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As per the given figure, two plates \mathrm{A}$ and $\mathrm{B} of thermal conductivity \mathrm{K}$ and $2 \mathrm{~K} are joined together to form a compound plate. The thickness of plates are 4.0 \mathrm{~cm}$ and $2.5 \mathrm{~cm} respectively and the area of cross-section is 120 \mathrm{~cm}^{2} for each plate. The equivalent thermal conductivity of the compound plate is \left(1+\frac{5}{\alpha}\right) \mathrm{K}, then the value of \alpha will be_____________.

Option: 1

21


Option: 2

56


Option: 3

47


Option: 4

6


Answers (1)

best_answer

\text{Thermal resistance}=\mathrm{R=\frac{1}{K A}}

\mathrm{\begin{aligned} & \\\\ &\mathrm{R_{e f f}=\left(R_{1}+R_{2}\right)} \\\\ &\mathrm{\frac{\left(L_{1}+L_{2}\right)}{K e f f \times A}=\frac{L_{1}}{K_{1} A}+\frac{L_{2}}{ K_{2} A}} \\\\ &\mathrm{\frac{6.5}{K e f f}=\frac{4}{k}+\frac{2.5}{2 K} }\\\\ &\text { Keff }=\frac{13 k}{10.5}=\left[1+\frac{5}{21}\right] \text{k} \\ &\alpha=21 \end{aligned} }

Hence correct answer is 21

Posted by

Pankaj

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