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(C) Let the length of each conductor be $L$ and the area of cross-section of the $1^{	ext{st}}$ and $2^{	ext{nd}}$ conductors be $A$. Then the area

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Let the length of each conductor be $L$ and the area of cross-section of the $1^{	ext{st}}$ and $2^{	ext{nd}}$ conductors be $A$. Then the area of the $3^{	ext{rd}}$ conductor is $2A$.Thermal resistance is given by $R = \frac{L}{kA}$.$R_1 = \frac{L}{k_1 A} = \frac{L}{60A}$,$R_2 = \frac{L}{k_2 A} = \frac{L}{120A}$,$R_3 = \frac{L}{k_3 (2A)} = \frac{L}{135 	imes 2A} = \frac{L}{270A}$.Conductors $1$ and $2$ are in parallel,so their equivalent resistance $R_{12}$ is:$\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{60A}{L} + \frac{120A}{L} = \frac{180A}{L} \implies R_{12} = \frac{L}{180A}$.In steady state,the heat current $H$ flowing through the combination is constant:$H = \frac{100 - 	heta}{R_{12}} = \frac{	heta - 0}{R_3}$$\frac{100 - 	heta}{L / 180A} = \frac{	heta}{L / 270A}$$180(100 - 	heta) = 270	heta$$2(100 - 	heta) = 3	heta$$200 - 2	heta = 3	heta$$5	heta = 200 \implies 	heta = 40^{\circ}C$.

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