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On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^\circ W$ and $239^\circ W$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^\circ C$ on the Celsius scale ?
A
$200^\circ W$
B
$139^\circ W$
C
$78^\circ W$
D
$117^\circ W$
Detailed Solution
For any two linear scales: $\dfrac{\text{reading} - \text{freezing point}}{\text{boiling point} - \text{freezing point}}$ is the same.
Celsius scale: freezing point $0^\circ C$, boiling point $100^\circ C$. W scale: freezing point $39^\circ W$, boiling point $239^\circ W$.
$\dfrac{C - 0}{100 - 0} = \dfrac{W - 39}{239 - 39}$
$\dfrac{C}{100} = \dfrac{W - 39}{200}$
$W = \dfrac{C}{100} \times 200 + 39$
For $C = 39$: $W = \dfrac{39}{100} \times 200 + 39 = 78 + 39 = 117$
So the temperature on the new scale is $117^\circ W$, corresponding to $39^\circ C$.
Celsius scale: freezing point $0^\circ C$, boiling point $100^\circ C$. W scale: freezing point $39^\circ W$, boiling point $239^\circ W$.
$\dfrac{C - 0}{100 - 0} = \dfrac{W - 39}{239 - 39}$
$\dfrac{C}{100} = \dfrac{W - 39}{200}$
$W = \dfrac{C}{100} \times 200 + 39$
For $C = 39$: $W = \dfrac{39}{100} \times 200 + 39 = 78 + 39 = 117$
So the temperature on the new scale is $117^\circ W$, corresponding to $39^\circ C$.