Two cylindrical rods A and B made of differen | vedclass

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(C) The thermal resistance $R$ of a rod is given by $R = \frac{L}{KA}$,where $L$ is the length,$K$ is the thermal conductivity,and $A = \pi r^2$ is th

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$360$

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(C) The thermal resistance $R$ of a rod is given by $R = \frac{L}{KA}$,where $L$ is the length,$K$ is the thermal conductivity,and $A = \pi r^2$ is the cross-sectional area.For the two rods in series,the rate of heat flow $\frac{dQ}{dt}$ is the same in steady state:$\frac{dQ}{dt} = \frac{\Delta T}{R} = \frac{400 - T}{R_A} = \frac{T - 200}{R_B}$$\frac{400 - T}{T - 200} = \frac{R_A}{R_B} = \left( \frac{L_A}{L_B} ight) \left( \frac{K_B}{K_A} ight) \left( \frac{r_B}{r_A} ight)^2$Given $\frac{L_A}{L_B} = \frac{1}{2}$,$\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$,and $\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$.Substituting these values:$\frac{R_A}{R_B} = \left( \frac{1}{2} ight) 	imes (2) 	imes \left( \frac{1}{2} ight)^2 = 1 	imes \frac{1}{4} = \frac{1}{4}$Now,$\frac{400 - T}{T - 200} = \frac{1}{4}$$4(400 - T) = T - 200$$1600 - 4T = T - 200$$5T = 1800$$T = 360 \ K$

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