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

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