Two identical square rods of metal are welded end to end as shown in figure (i), 20 calories of heat flows through it in 4 minutes. If the rods are welded as shown in figure (ii), the same amount of heat will flow through the rods in
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$\frac{Q}{t} = \frac{KA \Delta \theta}{l} = \frac{\Delta \theta}{(l/KA)} = \frac{\Delta \theta}{R} \quad (\mathrm{R = Thermal\ resistance})$ $\Rightarrow t \propto R \quad (\because Q \text{ and } \Delta \theta \text{ are same})$ $\Rightarrow \frac{t_P}{t_S} = \frac{R_P}{R_S} = \frac{R/2}{2R} = \frac{1}{4} \Rightarrow t_P = \frac{t_S}{4} = \frac{4}{4} = 1 \mathrm{min.}$ $(\text{Series resistance } R_S = R_1 + R_2 \text{ and parallel resistance } R_P = \frac{R_1 R_2}{R_1 + R_2})$
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