Thermal Properties of Matter

JEE Main level, single correct, medium

8 of 126

Question 8 of 126

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 :
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Answer: (B) tQ100=tP−30150\dfrac{{t}_{Q}}{100}=\dfrac{{t}_{P}-30}{150}

From the graph
For scale PP,
Lower fixed point=180−150=30∘P=180-150=30^{\circ}P
Upper fixed point=180∘P=180^{\circ}P
For scale QQ,
Lower fixed point=0∘Q=0^{\circ}Q
Upper fixed point=100∘Q=100^{\circ}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\Rightarrow \dfrac{{t}_{P}-30}{150}=\dfrac{{t}_{Q}}{100}

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