The resistance of the wire in the platinum resistance thermometer at ice point is `5Omega` and at steam point is `5.25Omega`. When the thermometer is inserted in an unknown hot bath its resistance is found to be `5.5Omega`. The temperature of the hot bath is

Options

A. `100^(@)C`
B. `200^(@)C`
C. `300^(@)C`
D. `350^(@)C`

Explanation

To solve the problem, we need to determine the temperature of the hot bath using the resistance values provided for the platinum resistance thermometer.

### Step-by-Step Solution:

1. **Identify Known Values:**
- Resistance at ice point (R₀) = 5 Ω (at 0 °C)
- Resistance at steam point (R₁₀₀) = 5.25 Ω (at 100 °C)
- Resistance in the hot bath (Rₜ) = 5.5 Ω (unknown temperature T)

2. **Use the Temperature Coefficient of Resistance Formula:**
The relationship between resistance and temperature for a platinum resistance thermometer can be expressed as:
\[ R_T = R_0 \left(1 + \alpha (T - T_0)\right) \]
where:
- \( R_T \) = resistance at temperature T
- \( R_0 \) = resistance at 0 °C
- \( \alpha \) = temperature coefficient of resistance
- \( T_0 \) = reference temperature (0 °C)

3. **Calculate the Temperature Coefficient (α):**
We first need to find α using the known values at the ice and steam points:
\[ R_{100} = R_0 \left(1 + \alpha (100 - 0)\right) \]
Substitute the known values:
\[ 5.25 = 5 \left(1 + 100\alpha\right) \]
Rearranging gives:
\[ 1 + 100\alpha = \frac{5.25}{5} = 1.05 \]
Thus,
\[ 100\alpha = 1.05 - 1 = 0.05 \]
Therefore,
\[ \alpha = \frac{0.05}{100} = 5 \times 10^{-4} \, \text{°C}^{-1} \]

4. **Substitute Values to Find Temperature (T):**
Now we can use the resistance at the hot bath to find the temperature:
\[ R_T = R_0 \left(1 + \alpha (T - T_0)\right) \]
Substitute \( R_T = 5.5 \, \Omega \), \( R_0 = 5 \, \Omega \), \( \alpha = 5 \times 10^{-4} \, \text{°C}^{-1} \), and \( T_0 = 0 \):
\[ 5.5 = 5 \left(1 + 5 \times 10^{-4} (T - 0)\right) \]
Simplifying gives:
\[ 5.5 = 5 + 5 \times 10^{-4} \cdot 5 T \]
\[ 5.5 - 5 = 5 \times 10^{-4} \cdot 5 T \]
\[ 0.5 = 25 \times 10^{-4} T \]
\[ T = \frac{0.5}{25 \times 10^{-4}} = \frac{0.5}{0.0025} = 200 \, \text{°C} \]

5. **Conclusion:**
The temperature of the hot bath is **200 °C**.