A piece of metal weighs 46 g in air. When it is immersed in a liquid of specific gravity 1.24 at 27^(@)C, it weight 30 g. When the temperature of the liquid is raised to 42^(@)C, the metal piece weighs 30.5 g. Specific gravity of liquid at 42^(@)C is 1.2. Calculate the coefficient of linear expansion of the metal.

A piece of metal weighs 46 g in air. When it is immersed in a liquid o

1.	A piece of metal weighs 46 g in air. When it is immersed in a liquid of specific gravity 1.24 at 27^(@)C, it weight 30 g. When the temperature of the liquid is raised to 42^(@)C, the metal piece weighs 30.5 g. Specific gravity of liquid at 42^(@)C is 1.2. Calculate the coefficient of linear expansion of the metal.
Answer» Solution :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`