Three rods A, B, and C of thermal conductivit | vedclass

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(B) Let the temperature of the junction be $	heta$. The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.For rod $A$: $H_A = \frac{K A (100

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$\frac{300}{13}\ ^oC$

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(B) Let the temperature of the junction be $	heta$. The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.For rod $A$: $H_A = \frac{K A (100 - 	heta)}{2l}$.For rod $B$: $H_B = \frac{(2K) A (	heta - 50)}{l}$.For rod $C$: $H_C = \frac{(4K) A (	heta - 0)}{l}$.At the junction,the sum of heat currents must be zero: $H_A + H_B + H_C = 0$.$\frac{KA(100 - 	heta)}{2l} + \frac{2KA(	heta - 50)}{l} + \frac{4KA(	heta - 0)}{l} = 0$.Dividing by $\frac{KA}{2l}$,we get: $(100 - 	heta) + 4(	heta - 50) + 8(	heta) = 0$.$100 - 	heta + 4	heta - 200 + 8	heta = 0$.$11	heta - 100 = 0$.$	heta = \frac{100}{11}\ ^oC$. Wait,re-evaluating the junction currents: The heat flows from $100\ ^oC$ to $	heta$,and from $	heta$ to $50\ ^oC$ and $0\ ^oC$. So,$H_A = H_B + H_C$.$\frac{KA(100 - 	heta)}{2l} = \frac{2KA(	heta - 50)}{l} + \frac{4KA(	heta - 0)}{l}$.$\frac{100 - 	heta}{2} = 2(	heta - 50) + 4	heta$.$100 - 	heta = 4(	heta - 50) + 8	heta$.$100 - 	heta = 4	heta - 200 + 8	heta$.$100 - 	heta = 12	heta - 200$.$300 = 13	heta$.$	heta = \frac{300}{13}\ ^oC$.

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