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(B) The rate of heat flow $(H)$ through a rod is given by the formula: $H = \frac{KA \Delta T}{L}$ where $K$ is the thermal conductivity,$A$ is the cr

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$K_1 A_1 = K_2 A_2$

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(B) The rate of heat flow $(H)$ through a rod is given by the formula: $H = \frac{KA \Delta T}{L}$ where $K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $L$ is the length of the rod. Given that the rate of heat flow is the same for both rods $(H_1 = H_2)$,the lengths are equal $(L_1 = L_2 = L)$,and the temperature differences are equal $(\Delta T_1 = \Delta T_2 = \Delta T)$. Equating the heat flow rates: $\frac{K_1 A_1 \Delta T}{L} = \frac{K_2 A_2 \Delta T}{L}$ Canceling the common terms $\Delta T$ and $L$ from both sides,we get: $K_1 A_1 = K_2 A_2$ Therefore,the required condition is $K_1 A_1 = K_2 A_2$.

Two metal rods $1$ and $2$ have the same length and the same temperature difference at their ends. Their thermal conductivities are $K_1$ and $K_2$ and their cross-sectional areas are $A_1$ and $A_2$ respectively. The condition for the same rate of heat flow in them is ........

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