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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Units and Dimensions

BITSAT 2023 · Paper 2 · Physics · Units and Dimensions

On a temperature scale ' X '. The boiling point of water is 65^ ^ X...

On a temperature scale ' XX '. The boiling point of water is 65∘X65^{{}^{\circ}} \mathrm{X} and the freezing point is −15∘X-15^{{}^{\circ}} \mathrm{X}. Assume that the X\mathrm{X} scale is linear. The equivalent temperature corresponding to −95∘X-95^{{}^{\circ}} \mathrm{X} on the Farenheit scale would be:

Options

  1. A−63∘F-63^{{}^{\circ}} \mathrm{F}
  2. B−112∘F-112^{{}^{\circ}} \mathrm{F}
  3. C−48∘F-48^{{}^{\circ}} \mathrm{F}
  4. D−148∘F-148^{{}^{\circ}} \mathrm{F}

Correct answer

D. −148∘F-148^{{}^{\circ}} \mathrm{F}

Step-by-step solution

Boiling point of water, = UFP 65∘65^{{}^{\circ}}
Freezing point of water, =LFP−15∘=\mathrm{LFP}-15^{{}^{\circ}}
X−LFPUFP−LFP=TF−32212−32\frac{\mathrm{X}-\mathrm{LFP}}{\mathrm{UFP}-\mathrm{LFP}}=\frac{\mathrm{T}_{\mathrm{F}}-32}{212-32}
⇒−95−(−15)65−(−15)=TF−32180\Rightarrow \frac{-95-(-15)}{65-(-15)}=\frac{\mathrm{T}_{\mathrm{F}}-32}{180}
⇒−8080=TF−32180⇒TF=−180+32\Rightarrow \frac{-80}{80}=\frac{T_F-32}{180} \Rightarrow T_F=-180+32
⇒TF=−148∘F\Rightarrow \mathrm{T}_{\mathrm{F}}=-148^{{}^{\circ}} \mathrm{F}
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