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. 10.34 cm
B. `16.48 cm`
C. `21.68 cm`
D. `27.87 cm`

Answer

To solve the problem, we need to find the point along the copper rod where the temperature should be maintained at \(200^\circ C\) 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 Heat Transfer**:
- The heat flowing into the ice at \(0^\circ C\) can be expressed as:
\[ Q_{\text{ice}} = \frac{kA (T - T_{\text{ice}}) T}{x} \]
- The heat flowing into the steam at \(100^\circ C\) can be expressed as:
\[ Q_{\text{steam}} = \frac{kA (T - T_{\text{steam}}) T}{L - x} \]
where \(T\) is the temperature at the point we want to maintain at \(200^\circ C\), \(x\) is the distance from the ice end, and \(L\) is the total length of the rod (1.5 m).

2. **Setting Up the Mass Equalities**:
- Let \(m\) be the mass of ice melted and steam produced in time \(T\).
- The heat required to melt the ice is given by:
\[ m L_f = Q_{\text{ice}} = \frac{kA (200 - 0) T}{x} \]
- The heat required to produce steam is given by:
\[ m L_v = Q_{\text{steam}} = \frac{kA (200 - 100) T}{1.5 - x} \]
- Here, \(L_f = 80 \text{ cal/g}\) and \(L_v = 540 \text{ cal/g}\).

3. **Equating the Masses**:
- Since the mass of ice melting equals the mass of steam produced:
\[ \frac{kA (200 - 0) T}{x} \cdot \frac{1}{80} = \frac{kA (200 - 100) T}{1.5 - x} \cdot \frac{1}{540} \]
- Simplifying this, we can cancel out \(kA T\):
\[ \frac{200}{80x} = \frac{100}{540(1.5 - x)} \]

4. **Cross Multiplying**:
- Cross multiplying gives:
\[ 200 \cdot 540(1.5 - x) = 100 \cdot 80x \]
- Simplifying:
\[ 108000(1.5 - x) = 8000x \]

5. **Expanding and Rearranging**:
- Expanding the left side:
\[ 162000 - 108000x = 8000x \]
- Rearranging gives:
\[ 162000 = 108000x + 8000x \]
\[ 162000 = 116000x \]

6. **Solving for \(x\)**:
- Solving for \(x\):
\[ x = \frac{162000}{116000} \approx 1.396 m \]

7. **Finding the Distance from the Hot End**:
- The distance from the hot end (at \(100^\circ C\)) is:
\[ L - x = 1.5 - 1.396 \approx 0.104 m \approx 10.4 cm \]

### Final Answer:
The point along the length of the copper rod where the temperature should be maintained at \(200^\circ C\) is approximately \(10.4 cm\) from the end that is at \(100^\circ C\).