A cylinder of thermal conductivity k_1 and ra | vedclass

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(D) Since the two cylinders are arranged in parallel,the total thermal conductance is the sum of the individual thermal conductances.Thermal conductan

Vedclass · Accepted Answer

$1/4 (k_1 + 3k_2)$

Vedclass · Answer

(D) Since the two cylinders are arranged in parallel,the total thermal conductance is the sum of the individual thermal conductances.Thermal conductance $C = \frac{kA}{l}$.For parallel combination,$C_{eq} = C_1 + C_2$.$\frac{k_{eq} A_{total}}{l} = \frac{k_1 A_1}{l} + \frac{k_2 A_2}{l}$.Since the lengths $l$ are equal,$k_{eq} A_{total} = k_1 A_1 + k_2 A_2$.Here,$A_1 = \pi r^2$ and $A_2 = \pi ((2r)^2 - r^2) = 3\pi r^2$.The total area $A_{total} = A_1 + A_2 = 4\pi r^2$.Substituting these values: $k_{eq} (4\pi r^2) = k_1 (\pi r^2) + k_2 (3\pi r^2)$.$4 k_{eq} = k_1 + 3k_2$.$k_{eq} = \frac{1}{4} (k_1 + 3k_2)$.

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