There is a temperature difference of `1K` between two black patches of skin on patient's chest and each patch having area A The radiant heat emitted from them is differ by `2%` then temperature of two patches may be .

Options

A. `100 K, 101K`
B. `300K,301K`
C. `200K,201K`
D. `400K,401K`

Explanation

To solve the problem, we will use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have two black patches on a patient's chest with a temperature difference of 1 K. The radiant heat emitted from these patches differs by 2%. We need to find the temperatures of these two patches.

### Step 2: Define Variables
Let:
- \( T_2 \) = temperature of the cooler patch
- \( T_1 \) = temperature of the warmer patch
- Given that \( T_1 - T_2 = 1 \, K \)

### Step 3: Apply the Stefan-Boltzmann Law
According to the Stefan-Boltzmann law, the power radiated by each patch can be expressed as:
- \( P_1 = \sigma A T_1^4 \) (for the warmer patch)
- \( P_2 = \sigma A T_2^4 \) (for the cooler patch)

### Step 4: Set Up the Power Difference Equation
The difference in power radiated is given as:
\[ \frac{P_1 - P_2}{P_2} \times 100 = 2\% \]
This can be rewritten as:
\[ P_1 - P_2 = 0.02 P_2 \]
Substituting the expressions for \( P_1 \) and \( P_2 \):
\[ \sigma A T_1^4 - \sigma A T_2^4 = 0.02 \sigma A T_2^4 \]
We can cancel \( \sigma A \) from both sides:
\[ T_1^4 - T_2^4 = 0.02 T_2^4 \]

### Step 5: Rearrange the Equation
Rearranging gives us:
\[ T_1^4 = 1.02 T_2^4 \]

### Step 6: Substitute \( T_1 \) in Terms of \( T_2 \)
Since \( T_1 = T_2 + 1 \):
\[ (T_2 + 1)^4 = 1.02 T_2^4 \]

### Step 7: Expand the Left Side
Using the binomial expansion:
\[ T_2^4 + 4T_2^3 + 6T_2^2 + 4T_2 + 1 = 1.02 T_2^4 \]
Rearranging gives:
\[ 0.02 T_2^4 - 4T_2^3 - 6T_2^2 - 4T_2 - 1 = 0 \]

### Step 8: Solve the Polynomial
This is a polynomial equation in \( T_2 \). Solving this numerically or using approximation methods will yield the value of \( T_2 \).

### Step 9: Find \( T_1 \)
Once \( T_2 \) is found, use \( T_1 = T_2 + 1 \) to find \( T_1 \).

### Final Step: Calculate and State the Answer
After solving the polynomial, we find:
- \( T_2 \approx 200 \, K \)
- \( T_1 \approx 201 \, K \)

Thus, the temperatures of the patches are:
- \( T_1 = 201 \, K \)
- \( T_2 = 200 \, K \)