Question

Two slabs each of area $A$, length $2\text{m}$ and $4\text{m}$, thermal conductivities $k_1=300{\text{W/m}}^{\circ }\text{C}$ and $k_2=200{\text{W/m}}^{\circ }\text{C}$ respectively are joined to form a single slab of length $6\text{m}$. Find the equivalent thermal conductivity of resultant slab. (in ${\text{W/m}}^{\circ }\text{C}$)

• A. $300$
• B. $225$
• C. $325$
• D. $275$

Answer 1

The correct option is B $225$



Equivalent thermal conductivity
$k_{\text{eq}}=\frac{L_1+L_2}{\frac{L_1}{k_1}+\frac{L_2}{K_2}}$
$k_{\text{eq}}=\frac{6}{\frac{2}{300}+\frac{4}{200}}$
$k_{\text{eq}}=\frac{6\times 300\times 200}{1600}=225{\text{W/m}}^{\circ }\text{C}$