Question

On a new scale of temperature (which is linear) and called the $W$ scale, the freezing and boiling points of water are ${42}^{\circ }\text{W}$ and ${242}^{\circ }\text{W}$ respectively. What will be the temperature on the new scale, corresponding to the temperature of ${42}^{\circ }\text{C}$ on the Celsius scale ?

• A. ${200}^{\circ }\text{W}$
• B. ${142}^{\circ }\text{W}$
• C. ${84}^{\circ }\text{W}$
• D. ${126}^{\circ }\text{W}$

Answer 1

The correct option is D ${126}^{\circ }\text{W}$

We know,
$\frac{(x-LFP)}{(UFP-LFP)}=$ constant for every scale.
On Celsius scale,
$\frac{42-0}{100-0}$= constant
On $W$ scale,
$\frac{W-42}{242-42}$= constant
$\therefore \frac{42-0}{100-0}=\frac{W-42}{242-42}$
$W=\frac{42\times 200}{100}+42={126}^{\circ }\text{W}$