A metal rod $AB$ of length $10x$ has its one end $A$ in ice at $0^\circ\text{C}$, and the other end $B$ in water at $100^\circ\text{C}$. If a point $P$ on the rod is maintained at $400^\circ\text{C}$, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540\,\text{cal/g}$ and latent heat of melting of ice is $80\,\text{cal/g}$. If the point $P$ is at a distance of $\lambda x$ from the ice end $A$, find the value of $\lambda$. [Neglect any heat loss to the surrounding.]

Step 1: Define the lengths of the rod segments.

The point $P$ divides the rod $AB$ of length $10x$ into two segments.

$$AP = \lambda x$$ $$PB = (10 - \lambda)x$$

Step 2: Calculate the temperature differences across each segment.

The temperature differences from point $P$ to the ends $A$ and $B$ drive the heat flow.

$$\Delta T_A = 400^\circ\text{C} - 0^\circ\text{C} = 400^\circ\text{C}$$ $$\Delta T_B = 400^\circ\text{C} - 100^\circ\text{C} = 300^\circ\text{C}$$

Step 3: Express the rate of heat flow through each segment.

Apply the thermal conduction formula for each segment, where $k$ is the thermal conductivity and $A$ is the cross-sectional area.

$$Q_A = \frac{400 kA}{\lambda x}$$ $$Q_B = \frac{300 kA}{(10 - \lambda)x}$$

Step 4: Equate the mass rates of phase change.

The rate of ice melting at $A$ equals the rate of water evaporating at $B$.

$$\frac{Q_A}{L_{\text{melt}}} = \frac{Q_B}{L_{\text{evap}}}$$ $$\frac{Q_A}{80} = \frac{Q_B}{540}$$

Step 5: Solve for the parameter $\lambda$.

Substitute the heat flow expressions into the mass rate equation and simplify.

$$\frac{400 kA}{80 \lambda x} = \frac{300 kA}{540 (10 - \lambda)x}$$ $$\frac{5}{\lambda} = \frac{5}{9(10 - \lambda)}$$ $$9(10 - \lambda) = \lambda$$ $$90 - 9\lambda = \lambda$$ $$10\lambda = 90$$ $$\lambda = 9$$

Answer: $\lambda = 9$