In steady state, the heat current through both rods is equal. For rod A (length ℓ₁, thermal conductivity K₁) and rod B (length ℓ₂, conductivity K₂), the heat flow equation is:

$\frac{K_{1}A(100-\theta )}{{\ell }_1}=\frac{K_{2}A(\theta -0)}{{\ell }_2}$

Given ℓ₂ = 2ℓ₁ and K₁ = 2K₂, substitute and solve:

$\frac{2K_2(100-\theta )}{{\ell }_1}=\frac{K_2\theta }{2{\ell }_1}$

Simplify: 4(100 - θ) = θ → 400 - 4θ = θ → 400 = 5θ → θ = 80°C.

Final Answer: 80°C