Question

Two coaxial cylinders of radius r and 3r of thermal conductivity

Answer 1

$K_{eff}=\dfrac{K_{1}+8K_{2}}{9}$

Explanation:

$\dfrac{dQ}{dt} =\dfrac{K_{1}\pi r^{2}(t_{1}-t_{2)} }{L} +\dfrac{K_{2}\pi (3r)^{2}(t_{1}-t_{2)} }{L}$

$= \dfrac{(K_{1} +8K_{2})(t_{2}-t_{1} )\pi r^{2} }{L}$

∴ $\dfrac{dQ}{dt} =\dfrac{(K_{1} +8K_{2})(t_{2}-t_{1} )\pi r^{2} }{L}$

Therefore ,

$K_{eff}=\dfrac{K_{1}+8K_{2}}{9}$

Where, $K_{eff}$ is the thermal conductivity of the combined system.