Question

Two rods of equal length and diameter have thermal conductivities 3 and 4 units respectively. If they are joined in series, the thermal conductivity of the combination in the given units would be

• A. 3.43
• B. 4.43
• C. 5.43
• D. 2.43

Answer 1

Answer: (A)

3.43

Answer 2

Given $L_{1} = L_{2} = L,A_{1} = A_{2} = A$

$K_{1} = 3$unit and $K_{2} = 4$ unit

If $R_{1}$and $R_{2}$are the thermal resistances of the two rode then $R_{1} = \frac{L_{1}}{K_{1}A_{1}} = \frac{L}{3 \times A}andR_{2} = \frac{L_{2}}{K_{2}A_{2}} = \frac{L}{4 \times A}$

If $R_{eq}$is the thermal resistances of the two rods in series and $K_{eq}$is the equivalent thermal conductivity. Then

$R_{eq} = \frac{L_{1} + L_{2}}{K_{eq} \times A} = \frac{2L}{K_{eq}A}$

As $R_{eq} = R_{1} + R_{2}$So $\frac{2L}{K_{eq}A} = \frac{L}{3A} + \frac{L}{4A} = \frac{71}{12A}$Or $K_{eq} = \frac{12 \times 2}{7} = 3.43units$