question_answer

Three Carnot engines operate in series between a heat source at a temperature \[{{T}_{1}}\] and a heat sink at temperature \[{{T}_{4}}\] (see figure). There are two other reservoirs at temperature\[{{T}_{2}}and\text{ }{{T}_{3}}\], as shown, with\[{{T}_{1}}>{{T}_{2}}>{{T}_{3}}>{{T}_{4}}\]. The three engines are equally efficient if -                                                                           [JEE Main 10-Jan-2019 Morning]

A) \[{{T}_{2}}=\,\,{{({{T}_{1}}^{3}{{T}_{4}})}^{1/4}};\,\,=\,\,{{T}_{3}}\,\,=\,\,{{({{T}_{1}}{{T}_{4}}^{3})}^{1/4}}\]

B)    \[{{T}_{2}}=\,\,{{({{T}_{1}}{{T}_{4}})}^{1/2}};\,\,\,{{T}_{3}}\,\,=\,\,{{({{T}_{1}}^{2}{{T}_{4}})}^{1/3}}\]

C)    \[{{T}_{2}}=\,\,{{({{T}_{1}}{{T}_{4}}^{2})}^{1/3}};\,\,{{T}_{3}}\,\,=\,\,{{({{T}_{1}}^{2}{{T}_{4}})}^{1/3}}\]

D)    \[{{T}_{2}}=\,\,{{({{T}_{1}}^{2}{{T}_{4}})}^{1/3}};\,\,\,\,{{T}_{3}}\,\,=\,\,{{({{T}_{1}}{{T}_{4}}^{2})}^{1/3}}\]

Correct Answer: D

Solution :

[d]

\[As\,=\,{{\eta }_{1}}={{\eta }_{2}}\]

\[1-\frac{{{T}_{2}}}{{{T}_{1}}}\,=\,1-\frac{{{T}_{3}}}{{{T}_{2}}}\]

\[{{T}^{2}}\,=\,\sqrt{{{T}_{1}}{{T}_{3}}}\]

\[As\,{{\eta }_{2}}=\,{{\eta }_{3}}\]

\[1-\frac{{{T}^{3}}}{{{T}^{2}}}\,=\,\,1\,-\,\,\frac{{{T}^{4}}}{{{T}^{3}}}\]

\[{{T}_{3}}=\sqrt{{{T}_{2}}{{T}_{4}}}\]

\[As\,\,{{\eta }_{1}}={{\eta }_{3}}\]

\[1-\frac{{{T}_{2}}}{{{T}_{1}}}\,=\,1\,-\,\frac{{{T}^{4}}}{{{T}^{3}}}\,\,\,Hence\,\,{{T}_{2}}{{T}_{3}}\,=\,{{T}_{1}}{{T}_{4}}\,\]

\[\frac{{{T}_{1}}{{T}_{4}}}{{{T}_{3}}}\,=\,\sqrt{{{T}_{1}}{{T}_{3}}}\,\,\,\,\,\Rightarrow \,\,\,\frac{{{T}^{2}}_{1}{{T}^{2}}_{4}}{{{T}_{3}}^{2}}\,\,=\,\,{{T}_{1}}{{T}_{3}}\]

\[{{T}_{3}}=\,\,{{({{T}_{1}}{{T}_{4}}^{2})}^{1/3}}\]

\[{{({{T}_{1}}{{T}_{4}}^{2})}^{1/3}}\,=\,{{({{T}_{2}}{{T}_{4}})}^{1/2}}\]

\[{{T}_{1}}^{2/3}{{T}_{4}}^{4/3}\,=\,\,{{T}_{2}}{{T}_{4}}\]

\[{{T}_{2}}\,=\,{{({{T}_{1}}^{2}{{T}_{4}})}^{1/3}}\]