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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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