Two thermometers 'X' & 'Y' shows boiling point & freezing point of water as `220^(@)X` & `20^(@)X` and `120^(@)Y` & `-40^(@)Y` respectively. If 'X' shows `100^(@)X`. then find the reading in `Y` thermomter.

Options

A. `25^(@)Y`
B. `50^(@)Y`
C. `20^(@)Y`
D. `24^(@)Y`

Answer

To solve the problem, we will use the concept of linear interpolation between the freezing and boiling points of the two thermometers.

### Step-by-Step Solution:

1. **Identify the Fixed Points:**
- For thermometer X:
- Freezing point: \( 20^\circ X \)
- Boiling point: \( 220^\circ X \)
- For thermometer Y:
- Freezing point: \( -40^\circ Y \)
- Boiling point: \( 120^\circ Y \)

2. **Set Up the Formula:**
We can use the formula for converting temperatures between the two scales:
\[ \frac{T_X - T_{X_{min}}}{T_{X_{max}} - T_{X_{min}}} = \frac{T_Y - T_{Y_{min}}}{T_{Y_{max}} - T_{Y_{min}}} \]
Where:
- \( T_X \) is the temperature in thermometer X.
- \( T_Y \) is the temperature in thermometer Y.
- \( T_{X_{min}} = 20^\circ X \)
- \( T_{X_{max}} = 220^\circ X \)
- \( T_{Y_{min}} = -40^\circ Y \)
- \( T_{Y_{max}} = 120^\circ Y \)

3. **Substituting Known Values:**
Given that \( T_X = 100^\circ X \):
\[ \frac{100 - 20}{220 - 20} = \frac{T_Y - (-40)}{120 - (-40)} \]
Simplifying the left side:
\[ \frac{80}{200} = \frac{T_Y + 40}{160} \]
This simplifies to:
\[ \frac{2}{5} = \frac{T_Y + 40}{160} \]

4. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 2 \times 160 = 5(T_Y + 40) \]
Which simplifies to:
\[ 320 = 5T_Y + 200 \]

5. **Solving for \( T_Y \):**
Rearranging the equation:
\[ 5T_Y = 320 - 200 \]
\[ 5T_Y = 120 \]
\[ T_Y = \frac{120}{5} = 24 \]

6. **Final Answer:**
Therefore, the reading in thermometer Y is:
\[ T_Y = 24^\circ Y \]

### Conclusion:
The reading in thermometer Y when thermometer X shows \( 100^\circ X \) is \( 24^\circ Y \).