The resistances of a platinum resistance thermometer at the ice point, the steam point and the boiling point of sulphur are `2.50, 3.50 and 6.50 (Omega)` respectively. Find the boiling point of sulphur on the platinum scale. The ice point and the steam point measure `0^@ and 100^@`, respectively.

Explanation

To solve the problem, we need to find the boiling point of sulphur on the platinum scale using the given resistances at the ice point, steam point, and boiling point of sulphur.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance at ice point (R0) = 2.50 Ω
- Resistance at steam point (R100) = 3.50 Ω
- Resistance at boiling point of sulphur (R) = 6.50 Ω
- Ice point temperature (T0) = 0°C
- Steam point temperature (T100) = 100°C

2. **Use the Resistance Formula:**
The resistance of the thermometer at any temperature T can be expressed as:
\[ R = R_0 \cdot (1 + \alpha \cdot (T - T_0)) \]
For the boiling point of sulphur, we have:
\[ 6.50 = 2.50 \cdot (1 + \alpha \cdot (T - 0)) \]
Simplifying this gives:
\[ 6.50 = 2.50 \cdot (1 + \alpha T) \]

3. **Rearranging the Equation:**
Dividing both sides by 2.50:
\[ \frac{6.50}{2.50} = 1 + \alpha T \]
This simplifies to:
\[ 2.6 = 1 + \alpha T \]
Therefore:
\[ \alpha T = 2.6 - 1 = 1.6 \quad \text{(Equation 1)} \]

4. **Finding Alpha (α):**
Now, we will find α using the ice point and steam point:
\[ R_{100} = R_0 \cdot (1 + \alpha \cdot (T_{100} - T_0)) \]
Substituting the values:
\[ 3.50 = 2.50 \cdot (1 + \alpha \cdot (100 - 0)) \]
Simplifying gives:
\[ 3.50 = 2.50 \cdot (1 + 100\alpha) \]
Dividing both sides by 2.50:
\[ \frac{3.50}{2.50} = 1 + 100\alpha \]
This simplifies to:
\[ 1.4 = 1 + 100\alpha \]
Therefore:
\[ 100\alpha = 1.4 - 1 = 0.4 \]
Thus:
\[ \alpha = \frac{0.4}{100} = 0.004 \quad \text{(Equation 2)} \]

5. **Substituting α back into Equation 1:**
Now substitute α back into Equation 1:
\[ 0.004T = 1.6 \]
Solving for T gives:
\[ T = \frac{1.6}{0.004} = 400°C \]

### Conclusion:
The boiling point of sulphur on the platinum scale is **400°C**.