Apiece of metal weight $$46$$ gm in air, when it is immersed in the liquid of specific gravity $$1.24$$ at $$27oC$$ it weighs $$30$$ gm. When the temperature of liquid is raised to $$42oC$$ the metal piece weight $$30.5$$ gm, specific gravity of the liquid at $$42oC$$ is $$1.20$$, then the linear expansion of the metal will be

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Loss of weight at $$27$$"C is          $$=46$$−$$30=16=V1$$×$$1.24$$ρl×g …………..(i)Loss$$ of weight at $$42$$"C is          $$=46$$−$$30.5=15.5=V2$$×$$1.2$$ρl×g ………..$$(ii)Now$$ dividing $$(i) by (ii), we get $$1615.5=V1V2$$×$$1.241.2But$$ $$V1V2=1+3$$α$$t2$$−$$t1=15.5$$×$$1.2416$$×$$1.2=1.001042$$⇒ $$3$$α$$42$$∘−$$27$$∘$$=0.001042$$⇒α$$=2.316$$×$$10$$−$$5/$$∘C

Apiece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27oC it weighs 30 gm. When the temperature of liquid is raised to 42oC the metal piece weight 30.5 gm, specific gravity of the liquid at 42oC is 1.20, then the linear expansion of the metal will be

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