A platinum resistance thermometer reads `0^@ C` when its resistance is `80 Omega and 100^@ C` when its resistance is `90 Omega`. Find the temperature at which the resistance is `86 Omega`.

Explanation

To solve the problem, we will use the formula that relates the resistance of a platinum resistance thermometer to temperature. The formula is:

\[ T = \frac{R_T - R_0}{R_{100} - R_0} \times 100 \]

Where:
- \(T\) is the temperature in degrees Celsius.
- \(R_T\) is the resistance at temperature \(T\).
- \(R_0\) is the resistance at \(0^\circ C\).
- \(R_{100}\) is the resistance at \(100^\circ C\).

### Step-by-Step Solution:

1. **Identify the given values**:
- \(R_0 = 80 \, \Omega\) (Resistance at \(0^\circ C\))
- \(R_{100} = 90 \, \Omega\) (Resistance at \(100^\circ C\))
- \(R_T = 86 \, \Omega\) (Resistance at unknown temperature \(T\))

2. **Substitute the values into the formula**:
\[ T = \frac{R_T - R_0}{R_{100} - R_0} \times 100 \]
Plugging in the values:
\[ T = \frac{86 - 80}{90 - 80} \times 100 \]

3. **Calculate the numerator**:
\[ 86 - 80 = 6 \]

4. **Calculate the denominator**:
\[ 90 - 80 = 10 \]

5. **Substitute these results back into the equation**:
\[ T = \frac{6}{10} \times 100 \]

6. **Perform the division**:
\[ \frac{6}{10} = 0.6 \]

7. **Multiply by 100**:
\[ T = 0.6 \times 100 = 60 \]

8. **Final Result**:
The temperature at which the resistance is \(86 \, \Omega\) is \(60^\circ C\).