Question

Thermal conductances of two slabs of equal length are $2$ and $3$ units respectively. If they are joined in series. The conductance of the combination is

• A. $1$ unit
• B. $5$ units
• C. $2.4$ units
• D. $1.2$ units

Answer 1

The correct option is D $1.2$ units

Let the conductance of the combined unit be $\frac{kA}{L}$
The conductance of both the individual units is $\frac{k_{1}A}{L_1}$ and $\frac{k_{2}A}{L_2}$.
As they are connected in series, thus
$\frac{L}{kA}=\frac{L_1}{k_{1}A}+\frac{L_2}{k_{2}A}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$
or
$\frac{2l}{kA}=\frac{l}{k_{1}A}+\frac{l}{k_{2}A}=\frac{5}{6}$ (let $L_1=L_2=l$)

Thus we get: $\frac{kA}{2l}=\frac{6}{5}=1.2units$