Two thermometers X and Y have ice points mark | vedclass

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(C) For any temperature scale,the ratio $\frac{	ext{Reading} - 	ext{Ice Point}}{	ext{Steam Point} - 	ext{Ice Point}}$ is constant.For thermometer

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(C) For any temperature scale,the ratio $\frac{	ext{Reading} - 	ext{Ice Point}}{	ext{Steam Point} - 	ext{Ice Point}}$ is constant.For thermometer $X$:$	ext{Ice Point} = 15^o$,$	ext{Steam Point} = 75^o$,$	ext{Reading} = 60^o$.Ratio $= \frac{60 - 15}{75 - 15} = \frac{45}{60} = \frac{3}{4}$.For thermometer $Y$:$	ext{Ice Point} = 25^o$,$	ext{Steam Point} = 125^o$,$	ext{Reading} = T_Y$.Ratio $= \frac{T_Y - 25}{125 - 25} = \frac{T_Y - 25}{100}$.Equating the ratios:$\frac{3}{4} = \frac{T_Y - 25}{100}$.$T_Y - 25 = \frac{3}{4} 	imes 100 = 75$.$T_Y = 75 + 25 = 100^o$.

Two thermometers $X$ and $Y$ have ice points marked at $15^o$ and $25^o$ and steam points marked as $75^o$ and $125^o$ respectively. When thermometer $X$ measures the temperature of a bath as $60^o$ on it,what would thermometer $Y$ read when it is used to measure the temperature of the same bath?

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