Two identical square rods of metal are welded | vedclass

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(A) The rate of heat flow is given by $Q/t = \frac{KA(\Delta 	heta)}{L}$.For figure $(a)$,the rods are in series. The equivalent length is $L_{eq} =

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(A) The rate of heat flow is given by $Q/t = \frac{KA(\Delta 	heta)}{L}$.For figure $(a)$,the rods are in series. The equivalent length is $L_{eq} = 2l$ and the area of cross-section is $A$. The rate of heat flow is $H_a = \frac{KA(100-0)}{2l} = \frac{50KA}{l}$.Given $Q = 20 	ext{ cal}$ and $t_1 = 4 	ext{ min}$,so $H_a = 20/4 = 5 	ext{ cal/min}$.For figure $(b)$,the rods are in parallel. The equivalent area is $A_{eq} = 2A$ and the length is $l$. The rate of heat flow is $H_b = \frac{K(2A)(100-0)}{l} = \frac{200KA}{l}$.Comparing $H_a$ and $H_b$,we see $H_b = 4 	imes H_a$.Since $H_b = 4 	imes 5 = 20 	ext{ cal/min}$,the time taken for $20 	ext{ cal}$ to flow is $t_2 = Q/H_b = 20/20 = 1 	ext{ min}$.

Two identical square rods of metal are welded end to end as shown in figure $(a)$. $20 \text{ calories}$ of heat flows through them in $4 \text{ minutes}$. If the rods are welded as shown in figure $(b)$,the same amount of heat will flow through the rods in ...... $\text{min}$.

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