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 Online Paper (Held On 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 :

\[As\,=\,{{\eta }_{1}}={{\eta }_{2}}\] \[1-\frac{{{T}_{2}}}{{{T}_{1}}}\,=\,1-\frac{{{T}_{3}}}{{{T}_{2}}}\] \[{{T}^{2}}\,=\,\sqrt{{{T}_{1}}{{T}_{3}}}\] \[As\,{{\eta }_{2}}=\,{{\eta }_{3}}\] \[1-\frac{{{T}^{3}}}{{{T}^{2}}}\,=\,\,1\,-\,\,\frac{{{T}^{4}}}{{{T}^{3}}}\] \[{{T}_{3}}=\sqrt{{{T}_{2}}{{T}_{4}}}\] \[As\,\,{{\eta }_{1}}={{\eta }_{3}}\] \[1-\frac{{{T}_{2}}}{{{T}_{1}}}\,=\,1\,-\,\frac{{{T}^{4}}}{{{T}^{3}}}\,\,\,Hence\,\,{{T}_{2}}{{T}_{3}}\,=\,{{T}_{1}}{{T}_{4}}\,\] \[\frac{{{T}_{1}}{{T}_{4}}}{{{T}_{3}}}\,=\,\sqrt{{{T}_{1}}{{T}_{3}}}\,\,\,\,\,\Rightarrow \,\,\,\frac{{{T}^{2}}_{1}{{T}^{2}}_{4}}{{{T}_{3}}^{2}}\,\,=\,\,{{T}_{1}}{{T}_{3}}\] \[{{T}_{3}}=\,\,{{({{T}_{1}}{{T}_{4}}^{2})}^{1/3}}\] \[{{({{T}_{1}}{{T}_{4}}^{2})}^{1/3}}\,=\,{{({{T}_{2}}{{T}_{4}})}^{1/2}}\] \[{{T}_{1}}^{2/3}{{T}_{4}}^{4/3}\,=\,\,{{T}_{2}}{{T}_{4}}\] \[{{T}_{2}}\,=\,{{({{T}_{1}}^{2}{{T}_{4}})}^{1/3}}\]