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(B) Let the length of each rod be $l$,area of cross-section be $A$,and thermal conductivity be $K$. The heat transferred is $Q$ and the temperature di

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$3$

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(B) Let the length of each rod be $l$,area of cross-section be $A$,and thermal conductivity be $K$. The heat transferred is $Q$ and the temperature difference is $\Delta 	heta$.Case $1$: When joined end to end (series),the total length is $2l$ and the area is $A$. The equivalent thermal resistance is $R_{eq} = R_1 + R_2 = \frac{l}{KA} + \frac{l}{KA} = \frac{2l}{KA}$.The heat flow is $Q = \frac{\Delta 	heta}{R_{eq}} 	imes t_1 = \frac{\Delta 	heta \cdot KA}{2l} 	imes 12 = \frac{6 KA \Delta 	heta}{l} \dots (i)$.Case $2$: When joined lengthwise (parallel),the length is $l$ and the total area is $2A$. The equivalent thermal resistance is $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{KA}{l} + \frac{KA}{l} = \frac{2KA}{l}$,so $R_{eq} = \frac{l}{2KA}$.The heat flow is $Q = \frac{\Delta 	heta}{R_{eq}} 	imes t_2 = \frac{\Delta 	heta \cdot 2KA}{l} 	imes t_2 \dots (ii)$.Equating $(i)$ and $(ii)$:$\frac{6 KA \Delta 	heta}{l} = \frac{2 KA \Delta 	heta}{l} 	imes t_2$$6 = 2 t_2 \implies t_2 = 3 \, s$.

Two rods of same length and material transfer a given amount of heat in $12 \, s$ when they are joined end to end. But when they are joined lengthwise (parallel),they will transfer the same amount of heat under the same conditions in .......... $s$.

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