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Heat and Thermodynamics question

2023 · 25 Jan · Shift 2 · Q56
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  5. /2023 · 25 Jan · Shift 2 · Q56

Heat and Thermodynamics question

2023 · 25 Jan · Shift 2 · Q56

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
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 :- JEE Main 2023 (Online) 25th January Evening Shift Physics - Heat and Thermodynamics Question 128 English
  1. A
    tP100=tQ−180150{{{t_P}} \over {100}} = {{{t_Q} - 180} \over {150}}100tP​​=150tQ​−180​
  2. B
    tP180−tQ−40100{{{t_P}} \over {180}} - {{{t_Q} - 40} \over {100}}180tP​​−100tQ​−40​
  3. C
    tQ150=tP−180100{{{t_Q}} \over {150}} = {{{t_P} - 180} \over {100}}150tQ​​=100tP​−180​
  4. D
    tQ100=tP−30150{{{t_Q}} \over {100}} = {{{t_P} - 30} \over {150}}100tQ​​=150tP​−30​
View written solutionFree

Correct answer: D

  1. Use the fixed-point relation for linear temperature scales

For any two linear temperature scales, the relation is

tP−LPUP−LP=tQ−LQUQ−LQ\frac{t_P-L_P}{U_P-L_P}=\frac{t_Q-L_Q}{U_Q-L_Q}UP​−LP​tP​−LP​​=UQ​−LQ​tQ​−LQ​​

where:

  • LP,LQL_P, L_QLP​,LQ​ are the lower fixed points on scales PPP and QQQ
  • UP,UQU_P, U_QUP​,UQ​ are the upper fixed points on scales PPP and QQQ
  1. Extract information from the figure/options

The problem states:

  • Scale PPP has 150150150 equal divisions between lower and upper fixed points.
  • Scale QQQ has 100100100 equal divisions between lower and upper fixed points.

From the options, the lower fixed points indicated are:

  • For scale PPP: 303030
  • For scale QQQ: 000

Thus,

UP−LP=150,UQ−LQ=100U_P-L_P=150, \qquad U_Q-L_Q=100UP​−LP​=150,UQ​−LQ​=100

and the linear relation should be

tP−30150=tQ100\frac{t_P-30}{150}=\frac{t_Q}{100}150tP​−30​=100tQ​​

Rearranging,

tQ100=tP−30150\frac{t_Q}{100}=\frac{t_P-30}{150}100tQ​​=150tP​−30​
  1. Match with the options

This exactly matches Option D:

tQ100=tP−30150\frac{t_Q}{100}=\frac{t_P-30}{150}100tQ​​=150tP​−30​
  1. Check the other options briefly
  • A uses lower fixed points 000 and 180180180, inconsistent.
  • B is not even written as a proper equality.
  • C gives a different shift and reversed placement.
  • D correctly represents a linear mapping with 150150150 divisions on PPP and 100100100 on QQQ.

Therefore, the correct answer is D.

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