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 $12^{\circ} \mathrm{C}$ is maintained across the composite slab. If $k_1=\frac{k_2}{2}$, the temperature difference across slab $A$ is

• A

  $4^{\circ} \mathrm{C}$
• B

  $6^{\circ} \mathrm{C}$
• Correct answer
  C

  $8^{\circ} \mathrm{C}$
• D

  $10^{\circ} \mathrm{C}$

Solution

The given situation can be shown as
Rate of flow of heat will be equal in both the slabs $\begin{aligned} & \therefore & (12-x) K_1 & =K_2(x-0) \\ & & 12-x & =2 x \quad\left(\because K_1=\frac{K_2}{2}\right) \\ & & x & =4^{\circ} \mathrm{C} \end{aligned}$ The temperature difference across slab $\begin{aligned} A & =(12-x)=(12-4) \\ & =8^{\circ} \mathrm{C} \end{aligned}$

Hence, the correct answer is option: 'C'.