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Q.

Two metallic block $M_{1}$ and $M_{2}$ of same area of cross-section are connected to each other (as shown in figure). If the thermal conductivity of $M_{2}$ is K then the thermal conductivity of $M_{1}$ will be: [Assume steady state heat conduction]

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a

12.5 K

b

2 K

c

8 K

d

10 K

answer is B.

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Detailed Solution

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As Q=constant in series So, $Q_{M_{1}} = Q_{M_{2}}$
$\Rightarrow \frac{(100 - 80)}{\frac{16}{k_{1} A}} = \frac{80 - 0}{\frac{8}{k_{2} A}} [∵ Q = \frac{T_{1} - T_{2}}{l / k A}] \Rightarrow \frac{20}{16} \times k_{1} A = \frac{80}{8} k_{2} A \Rightarrow k_{2} = \frac{20 \times 8}{16 \times 80} k_{1}$

$\Rightarrow k_{2} = \frac{k_{1}}{8} \Rightarrow k_{1} = 8 k_{2} = 8 k$

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