The efficiency of carnot engine is 50% and temperature of sink is 500K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of the sink will be : -

Options

A. `600 K`
B. `500 K`
C. `400 K`
D. `100 K`

Answer

To solve the problem, we will use the formula for the efficiency of a Carnot engine, which is given by:

\[ \eta = 1 - \frac{T_2}{T_1} \]

where:
- \(\eta\) is the efficiency,
- \(T_1\) is the temperature of the source,
- \(T_2\) is the temperature of the sink.

### Step 1: Find the temperature of the source \(T_1\) using the initial efficiency

Given:
- Initial efficiency \(\eta = 50\% = \frac{50}{100} = 0.5\)
- Temperature of the sink \(T_2 = 500 \, K\)

Using the efficiency formula:

\[ 0.5 = 1 - \frac{500}{T_1} \]

Rearranging gives:

\[ \frac{500}{T_1} = 1 - 0.5 = 0.5 \]

Now, solving for \(T_1\):

\[ T_1 = \frac{500}{0.5} = 1000 \, K \]

### Step 2: Find the new temperature of the sink \(T_2\) for the new efficiency

Now, we need to find the new temperature of the sink when the efficiency is raised to 60%.

Given:
- New efficiency \(\eta = 60\% = \frac{60}{100} = 0.6\)
- Temperature of the source \(T_1 = 1000 \, K\)

Using the efficiency formula again:

\[ 0.6 = 1 - \frac{T_2}{1000} \]

Rearranging gives:

\[ \frac{T_2}{1000} = 1 - 0.6 = 0.4 \]

Now, solving for \(T_2\):

\[ T_2 = 0.4 \times 1000 = 400 \, K \]

### Conclusion

The required temperature of the sink when the efficiency is raised to 60% is:

\[ \boxed{400 \, K} \]