JEE Mains · Physics · STD 11 - Transmission of heat
- JEE Mains 2021
- Hard
- Numerical
A rod \(C D\) of thermal resistance \(10.0\; {KW}^{-1}\) is joined at the middle of an identical rod \({AB}\) as shown in figure, The end \(A, B\) and \(D\) are maintained at \(200^{\circ} {C}, 100^{\circ} {C}\) and \(125^{\circ} {C}\) respectively. The heat current in \({CD}\) is \({P}\) watt. The value of \({P}\) is ... .

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Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
Rods are identical so
\({R}_{{AB}}={R}_{{CD}}=10 {Kw}^{-1}\)
\({C}\) is mid-point of \({AB}\), so
\({R}_{{AC}}={R}_{{CB}}=5 {K} {w}^{-1}\)
at point \({C}\)
\(\frac{200-{T}}{5}=\frac{{T}-125}{10}+\frac{{T}-100}{5}\)
\(2(200-{T})={T}-125+2({T}-100)\)
\(400-2 {T}={T}-125+2 {T}-200\)
\({T}=\frac{725}{5}=145^{\circ} {C}\) \({I}_{{b}}=\frac{145-125}{10} {w}=\frac{20}{10} {w}\)
\({I}_{{h}}=2 {w}\)
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