Question

On an imaginary linear scale of temperature (called 'W' scale) the freezing and boiling points of water are $39^\circ\text{W}$ and $239^\circ\text{W}$ respectively. The temperature on the new scale corresponding to $39^\circ\text{C}$ temperature on Celsius scale will be

• A. $139^\circ\text{W}$
• B. $78^\circ\text{W}$
• C. $117^\circ\text{W}$
• D. $200^\circ\text{W}$

Hint:

Notice that the total range of the 'W' scale ($239 - 39 = 200$ divisions) is exactly twice as large as the Celsius scale (100 divisions). This means each $1^\circ\text{C}$ step corresponds to exactly a $2^\circ\text{W}$ step! So, a change of $39^\circ\text{C}$ means an increase of $39 \times 2 = 78$ units above the base freezing point: $39 + 78 = 117^\circ\text{W}$. A very fast mental verification!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given an arbitrary linear temperature scale, the 'W' scale, defined by specific calibration boundaries for the freezing and boiling points of water. We need to find the equivalent value on this new scale for a given Celsius temperature ($39^\circ\text{C}$).

Step 2: Key Formula or Approach:
For any two linear temperature scales, the relative reading ratio between a temperature value, its lower fixed point (LFP), and its upper fixed point (UFP) remains constant: $$ \frac{\text{Reading} - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{constant} $$ Setting up this linear conversion relationship between the Celsius scale and the new 'W' scale: $$ \frac{C - 0}{100 - 0} = \frac{W - \text{LFP}_\text{W}}{\text{UFP}_\text{W} - \text{LFP}_\text{W}} $$

Step 3: Detailed Explanation:
Let's plug in the given fixed reference parameters:

• For the Celsius scale: $\text{LFP} = 0^\circ\text{C}$, $\text{UFP} = 100^\circ\text{C}$

• For the 'W' scale: $\text{LFP}_\text{W} = 39^\circ\text{W}$, $\text{UFP}_\text{W} = 239^\circ\text{W}$
Substituting these values into our linear calibration equation: $$ \frac{C}{100} = \frac{W - 39}{239 - 39} $$ $$ \frac{C}{100} = \frac{W - 39}{200} $$ Now, let's substitute the given target temperature $C = 39^\circ\text{C}$ to solve for $W$: $$ \frac{39}{100} = \frac{W - 39}{200} $$ Multiply both sides by 200: $$ 39 \times 2 = W - 39 $$ $$ 78 = W - 39 $$ Isolating the variable $W$: $$ W = 78 + 39 = 117^\circ\text{W} $$

Step 4: Final Answer:
The temperature on the new scale is $117^\circ\text{W}$, which corresponds to option (C).