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(B) Let the thermal resistance of each rod be $R$. In figure $(a)$,the rods are in series,so the equivalent thermal resistance is $R_{eq1} = R + R = 2

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$0.5$

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(B) Let the thermal resistance of each rod be $R$. In figure $(a)$,the rods are in series,so the equivalent thermal resistance is $R_{eq1} = R + R = 2R$. The rate of heat flow is $\frac{Q_1}{t_1} = \frac{\Delta T}{R_{eq1}} = \frac{100 - 0}{2R} = \frac{50}{R}$. Given $Q_1 = 10 \, cal$ and $t_1 = 2 \, min$,we have $\frac{10}{2} = \frac{50}{R}$,which gives $R = 10 \, K/cal \cdot min$. In figure $(b)$,the rods are in parallel,so the equivalent thermal resistance is $R_{eq2} = \frac{R \cdot R}{R + R} = \frac{R}{2}$. The rate of heat flow is $\frac{Q_2}{t_2} = \frac{\Delta T}{R_{eq2}} = \frac{100 - 0}{R/2} = \frac{200}{R}$. For the same amount of heat $Q_2 = 10 \, cal$,we have $\frac{10}{t_2} = \frac{200}{R}$. Substituting $R = 10$,we get $\frac{10}{t_2} = \frac{200}{10} = 20$. Thus,$t_2 = \frac{10}{20} = 0.5 \, min$.

Two identical square rods of metal are welded end to end as shown in figure $(a)$. Assume that $10 \, cal$ of heat flows through the rods in $2 \, min$. Now the rods are welded as shown in figure $(b)$. The time it would take for $10 \, cal$ to flow through the rods now,is ........ $\min$.

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