Question

Two sheets of thickness $d$ and $3d$ , are touching each other. The temperature just outside the thinner sheet is $T_1$ and on the side of the thicker sheet is $T_3$. The interface temperature is $T_2,T_1,T_2,and{T}_3$ are in arithmetic progression. The ratio of thermal conductivity of thinner sheet to thicker sheet is

• A. 1 : 3
• B. 3 : 1
• C. 2 : 3
• D. 3 : 2

Answer 1

The correct option is A 1 : 3

Given : $L_A=d$ $L_B=3d$

As $T_2,$ $T_1,T_2$ & $T_3$ are in arithmetic progression.

$\therefore$ $2T_1=T_2+T_2$ $⟹T_1=T_2$

Similarly, $2T_2=T_1+T_3$

OR $2T_2=T_2+T_3$ $T_2=T_3$

Using: $\frac{d{Q}_A}{dt}=\frac{d{Q}_B}{dt}$

$\therefore$ $\frac{K_{A}A(T_1-T_2)}{L_A}=\frac{K_{B}A(T_2-T_3)}{L_B}$

OR $\frac{K_A}{d}=\frac{K_B}{3d}$

$⟹K_A:K_B=1:3$