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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 \(373\, K\))
- A \(-73.7\)
- B \(-233.7\)
- C \(-86.3\)
- D \(-106.3\)
Answer & Solution
Correct Answer
(C) \(-86.3\)
Step-by-step Solution
Detailed explanation
\(Reading\,on\,any\,scale-LFP\) \(UFP-LFP\) \(=constant\,for\,all\,scales\) \(\frac{{340 - 273}}{{373\left| { - 273} \right.}} = \frac{{^ \circ y - \left( { - 160} \right)}}{{ - 50 - \left( { - 160} \right)}}\) \( \Rightarrow \frac{{67}}{{100}} = \frac{{y + 160}}{{110}}\)…
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