question_answer

Temperature difference of \[120{}^\circ \,C\] is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length \[\frac{3L}{2}\], is connected across AB (See figure). In steady  state,  temperature  difference between P and Q will be close to:   [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

A) \[45{}^\circ \,C\]

B) \[75{}^\circ C\]

C) \[35{}^\circ C\]

D) \[60{}^\circ C\]

Correct Answer: A

Solution :

\[{{T}_{A}}\,\,-\,\,{{T}_{B}}\,\,=\,\,120{}^\circ C\] \[{{T}_{P}}-{{T}_{Q}}\,\,=\,\,?\] \[\because \,\,{{R}_{th}}\,\,=\,\,\frac{1}{K}\,\,\,\frac{L}{A}\] \[\because \,\,\,\,{{R}_{th}}\,\,=\,\,\frac{1}{K\,A}\,\,\,\left( \frac{L}{4} \right)\,\,=\,\,\,R\,\,(Let\,\,us\,\,say)\] \[{{(Red.)}_{AB}}\,=\,6.4\,\,R\] \[{{1}_{th}}\,=\,\,\frac{120}{6.4\,R};\,\,\,\,\,\,\,\,\,\,\,{{V}_{PQ}}\,\,=\,\,\frac{6\,R}{10\,R}\,.\,\frac{120\,R}{6.4\,R}\,\,=\,\,\frac{72}{6.4\,R}\] \[45{}^\circ C\]