`A` and `B` are two points on uniform metal ring whose centre is `O` The angle `AOB =theta` A and `B` are maintaind at two different constant temperatures When `theta =180^(@)` the rate of total heat flow from `A` to `B` is `1.2W` When `theta =90^(@)` this rate will be .

Options

A. `0.6` watt
B. `0.9` watt
C. `1.6` watt
D. `1.8` watt

Explanation

To solve the problem, we will use Fourier's law of heat conduction, which states that the rate of heat transfer (Q) through a material is proportional to the temperature difference (ΔT) and the area (A) through which the heat is being transferred, and inversely proportional to the length (L) of the material.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two points A and B on a uniform metal ring with center O.
- The angle AOB is given as θ.
- The temperatures at points A and B are maintained at different constants.

2. **Using Fourier's Law**:
- Fourier's law states:
\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \]
- Where:
- \( Q \) = rate of heat transfer (in Watts)
- \( K \) = thermal conductivity of the material
- \( A \) = cross-sectional area
- \( \Delta T \) = temperature difference between A and B
- \( L \) = length of the path through which heat is conducted

3. **Case 1: θ = 180°**:
- For θ = 180°, the length of the path (L) is half the circumference of the ring:
\[ L = \frac{C}{2} \]
- Given that \( Q = 1.2 \, W \) for θ = 180°, we can write:
\[ 1.2 = \frac{K \cdot A \cdot \Delta T}{\frac{C}{2}} \]
- Rearranging gives:
\[ K \cdot A \cdot \Delta T = 1.2 \cdot \frac{C}{2} \]
- Thus:
\[ K \cdot A \cdot \Delta T = 0.6C \]

4. **Case 2: θ = 90°**:
- For θ = 90°, the length of the path (L) is a quarter of the circumference:
\[ L = \frac{C}{4} \]
- We need to find the new rate of heat transfer \( Q' \):
\[ Q' = \frac{K \cdot A \cdot \Delta T}{\frac{C}{4}} \]
- Rearranging gives:
\[ Q' = K \cdot A \cdot \Delta T \cdot \frac{4}{C} \]

5. **Relating Q' to Q**:
- From the first case, we know \( K \cdot A \cdot \Delta T = 0.6C \).
- Substituting this into the equation for \( Q' \):
\[ Q' = (0.6C) \cdot \frac{4}{C} \]
- Simplifying:
\[ Q' = 0.6 \cdot 4 = 2.4 \, W \]

### Final Answer:
When θ = 90°, the rate of heat flow from A to B will be **2.4 W**.