meter rod of area of cross section `4mc^(2)` with `K =0.5 cal g^(-1) C^(-1)` is observed that at steady state `360` cal of heat flows per minute The temperature gradient along the rod is .

Options

A. `3^(@)C//cm`
B. `6^(@)C//an`
C. `12^(@)C//m`
D. `20^(@)C//cm`

Explanation

To solve the problem, we will use the formula for heat conduction, which relates the heat flow (Q) through a rod to the temperature gradient (dT/dx), the thermal conductivity (K), and the cross-sectional area (A).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Cross-sectional area, \( A = 4 \, \text{cm}^2 \)
- Thermal conductivity, \( K = 0.5 \, \text{kcal/g°C} \)
- Heat flow, \( Q = 360 \, \text{kcal/min} \)

2. **Convert Heat Flow to Seconds:**
- Since the heat flow is given per minute, we need to convert it to per second:
\[ Q = \frac{360 \, \text{kcal}}{60 \, \text{s}} = 6 \, \text{kcal/s} \]

3. **Use the Heat Conduction Formula:**
- The formula for heat conduction is given by:
\[ Q = K \cdot A \cdot \frac{dT}{dx} \]
Rearranging this formula to find the temperature gradient \( \frac{dT}{dx} \):
\[ \frac{dT}{dx} = \frac{Q}{K \cdot A} \]

4. **Substitute the Values into the Formula:**
- Substitute \( Q = 6 \, \text{kcal/s} \), \( K = 0.5 \, \text{kcal/g°C} \), and \( A = 4 \, \text{cm}^2 \):
\[ \frac{dT}{dx} = \frac{6 \, \text{kcal/s}}{0.5 \, \text{kcal/g°C} \cdot 4 \, \text{cm}^2} \]

5. **Calculate the Temperature Gradient:**
- First, calculate the denominator:
\[ K \cdot A = 0.5 \cdot 4 = 2 \, \text{kcal/g°C} \]
- Now substitute back:
\[ \frac{dT}{dx} = \frac{6 \, \text{kcal/s}}{2 \, \text{kcal/g°C}} = 3 \, \text{g°C/s} \]

6. **Convert Units:**
- Since \( 1 \, \text{g°C/s} \) can be interpreted in terms of temperature gradient, we can express it as:
\[ \frac{dT}{dx} = 3 \, \text{°C/cm} \]

### Final Answer:
The temperature gradient along the rod is \( 3 \, \text{°C/cm} \).