Related Formula
Rth = (L)/(KA) = (L)/(K(π r²))$$R_{\text{th}} = \frac{L}{KA} = \frac{L}{K(\pi r^2)}$$ [cite: 817]
(dQ)/(dt) = Δ TRth$$\frac{dQ}{dt} = \frac{\Delta T}{R_{\text{th}}}$$ [cite: 818]
Core Logic
At steady state equilibrium, the rate of heat flow through both sections in series must be identical: [cite: 193, 819]
(400 - T)/(R₁) = (T - 200)/(R₂) (400 - T)/(T - 200) = (R₁)/(R₂)$$\frac{400 - T}{R_1} = \frac{T - 200}{R_2} \implies \frac{400 - T}{T - 200} = \frac{R_1}{R_2}$$ [cite: 192, 820]
Let's evaluate the resistance ratio (R₁)/(R₂)$\frac{R_1}{R_2}$ using the dimensional parameters: [cite: 820]
(R₁)/(R₂) = ((LA)/(LB)) · ((rB)/(rA))² · ((KB)/(KA))$$\frac{R_1}{R_2} = \left(\frac{L_A}{L_B}\right) \cdot \left(\frac{r_B}{r_A}\right)^2 \cdot \left(\frac{K_B}{K_A}\right)$$ [cite: 820]
Substitute the given values: (LA)/(LB) = (1)/(2)$\frac{L_A}{L_B} = \frac{1}{2}$, (rA)/(rB) = 2 (rB)/(rA) = (1)/(2)$\frac{r_A}{r_B} = 2 \implies \frac{r_B}{r_A} = \frac{1}{2}$, and (KA)/(KB) = (1)/(2) (KB)/(KA) = 2$\frac{K_A}{K_B} = \frac{1}{2} \implies \frac{K_B}{K_A} = 2$ [cite: 189, 190]:
(R₁)/(R₂) = (1)/(2) × ((1)/(2))² × 2 = (1)/(4)$$\frac{R_1}{R_2} = \frac{1}{2} \times \left(\frac{1}{2}\right)^2 \times 2 = \frac{1}{4}$$ [cite: 821]
Now link this back into the temperature equation: [cite: 822]
(400 - T)/(T - 200) = (1)/(4) 1600 - 4T = T - 200$$\frac{400 - T}{T - 200} = \frac{1}{4} \implies 1600 - 4T = T - 200$$ [cite: 822, 823]
5T = 1800 T = 360 K$$5T = 1800 \implies T = 360\ \text{K}$$ [cite: 824, 825]
Pattern Recognition
Thermal conduction processes behave exactly like electric current fields in series connections[cite: 818, 819]. Cross-sectional area scales squarely with radius parameters, which requires extra care during substitution.
Chapter Mix
Class 11 Physics: Thermal Properties of Matter
