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

For water the triple-point is `T_0 = 273.16 K`
`R_0=101.6 Omega ` Case(i): `T_0=(600.5 K), R_0 = 165.5 Omega `
As `R=R_0(1 + alpha(T-T_0))`
` therefore 165.5 = 101.6 [1 + alpha(600.5 - 273.16)]`
`( 165.5 - 101.6)/(101.6 xx (600.5 - 273.16) ) = alpha`
` (63.9)/(101.6 xx 327.34) = alpha`
`rArr alpha = 0.0019`
Case (ii) ` R = 123.4 Omega , T = ?`
` R = R_0 (1+ alpha (T - T_0) )`
` 123.4 = 101.6 [1+0.0019(T - 273.16) ]`
` (123.4 - 101.6)/(101.6 xx 0.0019 ) = T - 273.16`
` = 112.9 + 273.16 = T`
T = 386.06 K