Temperature of source is `327^(@)C`. Temperature of sink is changed in order to increase the efficiency of engine from `(1)/(5)` to `(1)/(4)`, by

Options

A. 30K
B. 303 K
C. 603 K
D. 60 K

Explanation

To solve the problem, we need to find the change in the temperature of the sink (T2) that increases the efficiency of the engine from \( \frac{1}{5} \) to \( \frac{1}{4} \). We will use the formula for the efficiency of a Carnot engine.

### Step-by-Step Solution:

1. **Identify the temperatures**:
- The temperature of the source (T1) is given as \( 327^\circ C \).
- Convert T1 to Kelvin:
\[ T1 = 327 + 273 = 600 \, K \]

2. **Set up the efficiency equations**:
- The efficiency of a Carnot engine is given by:
\[ \eta = 1 - \frac{T2}{T1} \]
- For the initial efficiency \( \eta_1 = \frac{1}{5} \):
\[ \frac{1}{5} = 1 - \frac{T2}{T1} \]
- Rearranging gives:
\[ \frac{T2}{T1} = 1 - \frac{1}{5} = \frac{4}{5} \]
- Therefore:
\[ T2 = \frac{4}{5} T1 = \frac{4}{5} \times 600 = 480 \, K \]

3. **For the new efficiency \( \eta_2 = \frac{1}{4} \)**:
- Set up the equation:
\[ \frac{1}{4} = 1 - \frac{T2'}{T1} \]
- Rearranging gives:
\[ \frac{T2'}{T1} = 1 - \frac{1}{4} = \frac{3}{4} \]
- Therefore:
\[ T2' = \frac{3}{4} T1 = \frac{3}{4} \times 600 = 450 \, K \]

4. **Calculate the change in temperature of the sink**:
- The change in temperature \( \Delta T \) is given by:
\[ \Delta T = T2 - T2' = 480 \, K - 450 \, K = 30 \, K \]

Thus, the change in the temperature of the sink is \( 30 \, K \).

### Final Answer:
The change in the temperature of the sink is \( 30 \, K \).