- Home
- Standard 11
- Physics
$A$ wall has two layers $A$ and $B$ made of different materials. The thickness of both the layers is the same. The thermal conductivity of $A$ and $B$ are $K_A$ and $K_B$ such that $K_A = 3K_B$. The temperature across the wall is $20°C$ . In thermal equilibrium
The temperature difference across $A = 15^\circ C$
The temperature difference across $A = 5^\circ C$
The temperature difference across $A$ is $10°C$
The rate of transfer of heat through $A$ is more than that through $B$ .
Solution
(b) In series rate of flow of heat is same
==> $\frac{{{K_A}A({\theta _1} – \theta )}}{l} = \frac{{{K_B}A(\theta – {\theta _2})}}{l}$
==> $3{K_B}({\theta _1} – \theta ) = {K_B}(\theta – {\theta _2})$
==> $3({\theta _1} – \theta ) = (\theta – {\theta _2})$
==> $3{\theta _1} – 3\theta = \theta – {\theta _2}$==> $4{\theta _1} – 4\theta = {\theta _1} – {\theta _2}$
==> $4({\theta _1} – \theta ) = ({\theta _1} – {\theta _2})$
==> $4({\theta _1} – \theta ) = 20$==> $({\theta _1} – \theta ) = 5^\circ C$
