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Step-by-Step Solution
Step 1: Identify the Known Physical Quantities
• Mass of the load, M = 31.4\,\text{kg}
• Radius of wire, r = 10^{-3}\,\text{m}
• Density of material, \rho = 9 \times 10^{3}\,\text{kg/m}^3
• Young’s modulus, Y = 9.8 \times 10^{10}\,\text{N/m}^2
• Specific heat capacity, s = 490\,\text{J/(kg·K)}
• Fraction of work converted to heat = 75\% = \frac{3}{4}
• Take \pi = 3.14
• Acceleration due to gravity, g = 9.8\,\text{m/s}^2
Step 2: Write the Formula for Work Done in Stretching
The work done W in stretching an elastic wire can be written as:
W = \frac{1}{2} \times (\text{stress}) \times (\text{strain}) \times V,
using the relationship \text{strain} = \frac{\text{stress}}{Y} , this becomes:
W = \frac{1}{2} \times \frac{(\text{stress})^2}{Y} \times V,
where V is the volume of the wire.
Step 3: Express the Stress in Terms of the Force
The stress on the wire is the force divided by its cross-sectional area A . The force is Mg , where M is the mass and g is the acceleration due to gravity:
\text{stress} = \frac{F}{A} = \frac{Mg}{A} = \frac{Mg}{\pi r^2}.
Step 4: Write Down the Work Done, W
Substituting \text{stress} = \frac{Mg}{\pi r^2} into W = \frac{1}{2} \cdot \frac{(\text{stress})^2}{Y} \cdot V :
W = \frac{1}{2} \times \frac{\left(\frac{Mg}{\pi r^2}\right)^2}{Y} \times V.
Step 5: Relate the Heat Produced to the Work Done
According to the problem, 75\% of the work done is converted into heat:
H = \frac{3}{4} W.
This heat ( H ) raises the temperature of the wire:
H = m\,s\,\Delta \theta,
where m is the mass of the wire, s is its specific heat capacity, and \Delta \theta is the increase in temperature.
Step 6: Express the Mass of the Wire
The mass of the wire is given by:
m = \rho \, V.
Step 7: Set Up the Heat Balance Equation
From Steps 5 and 6:
\rho\,V\,s\,\Delta \theta = \frac{3}{4} W.
Substituting for W :
\rho\,V\,s\,\Delta \theta = \frac{3}{4} \times \frac{1}{2} \times \frac{\left(\frac{Mg}{\pi r^2}\right)^2}{Y} \times V.
We can cancel out V on both sides.
Step 8: Solve for \Delta \theta
Simplifying, we get:
\rho\,s\,\Delta \theta = \frac{3}{8} \times \frac{(Mg)^2}{(\pi r^2)^2\,Y}.
Hence,
\Delta \theta = \frac{3}{8\,\rho\,s} \times \frac{(Mg)^2}{(\pi r^2)^2\,Y}.
Step 9: Substitute Numerical Values
Substitute the given numerical values carefully:
M = 31.4\,\text{kg}
g = 9.8\,\text{m/s}^2
\pi = 3.14
r = 10^{-3}\,\text{m}
Y = 9.8 \times 10^{10}\,\text{N/m}^2
\rho = 9 \times 10^{3}\,\text{kg/m}^3
s = 490\,\text{J/(kg·K)}
After simplification, the temperature rise is found to be:
\Delta \theta = \frac{1}{1120}\,\text{K}.
Step 10: Conclude the Value of n
From the question, the temperature rise is given as \frac{1}{n}\,\text{K} . Therefore,
\Delta \theta = \frac{1}{1120}\,\text{K} \quad \Rightarrow \quad n = 1120.
