Three conductors of same length having thermal conductivity $k_{1}, k_{2}$ and $k_{3}$ are connected

Question

Three conductors of same length having thermal conductivity $k_{1}, k_{2}$ and $k_{3}$ are connected as shown in figure. Area of cross sections of $1^{	ext {st }}$ and $2^{	ext {nd }}$ conductor are same and for $3^{	ext {rd }}$ conductor it is double of the $1^{	ext {st }}$ \ conductor. The temperature are given the figure. In steady state condition, the value of $	heta$ is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
(Given : $k_{1}=60 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, k_{2}=120 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$,
$
\left.k_{3}=135 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}ight)
$

Professor PC Thomas Classes · Answer

(40) : As conductors $1^{	ext {st }}$ and $2^{	ext {nd }}$ are in parallel.\ So, equivalent conductivity between $1^{	ext {st }}$ and $2^{	ext {nd }}$ is\ $k_{12}=k_{1}+k_{2}=120+60=180 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$\ Let, $T_{1}=100^{\circ} \mathrm{C}=373 \mathrm{~K}, T_{2}=	heta \mathrm{K}, T_{3}=0^{\circ} \mathrm{C}$\ $=273 \mathrm{~K}$.\ $	herefore \quad \frac{T_{1}-T_{2}}{\frac{l}{k_{12} A}}=\frac{T_{2}-T_{3}}{\frac{l}{k_{3} 2 A}}$\ $\Rightarrow \frac{\frac{373-T_{2}}{l}}{\frac{l}{(180) A}}=\frac{T_{2}-273}{\frac{l}{(135) 2 A}}$\ $\Rightarrow(373-	heta) 180=135(	heta-273) 2$\ $\Rightarrow 746-2 	heta=3 	heta-819$\ $\Rightarrow \quad 	heta=313 \mathrm{~K}=40^{\circ} \mathrm{C}$.\