On a new scale of temperature (which is linear) and called the `W` scale. The freezing and boiling points of water are `39^(@)W` and `239^(@)W` respectively. What will be the temperature on the new scale, corresponding to a temperature of `39^(@)C` on the Celsius scale?

Options

A. `78^(@)W`
B. `117^(@)W`
C. `200^(@)W`
D. `139^(@)W`

Explanation

To find the temperature on the W scale corresponding to a temperature of 39°C on the Celsius scale, we can follow these steps:

### Step 1: Identify the freezing and boiling points
- The freezing point of water on the W scale is 39°W.
- The boiling point of water on the W scale is 239°W.
- The freezing point of water on the Celsius scale is 0°C.
- The boiling point of water on the Celsius scale is 100°C.

### Step 2: Determine the range of both scales
- The range of the Celsius scale between the freezing and boiling points is:
\[ 100°C - 0°C = 100°C \]
- The range of the W scale between the freezing and boiling points is:
\[ 239°W - 39°W = 200°W \]

### Step 3: Calculate the conversion factor
- Since the ranges are linear, we can find the conversion factor between the two scales:
\[ \text{Conversion factor} = \frac{\text{Range of W scale}}{\text{Range of Celsius scale}} = \frac{200°W}{100°C} = 2 \]
- This means that 1°C corresponds to 2°W.

### Step 4: Convert 39°C to the W scale
- To convert 39°C to the W scale, we first find the equivalent temperature in W scale by using the conversion factor:
\[ \text{Temperature in W scale} = 39°C \times 2 = 78°W \]

### Step 5: Adjust for the starting point of the W scale
- Since the freezing point of water (0°C) corresponds to 39°W, we need to add this offset to the converted temperature:
\[ \text{Final temperature in W scale} = 39°W + 78°W = 117°W \]

### Conclusion
- Therefore, the temperature on the W scale corresponding to 39°C is:
\[ \boxed{117°W} \]