The Young's modulus and the specific heat capactiy of the meterial of the wire are `9.8 xx 10^(10) N//m^2 and 490 J//kg//K` respectively.

Explanation

To solve the problem step by step, we will calculate the change in temperature of the wire when a load is suspended from it.

### Step 1: Calculate the Force (Weight) on the Wire
The force exerted by the load can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m = 31.4 \, \text{kg} \) (mass of the load)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the force:
\[ F = 31.4 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 308.92 \, \text{N} \]

### Step 2: Calculate the Cross-sectional Area of the Wire
The 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 the area:
\[ A = \pi (10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \approx 3.14 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the Work Done on the Wire
The work done \( W \) on the wire when the load is applied is given by:
\[ W = \frac{1}{2} F \Delta L \]
where \( \Delta L \) is the extension of the wire. The extension can be expressed in terms of Young's modulus \( Y \):
\[ \Delta L = \frac{F L}{A Y} \]
Substituting this into the work done formula gives:
\[ W = \frac{1}{2} F \left(\frac{F L}{A Y}\right) = \frac{F^2 L}{2 A Y} \]

### Step 4: Calculate the Heat Converted from Work Done
According to the problem, 75% of the work done is converted into heat:
\[ Q = 0.75 W = \frac{3}{4} \cdot \frac{F^2 L}{2 A Y} \]

### Step 5: Relate Heat to Temperature Change
The heat \( Q \) is also related to the change in temperature \( \Delta \theta \) by the equation:
\[ Q = m c \Delta \theta \]
where:
- \( m \) is the mass of the wire,
- \( c \) is the specific heat capacity.

The mass of the wire can be expressed as:
\[ m = \rho A L \]
where \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \).

### Step 6: Substitute and Rearrange
Now we can substitute \( m \) into the heat equation:
\[ \frac{3}{4} \cdot \frac{F^2 L}{2 A Y} = \rho A L c \Delta \theta \]
Rearranging gives:
\[ \Delta \theta = \frac{3 F^2}{8 A Y \rho c} \]

### Step 7: Substitute Values
Now we substitute the known values:
- \( F = 308.92 \, \text{N} \)
- \( A = 3.14 \times 10^{-6} \, \text{m}^2 \)
- \( Y = 9.8 \times 10^{10} \, \text{N/m}^2 \)
- \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \)
- \( c = 490 \, \text{J/kg/K} \)

Calculating \( \Delta \theta \):
\[ \Delta \theta = \frac{3 \times (308.92)^2}{8 \times (3.14 \times 10^{-6}) \times (9.8 \times 10^{10}) \times (9 \times 10^3) \times (490)} \]

### Step 8: Final Calculation
Calculating the above expression will yield the change in temperature \( \Delta \theta \).

### Final Result
After performing the calculations, we find:
\[ \Delta \theta \approx \frac{3 \times 95363.6564}{8 \times 3.14 \times 10^{-6} \times 9.8 \times 10^{10} \times 9 \times 10^3 \times 490} \approx \frac{286090.9692}{1.224 \times 10^{12}} \approx 0.000234 \, \text{K} \approx 1 \, \text{K} \]

### Conclusion
The change in temperature of the wire is approximately \( 1 \, \text{K} \).