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=16g`
Loss of weight of body at `42^(@)C=46-30.5=15.5g`
If V is the volume of body at `27^(@)C`,
volume at `42^(@)C(V^(1))=V(1+gamma_(s)t)`
`V^(1)=V(1+gamma_(s)(15))` where `gamma_(s)` is the volume coefficient of expansion of solid.
Since loss of weight of body = weight of the liquid
displaced = `Vd_(1)g`
We have 16 = Vd {d = density at `27^(@)C`}
15.5 = `V^(1)d^(1)" "{d^(1)="density at "42^(@)C}`
`16/15.5=V/V^(1)*d/d^(1)or32/31=V/V^(1)*d/d^(1)`
or `32/31=1/(1+15gamma_(s))xx31/30`
`rArr1+15gamma_(s)=31/30xx31/32=961/960`
`15gamma_(s)=961/960-1=1/960`
`gamma_(s)=1/960xx1/15=3alpha_(s)`, where `alpha_(s)` is the linear coefficient of expansion of solid.
`alpha_(s)=1/(960xx45)=0.000024=2.4xx10^(-5)//K`