The electrical resistance of pure platinum increases linearly with increasing temperature over a small range of temperature . This property is used in a Platinum resistance thermometer. The relation between `R_(theta)` (Resistance at `theta` K) and `R_(0)` (Resistance at `theta_(0)K`) is given by `R_(theta)=R_(0)[1+a(theta-theta_(0))]`, where `alpha=` temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads `0^(@)C` when its resistance is `80Omega` and `100^(@)C` when its resistance is `90 Omega` find the temperature at which its resistance is `86Omega`.

Explanation

To solve the problem, we need to find the temperature at which the resistance of the platinum thermometer is 86 Ω. We will use the linear relationship between resistance and temperature given in the problem.

### Step-by-Step Solution:

1. **Identify Given Values:**
- At \(0^\circ C\) (or \(273.15 K\)), the resistance \(R_0 = 80 \, \Omega\).
- At \(100^\circ C\) (or \(373.15 K\)), the resistance \(R_{100} = 90 \, \Omega\).
- We need to find the temperature \(T\) when the resistance \(R_T = 86 \, \Omega\).
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