A piece of metal weighs 46 g in air and 30 g in lipuid 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 liqued at `42^0 C` is `1.20 xx 10^3 kg m^(-3)`. Calculate the coefficient of linear expandsion of the metal.

Answer

`alpha = (1)/(43200) //.^(@)C = 2.31 xx 10^(-5)//.^(@)C`

Explanation

`W_(0) = mg = 46 g wt`
at `theta_(1) = 27^(@)C`
`W_(1) = 30 gm wt = W_(0) - B_(1)`
`rArr B_(1) = (46 - 30) gm`
`rArr B_(1) = 16 gm wt = V_(1) rho_(1) g`
at `theta_(2) = 42^(@)C`
`W_(2) = 30.5 g wt = W_(0) - B_(2)`
`rArr B_(2) = 15.5 gm wt = V_(2) rho_(2)g`
`:. (B_(2))/(B_(1)) = (V_(2)rho_(2))/(V_(1)rho_(1))`
`(15.5)/(16) = (1 +3 alpha_(s) xx 15) xx (1.2)/(1.24)`
`alpha_(s) = [((15.5)/(16) xx (1.24)/(1.2))-1] xx (1)/(45)`
`alpha_(s) = 2.31 xx 10^(-5)//.^(@)C = (1)/(43200)//.^(@)C`