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 ... .
JEE MAIN 2021, Diffcult
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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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