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**Problem:**
Two rectangular blocks, having identical dimensions, can be arranged either in *configuration I* or in *configuration II* as shown in the figure. One of the blocks has thermal conductivity κ and the other 2κ. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the *configuration I*. The time to transport the same amount of heat in the *configuration II* is?
(IIT JEE 2013)

- 2 s
- 3 s
- 4.5 s
- 6.0 s

**Answer:** The answer is (A) i.e., 2 s.

**Solution:**
Let $T_h$ and $T_l$ be the temperatures of hot and cold ends and $T_m$ be the temperature at the middle point of configuration I. The rate of heat flow is equal in both the blocks of configuration I because the blocks are connected in series, i.e.,
\begin{align}
\frac{\Delta Q_{\text{I}}}{\Delta t_{\text{I}}}&=\kappa A \frac{(T_h-T_m)}{x}\nonumber \\
& =2\kappa A\frac{(T_m-T_l)}{x}.
\end{align}
Solve above equation to get $T_m={(T_h+2T_l)}/{3}$. Thus, above equation becomes
\begin{align}
&\frac{\Delta Q_{\text{I}}}{\Delta t_\text{I}}=2\kappa A\frac{(T_h-T_l)}{(3x)}.
\end{align}
In the configuration II, total rate of heat flow is the sum of heat flows through the two blocks because the blocks are joined in parallel i.e.,
\begin{align}
\frac{\Delta Q_{\text{II}}}{\Delta t_{\text{II}}}&=\kappa A \frac{(T_h-T_l)}{x}+2\kappa A\frac{(T_h-T_l)}{x}\nonumber\\
&=3\kappa A\frac{(T_h-T_l)}{x}.
\end{align}
Divide first equation by the second. Substitute $\Delta Q_{\text{I}}=\Delta Q_{\text{II}}$ and $\Delta t_{\text{I}}=9$ s to get $\Delta t_\text{II}=2$ s.

Alternately, the thermal resistance ($\frac{x}{\kappa A}$) of series combination in configuration I is $\frac{3}{2}\frac{x}{\kappa A}$ and that of parallel combination in configuration II is $R_\text{II}=\frac{1}{3}\frac{x}{\kappa A}$.

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