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(B) Let $L$ be the length of each rod.The rate of heat flow $(H)$ through a rod is given by the formula $H = \frac{K A \Delta T}{L}$.For rod $1$,the r

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$K_1 A_1 = 4 K_2 A_2$

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(B) Let $L$ be the length of each rod.The rate of heat flow $(H)$ through a rod is given by the formula $H = \frac{K A \Delta T}{L}$.For rod $1$,the rate of heat flow is $H_1 = \frac{K_1 A_1 \Delta T}{L}$.For rod $2$,the rate of heat flow is $H_2 = \frac{K_2 A_2 \Delta T}{L}$.According to the problem,the rate of heat conduction in rod $1$ is four times that in rod $2$,so $H_1 = 4 H_2$.Substituting the expressions for $H_1$ and $H_2$:$\frac{K_1 A_1 \Delta T}{L} = 4 \left( \frac{K_2 A_2 \Delta T}{L} \right)$.Canceling the common terms $\frac{\Delta T}{L}$ from both sides,we get:$K_1 A_1 = 4 K_2 A_2$.

Two metal rods $1$ and $2$ of same lengths have same temperature difference between their ends. Their thermal conductivities are $K_1$ and $K_2$ and cross-sectional areas are $A_1$ and $A_2$,respectively. If the rate of heat conduction in rod $1$ is four times that in rod $2$,then:

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