One end of a copper rod of uniform cross section and length 1.5 m is kept in contact with ice and the other end with water at `100^@C`. At what point along its length should a temperature of `200^@C` be maintained so that in the steady state, the mass of ice melting be equal to that of the steam produced in same interval of time. Assume that the whole system is insulated from surroundings:
`[L_("ice")=80 cal//g,L_("steam")=540 cal//g]`

Options

A. 8.59 cm from ice and
B. 10.34 cm from water end
C. 10.34 cm from ice end
D. 8.76 cm from water end

Explanation

To solve the problem, we need to find the point along the copper rod where a temperature of 200°C should be maintained so that the mass of ice melting is equal to the mass of steam produced in the same interval of time.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- One end of the copper rod is at 0°C (in contact with ice) and the other end is at 100°C (in contact with water).
- The length of the rod is 1.5 m.

2. **Defining Variables**:
- Let \( x \) be the distance from the ice end to the point where the temperature is 200°C.
- The remaining length of the rod from the 200°C point to the water end will be \( 1.5 - x \).

3. **Heat Flow Calculation**:
- The rate of heat flow from the 200°C point to the ice (0°C) is given by:
\[ I_1 = \frac{kA(200 - 0)}{x} = \frac{200kA}{x} \]
- The rate of heat flow from the 200°C point to the water (100°C) is given by:
\[ I_2 = \frac{kA(200 - 100)}{1.5 - x} = \frac{100kA}{1.5 - x} \]

4. **Using Latent Heat**:
- The heat absorbed by the melting ice (mass \( dm \)) is:
\[ I_1 = \frac{dm}{dt} \cdot L_{\text{ice}} = \frac{dm}{dt} \cdot 80 \text{ cal/g} \]
- The heat released by the steam (mass \( dm \)) is:
\[ I_2 = \frac{dm}{dt} \cdot L_{\text{steam}} = \frac{dm}{dt} \cdot 540 \text{ cal/g} \]

5. **Equating Heat Flows**:
- Since the mass of ice melting is equal to the mass of steam produced, we can set \( \frac{dm}{dt} \) from both sides equal:
\[ \frac{200kA}{x} = \frac{100kA}{1.5 - x} \cdot \frac{80}{540} \]

6. **Simplifying the Equation**:
- Cancel \( kA \) from both sides:
\[ \frac{200}{x} = \frac{100}{1.5 - x} \cdot \frac{8}{54} \]
- Rearranging gives:
\[ \frac{200}{x} = \frac{800}{54(1.5 - x)} \]

7. **Cross-Multiplying**:
- Cross-multiplying yields:
\[ 200 \cdot 54(1.5 - x) = 800x \]
- Simplifying gives:
\[ 10800 - 10800x = 800x \]
- Combine like terms:
\[ 10800 = 10800x + 800x \]
\[ 10800 = 11600x \]

8. **Solving for \( x \)**:
- Thus:
\[ x = \frac{10800}{11600} = \frac{81}{58} \approx 1.396 \text{ m} \]

9. **Finding the Distance from the Water End**:
- The distance from the water end is:
\[ 1.5 - x = 1.5 - 1.396 = 0.104 \text{ m} = 10.4 \text{ cm} \]

### Final Answer:
The point along the length of the copper rod where a temperature of 200°C should be maintained is approximately **10.4 cm from the water end**.