A piece of metal weighs 46 g in air and 30 g in liquid of density `1.24 xx 10^3 kg m^(-3)` kept at `27^0C`. When the temperature of the liquid is raised to `42^0C`, the metal piece weights 30.5 g . The density of the liquid at `42^0 C` is `1.20 xx 10^3 kg m^(-3)`. Calculate the coefficient of linear expansion of the metal.

Explanation

Loss of weight of a body at `27^(@)C = 46 - 30 =16 g`
Loss of weight of body at `42^(@)C = 46 - 30.5 = 15.5 g `
If `V _(27)` is the bolume of body at `27^(@)c,,` volume at `42^(@)C = V _(27) (1 + gamma _(s) t ) = V _(27) (1 + 15 gamma _(s)) = V _(42),`
where `gamma _(s)` is the volume coefficient of expansion of solid.
SInce loss of weight of body = weight of the liquid displaced `= V d _(1) g,`
We have ` 16 = V _(27) d _(27) implies 15.5 = V _(42) d _(42)`
`therefore (16)/(15.5) = (V _(27))/( V _(42)) . (d _(27))/(d _(42))(or) (32)/(31) = ( V _(27))/( V _(42)) . ( d _(27))/( d _(42)) (or) (32)/(31) = (V _(27))/( V _(27) {1 + 16 gamma _(s) })xx (124)/(120) (or) (32)/(31) = (1)/(1 + 15 gamma_(s)) xx (31)/(120)`
`therefore 1 + 15 gamma _(s) = (31)/(30) xx (31)/(32) = (961)/(960) therefore 15 gamma _(s) = (961)/(960) -1 = (1)/( 960) therefore gamma _(s) = (1)/(960) xx (1)/(15) = 3 alpha _(s),`
where `alpha _(s)` is the linear coefficient of expansion of solid .
`therefore alpha _(s) = (1)/( 960 xx 45) = 0.00024 = 2.4 cc xx 10 ^(5) //K`