Two thermometers A and B are calibrated in different scales. When both the thermometers are placed in a container filled with hot water the mercury thread in the thermometer A has moved through $10$ divisions whereas the mercury thread in thermometer B has moved through $15$ divisions. The reading on thermometers A and B are $20^{0}$ and $40^{0}$ , respectively. If the lower fixed point of the thermometer A is $0^{0}$ , determine the lower fixed point of the thermometer B.

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Solution

Step 1: Given data:
The reading on thermometers A and B are $20^{0}$ and $40^{0}$ .
Lower fixed point (LPF) of the thermometer A, $L_{1} = 0^{0}$
Mercury thread in thermometer A has moved through $10$ divisions.
Mercury thread in thermometer B has moved through $15$ divisions.
Step 2: Finding the lower fixed point of the thermometer B:
The difference of temperature readings on A and B are $10^{0}$ and $15^{0}$ respectively.
But $\frac{∆ A}{U_{1} - L_{1}} = \frac{∆ B}{U_{2} - L_{2}}$
Where $U_{1}$ , $U_{2}$ are the upper fixed points and $L_{1}$ , $L_{2}$ are the lower fixed points of the thermometers A and B.
$\frac{U_{1} - L_{1}}{U_{2} - L_{2}} = \frac{10}{15} = \frac{2}{3}$
The reading on thermometers A and B are $20^{0}$ and $40^{0}$ .
We know that, for any temperature scale, the ratio $\frac{T h e r m o m e t e r R e a d i n g - L F P}{U F P - L F P}$ remains a constant.
$\frac{20 - L_{1}}{U_{1} - L_{1}} = \frac{40 - L_{2}}{U_{2} - L_{2}} \Rightarrow \frac{U_{1} - L_{1}}{U_{2} - L_{2}} = \frac{20 - L_{1}}{40 - L_{2}} \frac{2}{3} = \frac{20 - L_{1}}{40 - L_{2}} \Rightarrow \frac{2}{3} = \frac{20 - 0}{40 - L_{2}} 80 - 2 L_{2} = 60 - 2 L_{2} = - 20 L_{2} = 10$
Hence, the lower fixed point of B is $10^{0}$ .

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