A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and area of cross-section 100 cm. sq. The thermal conductivity of copper is 400 W/mC. The two ends of this rod must be kept at a temperature difference of-

Options

A. `1^(@)C`
B. `10^(@)C`
C. `100^(@)C`
D. `1000^(@)C`

Explanation

To solve the problem, we need to find the temperature difference (ΔT) required to maintain a heat flux of 4000 J/s through a copper rod. We can use the formula for heat conduction, which is given by Fourier's law:

\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]

Where:
- \(Q\) is the heat transfer per unit time (in watts, W),
- \(k\) is the thermal conductivity of the material (in W/m°C),
- \(A\) is the cross-sectional area (in m²),
- \(\Delta T\) is the temperature difference across the rod (in °C or K),
- \(L\) is the length of the rod (in meters).

### Step-by-Step Solution:

1. **Convert the given values to SI units:**
- Length of the rod, \(L = 10 \, \text{cm} = 0.1 \, \text{m}\)
- Area of cross-section, \(A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2\)
- Heat flux, \(Q = 4000 \, \text{J/s} = 4000 \, \text{W}\) (since 1 J/s = 1 W)
- Thermal conductivity of copper, \(k = 400 \, \text{W/m°C}\)

2. **Rearranging the formula to solve for ΔT:**
\[ \Delta T = \frac{Q \cdot L}{k \cdot A} \]

3. **Substituting the values into the equation:**
\[ \Delta T = \frac{4000 \, \text{W} \cdot 0.1 \, \text{m}}{400 \, \text{W/m°C} \cdot 0.01 \, \text{m}^2} \]

4. **Calculating the denominator:**
\[ k \cdot A = 400 \, \text{W/m°C} \cdot 0.01 \, \text{m}^2 = 4 \, \text{W/°C} \]

5. **Calculating ΔT:**
\[ \Delta T = \frac{4000 \cdot 0.1}{4} = \frac{400}{4} = 100 \, \text{°C} \]

### Final Answer:
The temperature difference (ΔT) required across the copper rod is **100°C**.