A constant volume gas thermometer show pressure reading of 50cm and 99 cm of mercury at `0^(@)C` and `100^(@)C` respectively. When the pressure reading is 60 cm of mercury, the temperature is

Options

A. `25^(@)`
B. `40^(@)`
C. `15^(@)`
D. `12.5^(@)`

Explanation

To solve the problem, we will use the relationship between pressure and temperature for a constant volume gas thermometer. The formula we will use is derived from the linear relationship between pressure and temperature:

\[ T = \frac{(P - P_0)}{(P_{100} - P_0)} \times 100 \]

Where:
- \( T \) is the temperature in degrees Celsius.
- \( P \) is the pressure reading at the temperature we want to find.
- \( P_0 \) is the pressure reading at \( 0^\circ C \).
- \( P_{100} \) is the pressure reading at \( 100^\circ C \).

### Step-by-step Solution:

1. **Identify the given values:**
- \( P_0 = 50 \, \text{cm of Hg} \) (pressure at \( 0^\circ C \))
- \( P_{100} = 99 \, \text{cm of Hg} \) (pressure at \( 100^\circ C \))
- \( P = 60 \, \text{cm of Hg} \) (pressure at the temperature we want to find)

2. **Substitute the values into the formula:**
\[ T = \frac{(60 \, \text{cm} - 50 \, \text{cm})}{(99 \, \text{cm} - 50 \, \text{cm})} \times 100 \]

3. **Calculate the numerator:**
\[ 60 \, \text{cm} - 50 \, \text{cm} = 10 \, \text{cm} \]

4. **Calculate the denominator:**
\[ 99 \, \text{cm} - 50 \, \text{cm} = 49 \, \text{cm} \]

5. **Now substitute the calculated values back into the formula:**
\[ T = \frac{10 \, \text{cm}}{49 \, \text{cm}} \times 100 \]

6. **Perform the division:**
\[ \frac{10}{49} \approx 0.2041 \]

7. **Multiply by 100 to find the temperature:**
\[ T \approx 0.2041 \times 100 \approx 20.41^\circ C \]

8. **Round the answer:**
\[ T \approx 20.4^\circ C \]

### Final Answer:
The temperature when the pressure reading is 60 cm of mercury is approximately \( 20.4^\circ C \).