A faulty thermometer has ice point at `-5^circC` and steam point at `105^circC ` What would be the temperature shown by this thermometer of a person?(Assume correct temperature of person is `37^circC`)

Options

A. `42^circC`
B. `32^circC`
C. `35.7^circC`
D. `47^circC`

Answer

To find the temperature shown by the faulty thermometer when the actual temperature of a person is 37°C, we can use the concept of linear interpolation between the ice point and steam point of the thermometer.

### Step-by-Step Solution:

1. **Identify the Ice and Steam Points**:
- Ice point (0°C) corresponds to -5°C on the faulty thermometer.
- Steam point (100°C) corresponds to 105°C on the faulty thermometer.

2. **Determine the Scale of the Faulty Thermometer**:
- The range of the correct thermometer is from 0°C to 100°C.
- The range of the faulty thermometer is from -5°C to 105°C.
- The difference in the ice point is:
\[ 0 - (-5) = 5°C \]
- The difference in the steam point is:
\[ 105 - (-5) = 110°C \]

3. **Set Up the Proportion**:
- Let \( T' \) be the temperature shown by the faulty thermometer.
- We can set up the following proportion based on the actual temperature of 37°C:
\[ \frac{T'}{110} = \frac{37 + 5}{100} \]
- Here, \( 37 + 5 \) accounts for the offset of the faulty thermometer.

4. **Solve the Proportion**:
- Simplifying the right side:
\[ \frac{T'}{110} = \frac{42}{100} \]
- Cross-multiplying gives:
\[ 100T' = 110 \times 42 \]
- Calculating the right side:
\[ 110 \times 42 = 4620 \]
- Therefore:
\[ T' = \frac{4620}{100} = 46.2°C \]

5. **Adjust for the Faulty Scale**:
- Since the faulty thermometer reads higher than the actual temperature, we need to adjust for the faulty scale:
\[ T' = 46.2 - 5 = 41.2°C \]

6. **Final Calculation**:
- The final temperature shown by the faulty thermometer is:
\[ T' = 41.2°C \]

### Conclusion:
The temperature shown by the faulty thermometer for a person with a correct temperature of 37°C is **41.2°C**.