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 ?

The electrical resistance in ohms of a certain thermometer varies with

1.	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 ?
Answer» SOLUTION :`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`