Apiece of metal weighs 46 gin air. When it is im mersed in a liquid of specific gravity 1.24 " at " 27^(@)C, it weighs 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.

Apiece of metal weighs 46 gin air. When it is im mersed in a liquid of

1.	Apiece of metal weighs 46 gin air. When it is im mersed in a liquid of specific gravity 1.24 " at " 27^(@)C, it weighs 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 = 16 g ` Loss of weigth of body at `42^(@) C = 46 - 30.5= 15.5 ` g If V is the valume of body at `27^(@)` C , volume at `42^(@) C (V^(1)) = V (1 + gamma_(S) t )` `V^(1) = V (1 + gamma_(s)) ` where `gamma_(s)` is the volume coefficient of expansion of solid. Since loss of weigth of body = weight of the liquid displaced = `Vd_(j) g ` We have 16 = VD { d = DENSITY at `27^(0)` C } 15.5 = `V^(1) d^(1) {d^(1) = "density at " 42^(0) C }` `(16)/(15.5) = (V)/(V^(1)).(d)/(d^(1)) " or " (32)/(31) = (V)/(V^(1)) .(d)/(d^(1))` or `(32)/(31) = (1)/(1 + 15_(gamma_(s))) xx (31)/(30)` `rArr 1 + 15 gamma_(s) = (31)/(30) xx (31)/(32) = (961)/(960)` `15 gamma_(s) = (961)/(960) -1 = (1)/(960)` `gamma_(s) = (1)/(960) xx (1)/(15)= 3 alpha_(s) , " where " alpha_(s)` is the linear coefficient of expansion of solid. `alpha_(s) = (1)/(960 xx 45) = 0.000024 xx 10^(-5)`/R