For the figure shown, when arc $ACD$ and $ADB$ are made of same material, the heat carried between $A$ and $B$ is $H$ . If $ADB$ is replaced with another material, the heat carried becomes $2H$ . If the temperatures at $A$ and $B$ are fixed at $T_1$ and $T_2$ , what is the ratio of the new conductivity to the old one of $ADB$
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$\left(\frac{\mathrm{KA}(\Delta \mathrm{T})}{3 \ell}+\frac{(\mathrm{KA}) \Delta \mathrm{T}}{\ell}\right) \times 2$
$=\left(\frac{(\mathrm{KA}) \Delta \mathrm{T}}{3 \ell}+\frac{\left(\mathrm{K}^{\prime}\right) \mathrm{A} \Delta \mathrm{T}}{\ell}\right)$
$\frac{4 \mathrm{K}}{3} \times 2=\frac{\mathrm{K}}{3}+\mathrm{K}^{\prime}$
$K^{\prime}=\frac{7 K}{3}$
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