Question

In a given scale $X$, the ice point is $40{}^{\circ }\text{X}$ and the steam point is $120{}^{\circ }\text{X}$. For another scale $Y$, the ice point and steam point are $-30{}^{\circ }\text{Y}$ and $130{}^{\circ }\text{Y}$ respectively. If $X$ reads $50{}^{\circ }\text{X}$ temperature of a body, then $Y$ would read

• A. $-5{}^{\circ }\text{C}$
• B. $-8{}^{\circ }\text{C}$
• C. $-10{}^{\circ }\text{C}$
• D. $-12{}^{\circ }\text{C}$

Answer 1

The correct option is C $-10{}^{\circ }\text{C}$

Given:
For scale $X$,
Ice point, $T_{iX}=40{}^{\circ }\text{X}$
Steam point, $T_{sX}=120{}^{\circ }\text{X}$
For scale $Y$,
Ice point, $T_{iY}=-30{}^{\circ }\text{Y}$
Steam point, $T_{sY}=130{}^{\circ }\text{Y}$

Reading in scale $X$ of temperature of the body, $T_X=50{}^{\circ }\text{X}$
To find: Reading in scale $Y$ of temperature of the body, $T_Y=?$
We know relation between any two scales are given by -
$\frac{T_X-T_{iX}}{T_{sX}-T_{iX}}=\frac{T_Y-T_{iY}}{T_{sY}-T_{iY}}$
$\Rightarrow \frac{T_X-40}{120-40}=\frac{T_Y-(-30)}{130-(-30)}$
$\Rightarrow \frac{50-40}{80}=\frac{T_Y+30}{160}$
$\therefore T_Y=-10{}^{\circ }\text{Y}$