During a welding operation, thermal power of 2500 W is incident normally on a metallic surface. As shown in the figure below (figure is NOT to scale), the heated area is circular. Out of the incident power, 85% of the power is absorbed within a circle of radius 5 mm while 65% is absorbed within an inner concentric circle of radius 3 mm. The power density in the shaded area is Solution: We are given the following: Total thermal power incident: Ptotal = 2500W Wmm^-2 (rounded off to 2 decimal places).

The correct answer is 9.95 W/mm^-2 (or 9.95).

Step 1: Extract data from the problem description and the provided image
Based on the problem text and the labels visible in the image, we have:
- Total incident thermal power:
$P_{\text{total}}=2500\text{W}$
- Inner radius of the concentric circular region:
$r_{\text{inner}}=3\text{mm}$
- Outer radius of the concentric circular region:
$r_{\text{outer}}=5\text{mm}$
- Percentage of power absorbed within the outer circle ($r=5\text{mm}$): 85%
- Percentage of power absorbed within the inner circle ($r=3\text{mm}$): 65%
- The shaded area is the annular region between the inner circle and the outer circle.

Step 2: Calculate the thermal power absorbed in the shaded region
First, find the power absorbed within the outer circle:
$P_{\text{outer}}=0.85\times 2500\text{W}=2125\text{W}$
Next, find the power absorbed within the inner circle:
$P_{\text{inner}}=0.65\times 2500\text{W}=1625\text{W}$
Therefore, the power absorbed specifically within the shaded annular region is:
$P_{\text{shaded}}=P_{\text{outer}}-P_{\text{inner}}=2125\text{W}-1625\text{W}=500\text{W}$

Step 3: Calculate the area of the shaded region
The area of the shaded circular ring (annulus) is given by:
$A_{\text{shaded}}=\pi \left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)$
Substituting the given radii:
$A_{\text{shaded}}=\pi \left(5^2-3^2\right)=\pi \left(25-9\right)=16\pi {\text{mm}}^2$

Step 4: Calculate the power density in the shaded area
The power density is the power absorbed per unit area:
$\text{Power Density}=\frac{P_{\text{shaded}}}{A_{\text{shaded}}}=\frac{500}{16\pi }$
Using the approximation $\pi \approx 3.14159$:
$\text{Power Density}\approx \frac{500}{50.2655}\approx 9.9472{\text{W/mm}}^{-2}$
Rounding off to two decimal places, we get:
$\text{Power Density}=9.95{\text{W/mm}}^{-2}$