Question

Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or II as shown in the figure. One of the blocks has thermal conductivity $K$ and the other, $2K$. The temperature difference between the ends along the $x$-axis is the same in both the configuarations. It takes $9\text{s}$ to transport a certain amount of heat from the hot end to the cold end in configuration I. The time to transport the same amount of heat in the configuration II is:-

• A. $2.0\text{s}$
• B. $3.0\text{s}$
• C. $4.5\text{s}$
• D. $6.0\text{s}$

Answer 1

The correct option is A $2.0\text{s}$

Let $R_1$ be the thermal resistance of block I and $R_2$ be the thermal resistance of block II
Configuration I: In this configuration, the slabs are in series combination.
Thus, the equivalent thermal resistance of the slabs can be written as:
$R_{eq}=R_1+R_2$
$R_{eq}=R_I=\frac{L}{KA}+\frac{L}{2KA}=\frac{3}{2}\frac{L}{KA}$
Configuration II: In this configuration, the slabs are in parallel combination.
Thus the equivalent thermal resistance can be written as:
$R_{eq}^{\prime }=R_{II}=\frac{R_1{R}_2}{R_1+R_2}$
$=\frac{\frac{L}{KA}\times \frac{L}{2KA}}{\frac{L}{KA}+\frac{L}{2KA}}=\frac{L}{3KA}$

$\because \frac{\Delta Q}{t}=\frac{\Delta T}{R}$
So, we can write, for equal $\Delta Q$ and $\Delta T$
$\frac{t_I}{t_{II}}=\frac{R_I}{R_{II}}=\frac{\frac{3L}{2KA}}{\frac{L}{3KA}}=\frac{9}{2}$
From the data given in the question,
$\frac{9}{t_{II}}=\frac{9}{2}$ $\Rightarrow t_{II}=2\text{s}$.
Thus, option (a) is the correct answer.