The freezing point on a thermometer is marked as 20∘ and the boiling point at as 150∘. A temperature of 60∘C on this thermometer will be read as
A
40∘
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B
65∘
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C
98∘
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D
110∘
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Solution
The correct option is C98∘ Temperature on any scale can be converted into other scale by X−LFPUFP−LFP = Constant for all scales ∴X−20∘150∘−20∘=C−0∘100∘−0o⇒ X = C×130∘100∘+20∘ = 60∘×130∘100∘+20∘=98∘