The temperature of hot and cold end of a 20 cm long rod in thermal steady state are at `100^(@)C` and `20^(@)C` respectively. Temperature at the centre of the rod is

Options

A. `50^(@)C`
B. `60^(@)C`
C. `40^(@)C`
D. `30^(@)C`

Answer

To find the temperature at the center of a 20 cm long rod in thermal steady state with hot and cold ends at 100°C and 20°C respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Length of the rod (L) = 20 cm
- Temperature at the hot end (T_hot) = 100°C
- Temperature at the cold end (T_cold) = 20°C

2. **Understand the Concept of Thermal Steady State**:
- In thermal steady state, the temperature gradient along the rod is linear. This means that the temperature changes uniformly from the hot end to the cold end.

3. **Determine the Temperature at the Center of the Rod**:
- The center of the rod is located at half the length of the rod. Therefore, the center position (x) is:
\[ x = \frac{L}{2} = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm} \]
- The temperature at any point in a uniformly varying temperature gradient can be calculated using the formula:
\[ T(x) = T_{\text{cold}} + \left(\frac{T_{\text{hot}} - T_{\text{cold}}}{L}\right) \cdot x \]
- Substituting the known values into the equation:
\[ T(10 \, \text{cm}) = 20°C + \left(\frac{100°C - 20°C}{20 \, \text{cm}}\right) \cdot 10 \, \text{cm} \]

4. **Calculate the Temperature Gradient**:
- The temperature difference between the hot and cold ends is:
\[ T_{\text{hot}} - T_{\text{cold}} = 100°C - 20°C = 80°C \]
- The temperature gradient (slope) is:
\[ \frac{80°C}{20 \, \text{cm}} = 4°C/\text{cm} \]

5. **Calculate the Temperature at the Center**:
- Now, substituting the gradient back into the temperature equation:
\[ T(10 \, \text{cm}) = 20°C + (4°C/\text{cm}) \cdot 10 \, \text{cm} = 20°C + 40°C = 60°C \]

6. **Final Result**:
- The temperature at the center of the rod is:
\[ T(10 \, \text{cm}) = 60°C \]

### Summary:
The temperature at the center of the rod is **60°C**.