JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 \(..........^{\circ} F\)
- A \(-63\)
- B \(-112\)
- C \(-48\)
- D \(-148\)
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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