A ring consisting of two parts ADB and ACB of | vedclass

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(A) Let $l_1$ be the length of $ADB$ and $l_2$ be the length of $ACB$. Given $\frac{l_2}{l_1} = 3$,so $l_2 = 3l_1$. Let $A$ be the cross-sectional are

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$\frac{7}{3} k$

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(A) Let $l_1$ be the length of $ADB$ and $l_2$ be the length of $ACB$. Given $\frac{l_2}{l_1} = 3$,so $l_2 = 3l_1$. Let $A$ be the cross-sectional area.Initially,both parts have conductivity $k$. The thermal resistances are $R_1 = \frac{l_1}{kA}$ and $R_2 = \frac{l_2}{kA} = \frac{3l_1}{kA} = 3R_1$.The total heat flow $H$ is the sum of heat flows through parallel paths: $H = H_1 + H_2 = \frac{T_1 - T_2}{R_1} + \frac{T_1 - T_2}{R_2} = (T_1 - T_2) \left( \frac{kA}{l_1} + \frac{kA}{3l_1} \right) = (T_1 - T_2) \frac{4kA}{3l_1}$.When $ADB$ is replaced by a metal of conductivity $k'$,the new heat flow is $H' = 2H = 2 \left( \frac{4kA(T_1 - T_2)}{3l_1} \right) = \frac{8kA(T_1 - T_2)}{3l_1}$.The new heat flow is $H' = \frac{k'A(T_1 - T_2)}{l_1} + \frac{kA(T_1 - T_2)}{3l_1}$.Equating the two expressions for $H'$:$\frac{k'A(T_1 - T_2)}{l_1} + \frac{kA(T_1 - T_2)}{3l_1} = \frac{8kA(T_1 - T_2)}{3l_1}$.Dividing by $\frac{A(T_1 - T_2)}{l_1}$: $k' + \frac{k}{3} = \frac{8k}{3}$.$k' = \frac{8k}{3} - \frac{k}{3} = \frac{7k}{3}$.

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