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प्रश्न
Solve the following problem.
In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C.
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उत्तर
Here thermometric property P is temperature at some random scale X.
Using equation,
T = `(100 ("P"_"T" - "P"_1))/("P"_2 - "P"_1)`
For P1 = 20 °X,
P2 = 200 °X,
T = 62 °C
∴ `62 = (100 ("P"_"T" - 20))/(200 - 20)`
∴ `"P"_"T" = (62 xx (200 - 20))/100 + 20`
= 111.6 + 20
= 131.6 °X
The boiling point of a liquid in this scale is 131.6 °X.
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