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JEE Main 2013 (9 Apr, Shift 1) · Thermometry & Thermal Expansion

Thermometry & Thermal Expansion question from JEE Main 2013 (9 Apr, Shift 1)

Q69JEE Main 2013 · 9 APR · SHIFT 1Easy

Thermometry & Thermal Expansion

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

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Answer: (c)

  1. 340340 K =67∘=67^\circC.
  2. Y−(−160)−50−(−160)=67100⇒Y+160=110×0.67=73.7\frac{Y-(-160)}{-50-(-160)}=\frac{67}{100}\Rightarrow Y+160=110\times 0.67=73.7
  3. Y=−86.3∘Y=-86.3^\circY
  4. So the answer is (c) −86.3∘-86.3^\circY.
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