A current-carrying copper conductor generates heat at rate 35 W at `20^(@)C`. If the potential difference applied across the conductor remains constant, the rate at which heat is liberated in the conductor at `65^(@)C` is ___________W. (Temperature coefficient of resistivity of copper `= 0.004//^(@)C`) (Neglect the effects of thermal expansion of copper)

Explanation

To solve the problem, we need to determine the rate of heat liberated in a copper conductor at a higher temperature while keeping the potential difference constant. The key points to consider are the relationship between power, resistance, and temperature.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Power at \( T_1 = 20^\circ C \) (denoted as \( P_1 \)) = 35 W
- Temperature coefficient of resistivity (\( \alpha \)) = 0.004 \( ^\circ C^{-1} \)
- Initial temperature (\( T_1 \)) = 20 \( ^\circ C \)
- Final temperature (\( T_2 \)) = 65 \( ^\circ C \)

2. **Calculate the Resistance at \( T_1 \):**
- The resistance at \( T_1 \) can be expressed as:
\[ R_1 = R_0 \left(1 + \alpha (T_1 - 0)\right) = R_0 \left(1 + 0.004 \times 20\right) \]
- Simplifying gives:
\[ R_1 = R_0 (1 + 0.08) = 1.08 R_0 \]

3. **Calculate the Resistance at \( T_2 \):**
- The resistance at \( T_2 \) can be expressed as:
\[ R_2 = R_0 \left(1 + \alpha (T_2 - 0)\right) = R_0 \left(1 + 0.004 \times 65\right) \]
- Simplifying gives:
\[ R_2 = R_0 (1 + 0.26) = 1.26 R_0 \]

4. **Relate Power to Resistance:**
- Since the potential difference \( V \) is constant, the power is inversely proportional to resistance:
\[ P \propto \frac{1}{R} \]
- Therefore, we can write:
\[ \frac{P_1}{P_2} = \frac{R_2}{R_1} \]

5. **Substituting the Resistance Values:**
- Substituting \( R_1 \) and \( R_2 \):
\[ \frac{35}{P_2} = \frac{1.26 R_0}{1.08 R_0} \]
- The \( R_0 \) cancels out:
\[ \frac{35}{P_2} = \frac{1.26}{1.08} \]

6. **Calculate \( P_2 \):**
- Rearranging gives:
\[ P_2 = 35 \times \frac{1.08}{1.26} \]
- Performing the calculation:
\[ P_2 = 35 \times 0.8571 \approx 30 W \]

### Final Answer:
The rate at which heat is liberated in the conductor at \( 65^\circ C \) is approximately **30 W**.