Consider the setup depicted in the figure. Assume that the thermal...

Let the temperature difference across the parallel section be $\Delta T$.

The rate of heat flow through a conductor is given by $H = \dfrac{KA\Delta T}{L}$, where $K is the thermal conductivity,$A$is the cross-sectional area, and$L$is the length.<br><br>For the bent part:<br>Thermal conductivity,$K_b = 780 \text{ J s}^{-1} \text{ m}^{-1} ^\circ\textC^{-1}$<br>Length of the bent part,$L_b = 5 \text{ cm} + 60 \text{ cm} + 5 \text{ cm} = 70 \text{ cm}$<br>Rate of heat flow,$H_b = \dfrac{K_b A \Delta T}{L_b}$<br><br>For the straight part in parallel:<br>Thermal conductivity,$K_s = 390 \text{ J s}^{-1} \text{ m}^{-1} ^\circ\textC^{-1}$<br>Length of the straight part,$L_s = 60 \text{ cm}$<br>Rate of heat flow,$H_s = \dfrac{K_s A \Delta T}{L_s}$<br><br>The ratio of the rate of heat flow through the bent part to that through the straight part is:<br>$\dfrac{H_b}{H_s} = \dfrac{K_b}{K_s} \times \dfrac{L_s}{L_b}$<br><br>Substituting the given values:<br>$\dfrac{H_b}{H_s} = \left(\dfrac{780}{390}\right) \times \left(\dfrac{60}{70}\right)

\dfrac{H_b}{H_s} =$2 \times$\dfrac{6}{7} = \dfrac{12}{7}

Thus, the required ratio is $12 : 7$.