Two rods A and B of different materials but s | vedclass

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(A) In steady state,the rate of heat flow (heat current) through both rods must be the same.Let $A$ be the cross-sectional area of the rods.The therma

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) In steady state,the rate of heat flow (heat current) through both rods must be the same.Let $A$ be the cross-sectional area of the rods.The thermal resistance $R$ of a rod is given by $R = \frac{l}{KA}$.For rod $A$: $R_1 = \frac{l_1}{K_1 A}$.For rod $B$: $R_2 = \frac{l_2}{K_2 A}$.Given $l_2 = 2l_1$ and $K_1 = 2K_2$,we can write $K_2 = \frac{K_1}{2}$.Substituting these into $R_2$: $R_2 = \frac{2l_1}{(K_1/2)A} = \frac{4l_1}{K_1 A} = 4R_1$.Since the rods are in series,the heat current $H$ is the same:$H = \frac{100 - 	heta}{R_1} = \frac{	heta - 0}{R_2}$.Substituting $R_2 = 4R_1$:$\frac{100 - 	heta}{R_1} = \frac{	heta}{4R_1}$.$4(100 - 	heta) = 	heta$.$400 - 4	heta = 	heta$.$5	heta = 400$.$	heta = 80^{\circ}C$.

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