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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
  • JEE Mains 2013
  • Medium
  • MCQ

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

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Answer & Solution

Correct Answer

(C) \(-86.3\)

Step-by-step Solution

Detailed explanation

\(Reading\,on\,any\,scale-LFP\)

          \(UFP-LFP\)

\(=constant\,for\,all\,scales\)

\(\frac{{340 - 273}}{{373\left| { - 273} \right.}} = \frac{{^ \circ y - \left( { - 160} \right)}}{{ - 50 - \left( { - 160} \right)}}\)

\( \Rightarrow \frac{{67}}{{100}} = \frac{{y + 160}}{{110}}\)

\(\therefore \,\,\,y =  - {86.3^ \circ }\,y\)

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