Thermal Properties of Matter

JEE Main level, single correct, medium

6 of 126

Question 6 of 126

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

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Answer: (C) −148∘F-148^{\circ}F

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{matrix}\dfrac{65+95}{80} & = & \dfrac{212-F}{180} \\ & \Rightarrow & 2\times 180=212-F \\ & \Rightarrow & F=212-360 \\ & = & -148^{\circ}F\end{matrix}

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