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JEE Main 2023 · 25 Jan · Shift 2

JEE Main 2023 Physics question on Thermodynamics

The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by:-

  1. A.
    tQ100=tP−30150\frac{t_Q}{100} = \frac{t_P - 30}{150}
  2. B.
    tQ150=tP−180100\frac{t_Q}{150} = \frac{t_P - 180}{100}
  3. C.
    tP100=tQ−180150\frac{t_P}{100} = \frac{t_Q - 180}{150}
  4. D.
    tP180=tQ−40100\frac{t_P}{180} = \frac{t_Q - 40}{100}
Show answer and solution

Correct option: A

Solution

From the graph, scale P has its lower fixed point at 30 (i.e., the lower fixed point reads 30 on P) and scale Q starts at 0 for the lower fixed point. The upper fixed point reads 180 on P (30 + 150) and 100 on Q.

The standard conversion formula equates the fractional position above the lower fixed point on each scale:

(tQ - 0)/(100 - 0) = (tP - 30)/(180 - 30)

This gives tQ/100 = (tP - 30)/150, which is option 1.

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