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 09-Jan-2019 Morning] |
A) \[45{}^\circ \,C\]
B) \[75{}^\circ \,C\]
C) \[35{}^\circ \,C\]
D) \[60{}^\circ \,C\]
Correct Answer: A
Solution :
[a] |
\[{{T}_{A}}\,\,-\,\,{{T}_{B}}\,\,=\,\,120{}^\circ C\] |
\[{{T}_{P}}-{{T}_{Q}}\,\,=\,\,?\] |
\[\because \,\,{{R}_{th}}\,\,=\,\,\frac{1}{K}\,\,\,\frac{L}{A}\] |
\[\because \,\,\,\,{{R}_{th}}\,\,=\,\,\frac{1}{K\,A}\,\,\,\left( \frac{L}{4} \right)\,\,=\,\,\,R\,\,(Let\,\,us\,\,say)\] |
\[{{(Red.)}_{AB}}\,=\,6.4\,\,R\] |
\[{{1}_{th}}\,=\,\,\frac{120}{6.4\,R};\,\,\,\,\,\,\,\,\,\,\,{{V}_{PQ}}\,\,=\,\,\frac{6\,R}{10\,R}\,.\,\frac{120\,R}{6.4\,R}\,\,=\,\,\frac{72}{6.4\,R}\] |
\[=45{}^\circ C\] |
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