Question

Two slabs $A$ and $B$ of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of $A$ and $B$ are $K_1$ and $K_2$ respectively. A steady temperature difference of 12°C is maintained across the composite slab. If $K_1=\frac{K_2}{2}$ the temperature difference across slabs $A$ is

• A. $4^{\circ }C$
• B. $6^{\circ }C$
• C. $8^{\circ }C$
• D. $10^{\circ }C$

Answer 1

Answer: (C)

The given situation can be shown as

Rate of flow of heat will be equal in both the slabs
$\therefore (12-x)K_1=K_2(x-0)$
$12-x=2x\left(\because K_1=\frac{K_2}{2}\right)$
$\Rightarrow x=4^{\circ }C$
The temperature difference across slab
$A=(12-x)=(12-4)$
$=8^{\circ }C$