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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
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Verified Answer
The correct answer is:
$8^{\circ} \mathrm{C}$
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
$\begin{aligned}
A & =(12-x)=(12-4) \\
& =8^{\circ} \mathrm{C}
\end{aligned}$
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
$\begin{aligned}
A & =(12-x)=(12-4) \\
& =8^{\circ} \mathrm{C}
\end{aligned}$
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