On a temperature scale ' X ', the boiling point of water is 65 and the freezing point is -15. Assume that the X scale is linear. The equivalent temperature corresponding to -95 on the Fahrenheit scale would be: (a) -148 (b) -48 (c) -63 (d) -112

Answer: (a). (X-(-15))/(65-(-15))=C/100 (-95+15)/(80)=C/100⇒ C=-100° C F=9/5(-100)+32=-148° F So the answer is (a).

On a temperature scale ' X ', the boiling point of water is 65 and… | JEE Main 2023 (11 Apr, Shift 1)

Q34JEE Main 2023 · 11 APR · SHIFT 1Medium

Thermometry & Thermal Expansion

On a temperature scale ' $\text{X}$ ', the boiling point of water is $65^\circ\text{X}$ and the freezing point is $-15^\circ\text{X}$ . Assume that the $\text{X}$ scale is linear. The equivalent temperature corresponding to $-95^\circ\text{X}$ on the Fahrenheit scale would be:

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Answer: (a)

$\frac{X-(-15)}{65-(-15)}=\frac{C}{100}$
$\frac{-95+15}{80}=\frac{C}{100}\Rightarrow C=-100^\circ$ C
$F=\frac{9}{5}(-100)+32=-148^\circ$ F
So the answer is (a).

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