The freezing point on a thermometer is marked as `-20^(@)` and the boiling point as `130^(@)`. A temperature of human body `(34^(@)C)` on this thermometer will be read as

Options

A. A) `31^(@)`
B. B) `51^(@)`
C. C) `20^(@)`
D. D) None of these

Answer

To solve the problem, we need to determine how the temperature of the human body (34°C) is represented on a thermometer that has its freezing point at -20°C and boiling point at 130°C. We can use a linear interpolation method to find the corresponding reading on the thermometer.

### Step-by-step Solution:

1. **Identify the known points on the thermometer:**
- Freezing point: -20°C (Thermometer reading = 0)
- Boiling point: 130°C (Thermometer reading = 100)

2. **Set up the linear relationship:**
We can define a linear equation that relates the actual temperature (T) to the thermometer reading (R). The relationship can be expressed as:
\[ R = mT + c \]
where \(m\) is the slope and \(c\) is the intercept.

3. **Calculate the slope (m):**
The slope \(m\) can be calculated using the two known points:
\[ m = \frac{R_2 - R_1}{T_2 - T_1} \]
Here, \(R_1 = 0\) at \(T_1 = -20°C\) and \(R_2 = 100\) at \(T_2 = 130°C\):
\[ m = \frac{100 - 0}{130 - (-20)} = \frac{100}{150} = \frac{2}{3} \]

4. **Find the intercept (c):**
We can use one of the points to find \(c\). Using the freezing point:
\[ 0 = \frac{2}{3}(-20) + c \implies c = \frac{2}{3} \times 20 = \frac{40}{3} \]

5. **Formulate the final equation:**
Now we have:
\[ R = \frac{2}{3}T + \frac{40}{3} \]

6. **Substitute the body temperature (34°C) into the equation:**
\[ R = \frac{2}{3}(34) + \frac{40}{3} = \frac{68}{3} + \frac{40}{3} = \frac{108}{3} = 36 \]

7. **Conclusion:**
The reading on the thermometer for a human body temperature of 34°C will be 36.

### Final Answer:
The temperature of the human body (34°C) will be read as **36** on this thermometer.