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

Question

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

Infinity Learn · Accepted Answer

Loss of weight at $$27$$ºC $$is=$$ $$46$$ – $$30$$ $$=$$ $$16$$ $$=$$ $$V1$$ × $$1.24$$ l × g    …(i)Loss$$ of weight at $$42$$ºC $$is=$$ $$46$$ – $$30.5$$ $$=$$ $$15.5$$ $$=$$ $$V2$$ × $$1.2$$ l × g     …$$(ii)Now$$ dividing $$(i) by (ii), we get    But   $$=$$ $$1$$ $$+$$ $$3$$ (t2$$ – $$t1) $$=$$   $$=$$ $$1.001042$$ $$3$$ (42$$º –  $$27$$º$$) $$=$$ $$0.001042$$   $$=$$ $$2.316$$ × $$10$$–$$5/$$ºC

	Column - I		Column - II
a	$γ_{g} < γ_{R}$	p	liquid level rises continuously from beginning
b	$γ_{g} = γ_{R}$	q	liquid level falls continuously from beginning
c	$γ_{g} > γ_{R}$	r	liquid level remains same
d	$γ_{g} = 0$	s	liquid level first falls and then rises

A piece 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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