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A centimetre scale is attached with a th...

A centimetre scale is attached with a thermometer of uniform bore. The thermometer reads 7.3 cm in melting ice, 23.8 cm in boiling water and 3.5 cm in a freezing mixture. What is the temperature of this freezing mixture in `""^(@)C`?

Text Solution

Verified by Experts

The lower and the upper fixed points correspond to readings of 7.3 cm and 23.8 cm respectively. The temperature of the freezing mixture in this scale corresponds to a scale reading of 3.5 cm.
Let C be the freezing mixture.s temperature in degree celsius.
`therefore " " (3.5-7.3)/(23.8-7.3)=(C-0)/(100-0) or, " " C=(-3.8)/(16.5) times 100=-23.03`
`therefore` Temperature of the freezing mixture `=-23.03^(@)C`.
Alternative solution :
When the temperature increases from `0^(@)C " to " 100^(@)C`, the corresponding change in scale reading = 23.8 - 7.3 = 16.5 cm.
So when the temperature changes by `1^(@)C`, the corresponding changes `=(16.5)/(100)=0.165 cm`
Let the temperature of the freezing mixture `=-x^(@)C`
So change in temperature in Celcius scale `=0-(-x)=x^(@)C`
Corresponding change in scale reading = 0.165x cm.
According to the question,
0.165x = 7.3 - 3.5 or, `x=(3.8)/(0.165)=23.03`
So the temperature of the freezing mixture is `-23.03^(@)C`.
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