The steam point and the ice point of a Mercury thermometer are marked as `80^0 and 20^0`. What will be the temperature in centigrade mercury scale when this thermometer reads `32^0`?

Explanation

To solve the problem, we need to convert the reading of the thermometer (which is marked differently) into the Celsius scale. The thermometer has its ice point and steam point marked at 20°C and 80°C, respectively. We can use the formula for linear interpolation to find the equivalent temperature in the Celsius scale.

### Step-by-Step Solution:

1. **Identify the Known Values:**
- Ice point (L0) = 20°C
- Steam point (L100) = 80°C
- Thermometer reading (L1) = 32 (the reading we want to convert)

2. **Set Up the Formula:**
The formula to convert the thermometer reading to the Celsius scale is:
\[ T = \frac{L1 - L0}{L100 - L0} \times 100 \]
where:
- \( T \) is the temperature in Celsius,
- \( L1 \) is the thermometer reading,
- \( L0 \) is the ice point,
- \( L100 \) is the steam point.

3. **Substitute the Known Values:**
Substitute \( L1 = 32 \), \( L0 = 20 \), and \( L100 = 80 \) into the formula:
\[ T = \frac{32 - 20}{80 - 20} \times 100 \]

4. **Calculate the Numerator and Denominator:**
- Numerator: \( 32 - 20 = 12 \)
- Denominator: \( 80 - 20 = 60 \)

5. **Plug in the Values:**
\[ T = \frac{12}{60} \times 100 \]

6. **Simplify the Fraction:**
\[ T = \frac{1}{5} \times 100 = 20 \]

7. **Final Answer:**
The temperature in the centigrade scale when the thermometer reads 32 is \( 20°C \).

### Summary:
The temperature in the centigrade scale when the thermometer reads 32 is \( 20°C \).