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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 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, |
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Answer» Solution :Loss of weight of a body at `27^(@)C = 46 - 30 =16 g` Loss of weight of body at `42^(@)C = 46 - 30.5 = 15.5 g ` If `V _(27)` is the bolume of body at `27^(@)c,,` VOLUME at `42^(@)C = V _(27) (1 + gamma _(s) t ) = V _(27) (1 + 15 gamma _(s)) = V _(42),` where `gamma _(s)` is the volume coefficient of expansion of solid. SINCE loss of weight of body = weight of the liquid displaced `= V d _(1) g,` We have ` 16 = V _(27) d _(27) implies 15.5 = V _(42) d _(42)` `therefore (16)/(15.5) = (V _(27))/( V _(42)) . (d _(27))/(d _(42))(or) (32)/(31) = ( V _(27))/( V _(42)) . ( d _(27))/( d _(42)) (or) (32)/(31) = (V _(27))/( V _(27) {1 + 16 gamma _(s) })xx (124)/(120) (or) (32)/(31) = (1)/(1 + 15 gamma_(s)) xx (31)/(120)` `therefore 1 + 15 gamma _(s) = (31)/(30) xx (31)/(32) = (961)/(960) therefore 15 gamma _(s) = (961)/(960) -1 = (1)/( 960) therefore gamma _(s) = (1)/(960) xx (1)/(15) = 3 alpha _(s),` where `alpha _(s)` is the linear coefficient of expansion of solid . `therefore alpha _(s) = (1)/( 960 xx 45) = 0.00024 = 2.4 cc xx 10 ^(5) //K` |
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