A system of rods is assembled such that each rod has a length $l$ and cross – sectional areas S. The mode of heat transfer is conduction and the system is in steady state. The temperature of Junction A (sink) is T and that of C (source) is 2T. Now Answer the following question.

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The rate of heat flow along BD is $\frac{21}{5} \frac{k T S}{l}$

Temperature of junction D is 1.4 T.

Temperature of junction B is 1.6 T.

The rate of heat flow along BD is $\frac{1}{5} \frac{k T S}{l}$

answer is A, B, C.

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Detailed Solution

A system of identical rods conducts heat in steady state. Each rod has length l and cross-sectional area S. Junction A (sink) is at temperature T and junction C (source) is at 2T.

Thermal-Resistance Model

Replace every rod by a thermal resistance R₀ = l ⁄ (kS).

Solving the network (node temperatures) gives T_B = 1.6 T and T_D = 1.4 T.

Heat Flow along BD

BD is a single rod, so the heat rate through BD is

q_BD = (kS ⁄ l) × (T_B − T_D) = (kS ⁄ l) × (1.6T − 1.4T) = (kS ⁄ l) × (0.2T) = (1 ⁄ 5) × (kTS ⁄ l).

Options Evaluation

Option	Statement	Verdict
a	The rate of heat flow along BD is (21 ⁄ 5) × (kTS ⁄ l)	False
b	Temperature of junction D is 1.4 T	True
c	Temperature of junction B is 1.6 T	True
d	The rate of heat flow along BD is (1 ⁄ 5) × (kTS ⁄ l)	True