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 will use the formula for the efficiency of a Carnot cycle, which is given by:

\[ \text{Efficiency} = 1 - \frac{T_2}{T_1} \]

where \( T_1 \) is the temperature of the heat source and \( T_2 \) is the temperature of the heat sink.

### Step 1: Set up the equations based on the given efficiencies

1. **Initial Efficiency**:
Given that the initial efficiency is \( \frac{1}{6} \):
\[ 1 - \frac{T_2}{T_1} = \frac{1}{6} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
Therefore, we can express \( T_1 \) in terms of \( T_2 \):
\[ T_1 = \frac{6}{5} T_2 \quad \text{(Equation 1)} \]

2. **New Efficiency**:
When the sink temperature is reduced by \( 65^\circ C \), the new efficiency becomes \( \frac{1}{3} \):
\[ 1 - \frac{T_2 - 65}{T_1} = \frac{1}{3} \]
Rearranging gives:
\[ \frac{T_2 - 65}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Therefore, we can express this as:
\[ T_2 - 65 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

### Step 2: Substitute Equation 1 into Equation 2

Now we will substitute \( T_1 \) from Equation 1 into Equation 2:
\[ T_2 - 65 = \frac{2}{3} \left(\frac{6}{5} T_2\right) \]
This simplifies to:
\[ T_2 - 65 = \frac{12}{15} T_2 \]
\[ T_2 - 65 = \frac{4}{5} T_2 \]

### Step 3: Solve for \( T_2 \)

Now, we will rearrange the equation to isolate \( T_2 \):
\[ T_2 - \frac{4}{5} T_2 = 65 \]
\[ \frac{1}{5} T_2 = 65 \]
Multiplying both sides by 5 gives:
\[ T_2 = 325 \, \text{K} \]

### Step 4: Find \( T_1 \)

Now that we have \( T_2 \), we can find \( T_1 \) using Equation 1:
\[ T_1 = \frac{6}{5} T_2 = \frac{6}{5} \times 325 = 390 \, \text{K} \]

### Final Answer

The source temperature \( T_1 \) is \( 390 \, \text{K} \) and the sink temperature \( T_2 \) is \( 325 \, \text{K} \).

---