Thermal Properties of Matter

JEE Main level, single correct, medium

11 of 126

Question 11 of 126

On a linear temperature scale Y\mathrm{Y}, water freezes at −160∘Y-160^{\circ} \mathrm{Y} and boils at −50∘Y-50^{\circ} \mathrm{Y}. On this Y\mathrm{Y} scale, a temperature of 340 K340 \mathrm{~K} would be read as : (water freezes at 273 K273 \mathrm{~K} and boils at 373 K373 \mathrm{~K} )

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Answer: (C) −86.3∘Y-86.3^{\circ} \mathrm{Y}

Reading on any scale - LFP
UFP - LFP
== constant for all scales
340−273373−273=∘y−(−160)−50−(−160)\frac{340-273}{373-273}=\frac{{ }^{\circ} \mathrm{y}-(-160)}{-50-(-160)}
⇒67100=y+160110\Rightarrow \frac{67}{100}=\frac{\mathrm{y}+160}{110}
∴y=−86.3∘y\therefore \quad y=-86.3^{\circ} \mathrm{y}

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