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JEE Main 202311 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
11 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2023 · 11 April · Morning Shift · Physics · Thermal Properties of Matter

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 ° X and the freezing point is −15°X- 15 ° X. Assuming that the XX scale is linear. The equivalent temperature corresponding to −95°X- 95 ° X on the Fahrenheit scale would be

Options

  1. A−112°F- 112 ° F
  2. B−48°F- 48 ° F
  3. C−148°F- 148 ° F
  4. D−63°F- 63 ° F

Correct answer

C. −148°F- 148 ° F

Step-by-step solution

The relation between two temperature scales can be written as follows

65−x65−(−15)=212−F180...(1)\dfrac{65 - x}{65 - \left(- 15\right)} = \dfrac{212 - F}{180} . . . \left(1\right)

Substitute the values of the known parameters in to equation (1) and sole to calculate the required temperature in Fahrenheit scale.

{65+9580=212−F180⇒2×180=212−F⇒F=212−360=−148°F\begin{cases}\dfrac{65 + 95}{80} & = & \dfrac{212 - F}{180} \\ & ⇒ & 2 \times 180 = 212 - F \\ & ⇒ & F = 212 - 360 \\ & = & - 148 ° F\end{cases}

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