The wall of a chemical plant is modified with two additional layers having half the thermal conductivity of original wall material. The inner layer is of half the thickness and outer layer of same thickness as original wall. What will be the heat flow across the modified wall?

Correct option is D

​$$\begin{aligned}&\textbf{Total Thermal Resistance of Modified Wall:} \\&\text{The layers are in series, so the total thermal resistance } R_{\text{total}} \text{ is the sum of individual resistances:} \\&R_{\text{total}} = R_{\text{inner}} + R_{\text{middle}} + R_{\text{outer}} = \frac{L}{kA} + \frac{L}{kA} + \frac{2L}{kA} = \frac{4L}{kA} \\[1em]&\textbf{Heat Flow Calculation:} \\&\text{The heat flow } Q \text{ is given by Fourier's Law:} \\&Q = \frac{\Delta T}{R_{\text{total}}}, \\&\text{where } \Delta T \text{ is the temperature difference across the wall.} \\&\text{Substitute } R_{\text{total}}: \\&Q = \frac{\Delta T}{\frac{4L}{kA}} = \frac{kA \Delta T}{4L} \\[1em]&\textbf{Comparison with Original Wall:} \\&\text{For the original wall (single layer), the heat flow } Q_{\text{original}} \text{ was:} \\&Q_{\text{original}} = \frac{\Delta T}{\frac{L}{kA}} = \frac{kA \Delta T}{L} \\[1em]&\text{Thus, the ratio of the modified heat flow to the original heat flow is:} \\&\frac{Q}{Q_{\text{original}}} = \frac{\frac{kA \Delta T}{4L}}{\frac{kA \Delta T}{L}} = \frac{1}{4}\end{aligned}$$​​