Question

Two sheets of thickness $d$ and $3d$, are in contact with each other. The temperature just outside the thinner sheet is $T_1$ and thicker sheet is $T_3$ as shown in figure. The temperature $T_1,T_2$ and $T_3$ are in arithmetic progression such that $T_1>T_3$. The ratio of thermal conductivity of thinner sheet to thicker sheet is

• A. $1:3$
• B. $3:1$
• C. $2:3$
• D. $2:9$

Answer 1

The correct option is A $1:3$


Heat flow rate will be same across both the sheets.
$H_1=H_2$
$\therefore \frac{T_1-T_2}{\left(\frac{d}{K_{1}A}\right)}=\frac{T_2-T_3}{\left(\frac{3d}{K_{2}A}\right)}$
$\therefore \frac{K_1}{K_2}=\left(\frac{T_2-T_3}{T_1-T_2}\right).\frac{1}{3}$
But $T_1,T_2$ and $T_3$ are in $\text{A.P}$
$\therefore T_2-T_3=T_1-T_2$
$\therefore \frac{K_1}{K_2}=\frac{1}{3}$