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JEE Main 20139 AprilOnline
Exam
JEE Main
Year
2013
Date
9 April
Mode
Online
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2013 · 9 April · Online · Physics · Thermal Properties of Matter

On a linear temperature scale Y , water freezes at -160^ Y and...

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} )

Options

  1. A−73.7∘Y-73.7^{\circ} \mathrm{Y}
  2. B−233.7∘Y-233.7^{\circ} \mathrm{Y}
  3. C−86.3∘Y-86.3^{\circ} \mathrm{Y}
  4. D−106.3∘Y-106.3^{\circ} \mathrm{Y}

Correct answer

C. −86.3∘Y-86.3^{\circ} \mathrm{Y}

Step-by-step solution

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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