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

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(D) The thermal resistance $R$ of a rod is given by $R = \frac{l}{KA}$.For rod $A$: $R_A = \frac{2l}{K \cdot A} = \frac{2l}{KA}$.For rod $B$: $R_B = \

Vedclass · Accepted Answer

$\frac{200}{13}\,^oC$

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(D) The thermal resistance $R$ of a rod is given by $R = \frac{l}{KA}$.For rod $A$: $R_A = \frac{2l}{K \cdot A} = \frac{2l}{KA}$.For rod $B$: $R_B = \frac{l}{(2K) \cdot (2A)} = \frac{l}{4KA}$.For rod $C$: $R_C = \frac{l}{(4K) \cdot (2A)} = \frac{l}{8KA}$.Let the junction temperature be $	heta$. According to the principle of conservation of energy (Kirchhoff's junction law for heat flow),the sum of heat currents at the junction must be zero: $\frac{100 - 	heta}{R_A} + \frac{50 - 	heta}{R_B} + \frac{0 - 	heta}{R_C} = 0$.Substituting the values of resistances:$\frac{100 - 	heta}{2l/KA} + \frac{50 - 	heta}{l/4KA} + \frac{0 - 	heta}{l/8KA} = 0$.Multiplying by $\frac{l}{KA}$:$\frac{100 - 	heta}{2} + 4(50 - 	heta) + 8(0 - 	heta) = 0$.$50 - 0.5	heta + 200 - 4	heta - 8	heta = 0$.$250 = 12.5	heta$.$	heta = \frac{250}{12.5} = 20\,^oC$.Wait,let's re-evaluate the calculation: $50 - 0.5	heta + 200 - 4	heta - 8	heta = 250 - 12.5	heta = 0 \implies 	heta = 20\,^oC$. Checking the options,it seems there might be a typo in the provided options or the setup. Let's re-check the heat flow: $\frac{100-	heta}{R_A} = \frac{	heta-50}{R_B} + \frac{	heta-0}{R_C}$.$\frac{100-	heta}{2} = 4(	heta-50) + 8(	heta-0)$.$50 - 0.5	heta = 4	heta - 200 + 8	heta$.$250 = 12.5	heta \implies 	heta = 20\,^oC$. Given the options,if we assume the heat flows from $100$ and $50$ towards the junction and then to $0$,the calculation holds. If the question implies a different configuration,the result might vary. Based on standard interpretation,$	heta = 20\,^oC$ is the result.

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