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.

Explanation

Here we first caculate the volume of the metal piece at `27^(@)C` as well a the volume of metal piece at `42^(@)C`. If is given that weight of the piece of metal in air =46 gm
Weight of the piece of metal in liquid at `27^(@)C=30gm`
Thus loss of the weight of the piece of metal in liquid `=46-30`
`=16m=` weight of liquid displaced
Thus the volume of metal piece at `27^(@)C` is
`=("weight of liquid displaced")/("density")=16/1.24 cm^(3)`
`V_(1)^(')=16/1.24 cm^(3)`
Similarly, the volume of the peice of metal aet `42^(@)C` cna be calculated in the following way:
Weight of the piece of the metal in air =46 gm
Weight of the piece of metal in liquid at `42^(@)C=30.5 gm`
Loss of the weight of the piece of metal `=46-30.5=15.5gm`
Weight of the volume displaced of liquid at `42^(@)C=15.5gm`
Now the volume of the liquid displaced
`=("Weight of the liquid displaced")/("density")`
`=15.5/1.20 cm^(3)`
Thus the volume of piece of metal at `42^(@)C`
`V_(2)=15.5/1.20 cm^(3)`
Applying the formula `V_(2)=V_(1)(1+gamma DeltaT)`, we have
`15.5/1.20=16/1.24 (1+gamma.15)` [As `DeltaT=42-27=15`]
`implies 1+15gamma=15.5/1.20x1.24/16`
or `gamma=1/15[15.5/1.20xx1.24/16 -1]=[1/15xx1/960]`
`impliesalpha=1/(3xx15xx960)=2.3015xx10^(-5).^(@)C^(-1)`