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(B) The thermal resistance $R$ of a layer is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the are

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$\frac{150}{7}\ ^oC$

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(B) The thermal resistance $R$ of a layer is given by $R = \frac{L}{KA}$,where $L$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area.For the first layer (thin layer): $R_1 = \frac{1}{K \cdot A}$.For the second layer: $R_2 = \frac{4}{3K \cdot A}$.Since the layers are in series,the heat current $H$ is the same through both layers. The temperature difference $\Delta T$ across a layer is given by $\Delta T = H \cdot R$.Thus,the ratio of temperature differences is $\frac{\Delta T_1}{\Delta T_2} = \frac{R_1}{R_2} = \frac{1/K}{4/3K} = \frac{3}{4}$.Given $\Delta T_1 + \Delta T_2 = 50\ ^oC$.Substituting $\Delta T_2 = \frac{4}{3} \Delta T_1$ into the equation: $\Delta T_1 + \frac{4}{3} \Delta T_1 = 50$.$\frac{7}{3} \Delta T_1 = 50 \implies \Delta T_1 = \frac{150}{7}\ ^oC$.

$A$ wall is made of two layers of thickness $1\ cm$ and $4\ cm$ and thermal conductivities $K$ and $3K$ respectively. If the temperature difference across the wall is $50\ ^oC$,then the temperature difference across the thin layer will be

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