A metal rod AB of length 10x has its one end A in ice at `0^@C`, and the other end B in water at `100^@C`. If a point P one the rod is maintained at `400^@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 `540cal//g` and latent heat of melting of ice is `80cal//g`. If the point P is at a distance of `lambdax` from the ice end A, find the value `lambda`. [Neglect any heat loss to the surrounding.]

Explanation

To solve the problem, we need to analyze the heat flow through the metal rod and set up equations based on the heat transfer principles. Let's break it down step by step.

### Step 1: Understanding the Setup
We have a metal rod AB of length \(10x\). One end A is at \(0^\circ C\) (ice) and the other end B is at \(100^\circ C\) (water). A point P on the rod is maintained at \(400^\circ C\). The distance from A to P is \(\lambda x\), and the distance from P to B is \(10x - \lambda x\).

### Step 2: Heat Transfer to Ice
The heat flow from point P to the ice (end A) can be expressed as:
\[ Q_1 = \frac{dm_1}{dt} \cdot L_f \]
where \(L_f\) is the latent heat of fusion of ice (80 cal/g), and \(\frac{dm_1}{dt}\) is the mass of ice melted per unit time.

Using Fourier's law of heat conduction, we have:
\[ Q_1 = kA \cdot \frac{400 - 0}{\lambda x} = \frac{400kA}{\lambda x} \]

### Step 3: Heat Transfer to Water
Similarly, the heat flow from point P to the water (end B) can be expressed as:
\[ Q_2 = \frac{dm_2}{dt} \cdot L_v \]
where \(L_v\) is the latent heat of vaporization of water (540 cal/g), and \(\frac{dm_2}{dt}\) is the mass of water evaporated per unit time.

Using Fourier's law again, we have:
\[ Q_2 = kA \cdot \frac{400 - 100}{(10x - \lambda x)} = kA \cdot \frac{300}{(10 - \lambda)x} \]

### Step 4: Setting Up the Equation
Since it is given that equal amounts of water and ice evaporate and melt per unit time, we can set \(Q_1 = Q_2\):
\[ \frac{dm_1}{dt} \cdot L_f = \frac{dm_2}{dt} \cdot L_v \]
Substituting the expressions for \(Q_1\) and \(Q_2\):
\[ \frac{400kA}{\lambda x} \cdot 80 = \frac{300kA}{(10 - \lambda)x} \cdot 540 \]

### Step 5: Simplifying the Equation
We can cancel \(kA\) and \(x\) from both sides:
\[ \frac{400 \cdot 80}{\lambda} = \frac{300 \cdot 540}{10 - \lambda} \]

### Step 6: Cross-Multiplying
Cross-multiplying gives:
\[ 400 \cdot 80 \cdot (10 - \lambda) = 300 \cdot 540 \cdot \lambda \]
This simplifies to:
\[ 32000(10 - \lambda) = 162000\lambda \]

### Step 7: Expanding and Rearranging
Expanding the left side:
\[ 320000 - 32000\lambda = 162000\lambda \]
Rearranging gives:
\[ 320000 = 162000\lambda + 32000\lambda \]
\[ 320000 = 194000\lambda \]

### Step 8: Solving for \(\lambda\)
Now, we can solve for \(\lambda\):
\[ \lambda = \frac{320000}{194000} \approx 1.649 \]

### Final Answer
Thus, the value of \(\lambda\) is approximately \(1.649\).