The efficiency of a Carnot cycle is `1//6`. If on reducing the temperature of the sink by `65^@C`, the efficiency becomes `1//3`, find the source and sink temperatures between which the cycle is working.

Explanation

To solve the problem, we need to find the source temperature (T1) and the sink temperature (T2) of a Carnot cycle given the efficiencies and the change in sink temperature.

### Step-by-Step Solution:

1. **Understanding the Efficiency of a Carnot Cycle**:
The efficiency (η) of a Carnot cycle is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the source and \(T_2\) is the temperature of the sink.

2. **Setting Up the First Equation**:
We know that the initial efficiency is \( \frac{1}{6} \):
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \]
Rearranging this gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
This leads us to our first equation:
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

3. **Setting Up the Second Equation**:
After reducing the sink temperature by \(65^\circ C\), the new efficiency becomes \( \frac{1}{3} \):
\[ \frac{1}{3} = 1 - \frac{T_2 - 65}{T_1} \]
Rearranging this gives:
\[ \frac{T_2 - 65}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
This leads us to our second equation:
\[ T_2 - 65 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
Substitute \(T_2\) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_1 - 65 = \frac{2}{3} T_1 \]

5. **Solving for T1**:
To eliminate the fractions, we can multiply the entire equation by 6:
\[ 5T_1 - 390 = 4T_1 \]
Rearranging gives:
\[ 5T_1 - 4T_1 = 390 \]
Thus:
\[ T_1 = 390 \text{ K} \]

6. **Finding T2**:
Now substitute \(T_1\) back into Equation 1 to find \(T_2\):
\[ T_2 = \frac{5}{6} \times 390 = 325 \text{ K} \]

7. **Converting Temperatures to Celsius**:
To convert Kelvin to Celsius:
\[ T_1 = 390 - 273 = 117^\circ C \]
\[ T_2 = 325 - 273 = 52^\circ C \]

### Final Answer:
- Source Temperature (T1) = 117°C
- Sink Temperature (T2) = 52°C