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(B) Let the temperature at junction $B$ be $T_B$. The thermal resistance of a rod is given by $R = \frac{L}{kA}$. Since all rods have the same materia

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$\frac{7}{2}$

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(B) Let the temperature at junction $B$ be $T_B$. The thermal resistance of a rod is given by $R = \frac{L}{kA}$. Since all rods have the same material $(k)$ and same cross-section $(A)$,the thermal resistance is proportional to the length $(R \propto L)$.Let $R_{AB}, R_{BC}, R_{BD}$ be the resistances of rods $AB, BC, BD$ respectively.Given that there is no heat flow in $AB$,the temperature at $B$ must be equal to the temperature at $A$. Therefore,$T_B = T_A = 20^{\circ} C$.Since there is no heat flow in $AB$,all the heat flowing from $D$ to $B$ must flow from $B$ to $C$. Thus,the heat current $H_{DB} = H_{BC}$.Using the formula $H = \frac{\Delta T}{R}$,we have $\frac{T_D - T_B}{R_{BD}} = \frac{T_B - T_C}{R_{BC}}$.Substituting the values: $\frac{90 - 20}{R_{BD}} = \frac{20 - 0}{R_{BC}}$.$\frac{70}{R_{BD}} = \frac{20}{R_{BC}}$.$\frac{R_{BD}}{R_{BC}} = \frac{70}{20} = \frac{7}{2}$.Since $R \propto L$,we have $\frac{L_{BD}}{L_{BC}} = \frac{7}{2}$.

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