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 \(..........^{\circ} F\)
Pick an option to check your answer.
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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