A rod of 40 cm in length and temperature difference of $80^{\circ}C$ at its two ends. Another rod B of length 60 cm and of temperature difference $90^\circ C$ , having the same area of cross-section. If the rate of flow of heat is the same, then the ratio of their thermal conductivities will be
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$\frac{\mathrm{dQ}}{\mathrm{dt}} = \frac{\mathrm{KA}(\theta_1 - \theta_2)}{\mathrm{d}} \\ \Rightarrow \frac{\mathrm{K}_1 \Delta \theta_1}{l_1} = \frac{\mathrm{K}_2 \Delta \theta_2}{l_2} \quad (\because \frac{\mathrm{dQ}}{\mathrm{dt}} \text{ and A are same}) \\ \Rightarrow \frac{\mathrm{K}_1 \times 80}{40} = \frac{\mathrm{K}_2 \times 90}{60} \Rightarrow \frac{\mathrm{K}_1}{\mathrm{K}_2} = \frac{3}{4}$
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