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Physics Question on thermal properties of matter

On a linear temperature scale YY, water freezes at −160∘-160^{\circ} YY and boils at −50∘ Y-50^{\circ}\, Y. On this YY scale, a temperature of 340 K340\, K would be read as :
(water freezes at 273 K273\, K and boils at 373 K373\, K)

A

−73.7∘ Y-73.7^{\circ}\, Y

B

−233.7∘ Y-233.7^{\circ}\, Y

C

−86.3∘ Y-86.3^{\circ}\, Y

D

−106.3∘ Y-106.3^{\circ}\, Y

Answer

−86.3∘ Y-86.3^{\circ}\, Y

Explanation

Solution

The correct answer is (C) : −86.3°-86.3°
Reading on any scale − LFPUFP−LFP\frac{Reading \,on \,any \,scale\, -\, LFP}{UFP-LFP}
= constant for all scales 340−273373−273=∘y−(−160)−50−(−160)\frac{340-273}{373-273} = \frac{^{\circ}y-\left(-160\right)}{-50-\left(-160\right)}
⇒67100=y+160110\Rightarrow \frac{67}{100} = \frac{y + 160 }{110}
∴y=−86.3∘y\therefore\quad y=-86.3^{\circ}y