For the figure shown,when arcs ACD and ADB ar | vedclass

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(A) Let $R_1$ be the thermal resistance of arc $ADB$ and $R_2$ be the thermal resistance of arc $ACD$. The angle subtended by $ADB$ at the center is $

Vedclass · Accepted Answer

$\frac{7}{3}$

Vedclass · Answer

(A) Let $R_1$ be the thermal resistance of arc $ADB$ and $R_2$ be the thermal resistance of arc $ACD$. The angle subtended by $ADB$ at the center is $90^\circ$ (or $\pi/2$ radians),and the angle subtended by $ACD$ is $270^\circ$ (or $3\pi/2$ radians).Since the length of an arc is $L = r	heta$,the length of $ADB$ is $L_1 = r(\pi/2)$ and the length of $ACD$ is $L_2 = r(3\pi/2) = 3L_1$.Thermal resistance $R = \frac{L}{KA}$. Assuming same cross-sectional area $A$ and same initial conductivity $K$:$R_1 = \frac{L_1}{KA}$ and $R_2 = \frac{3L_1}{KA} = 3R_1$.The arcs are in parallel between $A$ and $B$,so the total heat current $H = \frac{\Delta T}{R_{eq}} = \Delta T \left(\frac{1}{R_1} + \frac{1}{R_2}ight) = \Delta T \left(\frac{1}{R_1} + \frac{1}{3R_1}ight) = \frac{4\Delta T}{3R_1}$.When $ADB$ is replaced with material of conductivity $K'$,the new resistance is $R_1' = \frac{L_1}{K'A} = R_1 \frac{K}{K'}$.The new heat current is $2H = \Delta T \left(\frac{1}{R_1'} + \frac{1}{R_2}ight) = \Delta T \left(\frac{K'}{R_1 K} + \frac{1}{3R_1}ight)$.Substituting $H = \frac{4\Delta T}{3R_1}$,we get $2 \left(\frac{4\Delta T}{3R_1}ight) = \Delta T \left(\frac{K'}{R_1 K} + \frac{1}{3R_1}ight)$.$\frac{8}{3R_1} = \frac{3K' + K}{3KR_1} \implies 8K = 3K' + K \implies 3K' = 7K \implies \frac{K'}{K} = \frac{7}{3}$.

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