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

Explanation

`R=R_(0)[1+alpha(T-T_(0))]`
`:.R-R_(0)=R_(0)xx5xx10^(-3)(T-T_(0))`
`:.T-T_(0)=(R-R_(0))/(5xx10^(-3)R_(0))`
At triple point of water :
`T_(1)-T_(0)=(R_(1)-R_(0))/(5xx10^(-3)R_(0))` . . .(1)
and at triple point of leat :
`T_(2)-T_(0)=(R_(2)-R_(0))/(5xx10^(-3)R_(0))` . . . (2)
By taking ratio of eauation (2) to equation (1),
`(T_(2)-T_(0))/(T_(1)-T_(0))=(R_(2)-R_(0))/(R_(1)-R_(0))`
Here `R_(0)=101.6Omega to` resistance at temp. of triple point
`T_(0)=273.16K`
`R_(1)=165.5Omega to` temp. `T_(1)=600.5K`
`R_(2)=123.4Omega to T_(2)=?`
`:.(T_(2)-273.16)/(600.5-273.16)=(123.4-101.6)/(165.5-101.6)`
`:.(T_(2)-273.16)/(327.34)=(21.8)/(63.9)` `:.T_(2)~~385K`