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

Explanation

To solve the problem step by step, 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,
- \(T_2\) is the temperature of the sink,
- \(T_1\) is the temperature of the source.

### Step 1: Identify the initial conditions
We are given:
- Initial efficiency \(\eta_1 = 50\% = \frac{1}{2}\)
- Temperature of the sink \(T_2 = 500 \, K\)

### Step 2: Calculate the temperature of the source \(T_1\) using the initial efficiency
Using the efficiency formula:

\[ \frac{1}{2} = 1 - \frac{500}{T_1} \]

Rearranging gives:

\[ \frac{500}{T_1} = 1 - \frac{1}{2} = \frac{1}{2} \]

Multiplying both sides by \(T_1\):

\[ 500 = \frac{1}{2} T_1 \]

Now, multiplying both sides by 2:

\[ T_1 = 1000 \, K \]

### Step 3: Set up the equation for the new efficiency
Now, we have the new efficiency \(\eta_2 = 60\% = \frac{6}{10}\). The temperature of the source remains constant at \(T_1 = 1000 \, K\).

Using the efficiency formula again:

\[ \frac{6}{10} = 1 - \frac{T_2}{1000} \]

### Step 4: Solve for the new sink temperature \(T_2\)
Rearranging the equation gives:

\[ \frac{T_2}{1000} = 1 - \frac{6}{10} = \frac{4}{10} = \frac{2}{5} \]

Now, multiplying both sides by 1000:

\[ T_2 = 1000 \times \frac{2}{5} = 400 \, K \]

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

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