On a thermometer, the freezing points of water is marked as `20^(@)C` and the boiling points of water is marked as `150^(@)C` . A temperature of `60^(@)C` will be read on this thermometer as

Options

A. `40^(@)C`
B. `65^(@)C`
C. `98 ^(@)C`
D. `110^(@)C`

Explanation

To solve the problem, we need to find out how the temperature of `60°C` will be read on a thermometer that has its freezing point of water marked at `20°C` and boiling point at `150°C`.

### Step-by-Step Solution:

1. **Identify the given points on the thermometer:**
- Freezing point of water (T_f) = `20°C`
- Boiling point of water (T_b) = `150°C`
- The temperature we want to convert (T) = `60°C`

2. **Determine the range of the thermometer:**
- The range of the thermometer is from `20°C` to `150°C`.
- The total range (R) = T_b - T_f = `150°C - 20°C = 130°C`.

3. **Determine the equivalent temperature scale:**
- The standard temperature scale for water is from `0°C` to `100°C`.
- The equivalent range for the standard scale is from `0°C` to `100°C`.

4. **Set up the proportion:**
- We can set up a proportion to find the equivalent temperature (X) on the thermometer:
\[ \frac{X - 20}{130} = \frac{60 - 0}{100} \]

5. **Cross-multiply to solve for X:**
\[ (X - 20) \cdot 100 = (60) \cdot (130) \]
\[ 100(X - 20) = 7800 \]

6. **Distribute and simplify:**
\[ 100X - 2000 = 7800 \]
\[ 100X = 7800 + 2000 \]
\[ 100X = 9800 \]

7. **Solve for X:**
\[ X = \frac{9800}{100} = 98 \]

8. **Conclusion:**
- The temperature of `60°C` will be read as `98°C` on the thermometer.

### Final Answer:
The temperature of `60°C` will be read as `98°C` on the thermometer.