Question

The graph shown gives the temperature along $\text{x-}$axis that extends directly through a wall consisting of three layers $\text{A, B}$ and $\text{C}$. The air temperature on one side of the wall differ from that on the other side. Thermal conduction through the wall is steady. Out of the three layers $\text{A, B}$ and $\text{C}$ thermal conductivity is greatest for layer

• A. $\text{A}$
• B. $\text{B}$
• C. $\text{C}$
• D. same for all

Answer 1

The correct option is A $\text{A}$

Applying law of conduction for $\text{A}$:
Heat flow rate through the wall,
$H=\frac{Q}{t}=k_A\left(\frac{\Delta T_A}{l_A}\right)$
where,
$k_A$ is conductivity of the wall,
$\Delta T$ is the temperature difference between two surface of the wall.

Substituting the values, from figure
$\frac{Q}{t}=k_A\left[\frac{150-140}{40}\right]=\frac{k_A}{4}$

$\Rightarrow k_A=4\left(\frac{Q}{t}\right)=4H...\left(1\right)$

Similarly for $\text{B}$:
$\frac{Q}{t}=k_B\left[\frac{140-120}{60-40}\right]=k_B$

$\Rightarrow k_B=\left(\frac{Q}{t}\right)=H...\left(2\right)$

For $\text{C}$:

$\frac{Q}{t}=k_C\left[\frac{120-80}{70-60}\right]=4k_C$

$k_C=\frac{1}{4}\left(\frac{Q}{t}\right)=\frac{H}{4}...\left(3\right)$

For steady flow $\frac{Q}{t}$ is same for all the layers, so

from $1,2$ and $3$, we have
$k_A>k_B>k_C$

Hence, option $\left(A\right)$ is correct.