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JEE Main 202325 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2023 · 25 January · Evening Shift · Physics · Thermal Properties of Matter

The graph between two temperature scales P and Q is shown in the...

The graph between two temperature scales PP and QQ is shown in the figure. Between upper fixed point and lower fixed point there are 150150 equal divisions of scale PP and 100100 divisions on scale QQ. The relationship for conversion between the two scales is given by :

The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed.

Options

  1. AtQ150=tP−180100\dfrac{{t}_{Q}}{150} = \dfrac{{t}_{P} - 180}{100}
  2. BtQ100=tP−30150\dfrac{{t}_{Q}}{100} = \dfrac{{t}_{P} - 30}{150}
  3. CtP180=tQ−40100\dfrac{{t}_{P}}{180} = \dfrac{{t}_{Q} - 40}{100}
  4. DtQ100=tP−180150\dfrac{{t}_{Q}}{100} = \dfrac{{t}_{P} - 180}{150}

Correct answer

B. tQ100=tP−30150\dfrac{{t}_{Q}}{100} = \dfrac{{t}_{P} - 30}{150}

Step-by-step solution

From the graph

For scale PP,
Lower fixed point=180−150=30°P= 180 - 150 = 30 ° P

Upper fixed point=180°P= 180 ° P

For scale QQ,
Lower fixed point=0°Q= 0 ° Q

Upper fixed point=100°Q= 100 ° Q

Thus, the relationship between the two scales PP and QQ is given by

tP−30180−30=tQ−0100−0\dfrac{{t}_{P} - 30}{180 - 30} = \dfrac{{t}_{Q} - 0}{100 - 0}

⇒tP−30150=tQ100⇒ \dfrac{{t}_{P} - 30}{150} = \dfrac{{t}_{Q}}{100}

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