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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
  • JEE Mains 2023
  • Medium
  • MCQ

On a temperature scale \(X\). The boiling point of water is \(65^{\circ} X\) and the freezing point is \(-15^{\circ} X\). Assume that the \(X\) scale is linear. The equivalent temperature corresponding to \(-95^{\circ} X\) on the Farenheit scale would be \(..........^{\circ} F\)

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Answer & Solution

Correct Answer

(D) \(-148\)

Step-by-step Solution

Detailed explanation

\(\frac{X-X_{\text {freez }}}{X_{\text {boil }}-X_{\text {freez }}}=\frac{t-32}{212-32}\)

\(\frac{-95-(-15)}{65-(-15)}=\frac{t-32}{180}\)

\(\frac{-80}{80}=\frac{t-32}{180}\)

\(t=-180+32\)

\(t=-148^{\circ} f\)

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