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Thermodynamics

# Rods of equal length & equal cross - section are arranged as shown - The thermal conductivity of the rods are written correspondingly . The equivalent thermal conductivity of the system is -

$(a)\;\large\frac{20}{3} K\qquad(b)\;\large\frac{10}{3} K\qquad(c)\;5K\qquad(d)\;None\;of\;these$

Answer : $\;\large\frac{20}{3} K$
Explanation :
For series connection ---
$K_{s}=\large\frac{K_{1}K_{2}}{K_{1}+K_{2}} \quad$ { For equal lengths & cross - section }
For parallel combination ---
$K_{p}=K_{1}+K_{2} \quad$ { For equal lengths & cross - section }
Therefore , The system reduces to
Now , $\; \bigtriangleup Q= \bigtriangleup Q_{1} + \bigtriangleup Q_{2} + \bigtriangleup Q_{3}$
$\large\frac{K_{eq} A (3T_{0})}{4L}=\large\frac{KAT_{0}}{L}+\large\frac{(3K)(2T_{0})}{2L}+\large\frac{KAT_{0}}{L}$
$K_{eq}(\large\frac{3}{4})=K + 3K +k$
$K_{eq}=\large\frac{20K}{3}$