JEE Main 2025 Jan 24 Shift 2, Physics Q10: Measurements

The conversion between the two scales is $\frac{C}{5}=\frac{F-32}{9}$, which rearranges to

$C=\frac{5}{9}F-\frac{160}{9}$

This is a straight line in the $C$ vs $F$ plane with:

• slope $=\frac{5}{9}>0$ — line goes up to the right;
• $C$-intercept (set $F=0$): $C=-\frac{160}{9}\approx -17.8$ — negative;
• $F$-intercept (set $C=0$): $F=32$ — positive.

So the graph passes through $(32,0)$ and $(0,-160/9)$ with positive slope: the line crosses the $F$ axis to the right of the origin and the $C$ axis below the origin.

Correct option: (2)