The electrical resistance in ohms of a certain thermometer varies with temperature ac cording to the approximate law: `R =R_(0)[1+alpha(T-T_(0))]`
The resistances is `101.6 Omega` at the triple-point of water `273.16K`, and `165.5 Omega` at the normal melting point of lead `(600.5K)`. What is the temperature when the resistance is `123.4 Omega`?

Explanation

It is given that :
`R=R_(0)[1+a(T-T_(0)]`…(i)
Where,
`R_(0)" and "T_(0)` are the initial resistance and temperature respectively
R and T are the final resistance and temperature respectively
α is a constant
At the triple point of water, `T_(0)` = 273.15 K
Resistance of lead, `R_(0) = 101.6 Omega`
At normal melting point of lead, T = 600.5 K
Resistance of lead, `R = 165.5 Omega`
Substituting these values in equation (i), we get:
`R=R_(0)[1+a(T-T_(0))]`
`165.5=101.6[1+a(600.5-273.15)]`
` 1.629=1+a(327.35)`
`:.a=(0.629)/(327.35)=1.92xx10^(-3)K^(-1)`
For resistance, `R_(1) = 123.4 Omega`
`R_(1)=R_(0)[1+a(T-T_(0))]`
Where, T is the temperature when the resis tan ce of lead is `123.4 Omega`
`123.4=101.6[1+1.92xx10^(-3)(T-273.15)]`
`1.214=1+1.92xx10^(-3)(T-273.15)`
`(0.214)/(1.92xx10^(-3))=T-273.15
`:.T=384.61 K`