Heat Conduction Time Through Blocks in Series and Parallel Configurations

To find the time required to transport the same amount of heat in Configuration II, we can use the concept of thermal resistance.

Let each rectangular block have a length $L$ along the direction of heat flow (the $x$-axis) and a cross-sectional area $A$ perpendicular to the $x$-axis.

The thermal resistance of a conductor is given by: $R = \frac{L}{K A}$ where $K$ is the thermal conductivity, $L$ is the length along the heat flow, and $A$ is the cross-sectional area.

Configuration I (Series Connection):

The two blocks are connected in series along the $x$-axis.

• Thermal resistance of the first block: $R_{\kappa} = \frac{L}{\kappa A}$
• Thermal resistance of the second block: $R_{2\kappa} = \frac{L}{2\kappa A}$

The equivalent thermal resistance for Configuration I is: $R_{\text{I}} = R_{\kappa} + R_{2\kappa} = \frac{L}{\kappa A} + \frac{L}{2\kappa A} = \frac{3L}{2\kappa A}$

Configuration II (Parallel Connection):

The two blocks are arranged in parallel along the $x$-axis.

• The equivalent thermal resistance for Configuration II is given by: $\frac{1}{R_{\text{II}}} = \frac{1}{R_{\kappa}} + \frac{1}{R_{2\kappa}} = \frac{\kappa A}{L} + \frac{2\kappa A}{L} = \frac{3\kappa A}{L}$ $R_{\text{II}} = \frac{L}{3\kappa A}$

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Relation between Heat, Time, and Thermal Resistance:

The rate of heat flow is given by: $\frac{dQ}{dt} = \frac{\Delta T}{R}$

For a constant temperature difference $\Delta T$, the amount of heat $Q$ transported in time $t$ is: $Q = \left(\frac{\Delta T}{R}\right) t \implies t = \frac{Q \cdot R}{\Delta T}$

Since $Q$ and $\Delta T$ are the same in both configurations: $t \propto R$

Therefore, the ratio of the times taken is: $\frac{t_{\text{II}}}{t_{\text{I}}} = \frac{R_{\text{II}}}{R_{\text{I}}} = \frac{\frac{L}{3\kappa A}}{\frac{3L}{2\kappa A}} = \frac{2}{9}$

Given $t_{\text{I}} = 9\text{ s}$: $t_{\text{II}} = 9\text{ s} \times \frac{2}{9} = 2.0\text{ s}$

Thus, the correct option is (A).