A cannot engine has efficiency `(1)/(6)`. If temperature of sink is decreased by `62^(@)C` then its efficiency becomes `(1)/(3)` then the temperature of source and sink:

Options

A. `T_(H)=98^(@)C,T_(C)=36^(@)C`
B. `T_(H)=99^(@)C,T_(C)=37^(@)C`
C. `T_(H)=100^(@)C,T_(C)=38^(@)C`
D. `T_(H)=102^(@)C,T_(C)=40^(@)C`

Answer

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

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

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

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

1. From the first condition, we have:
\[ 1 - \frac{T_2}{T_1} = \frac{1}{6} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \]
Therefore, we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = \frac{5}{6} T_1 \quad \text{(Equation 1)} \]

2. From the second condition, when the temperature of the sink is decreased by \( 62^\circ C \), the efficiency becomes \( \frac{1}{3} \):
\[ 1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \]
Rearranging gives:
\[ \frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
Therefore, we can express \( T_2 - 62 \) in terms of \( T_1 \):
\[ T_2 - 62 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

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

Now we will substitute \( T_2 \) from Equation 1 into Equation 2:
\[ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 \]

### Step 3: Solve for \( T_1 \)

To eliminate the fractions, we can multiply the entire equation by 6:
\[ 5 T_1 - 372 = 4 T_1 \]
Now, rearranging gives:
\[ 5 T_1 - 4 T_1 = 372 \]
\[ T_1 = 372 \, \text{K} \]

### Step 4: Find \( T_2 \)

Now, we can find \( T_2 \) using Equation 1:
\[ T_2 = \frac{5}{6} T_1 = \frac{5}{6} \times 372 = 310 \, \text{K} \]

### Final Answer

Thus, the temperatures of the source and sink are:
- Temperature of the source \( T_1 = 372 \, \text{K} \)
- Temperature of the sink \( T_2 = 310 \, \text{K} \)