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

Question

Infinity Learn · Accepted Answer

Loss of weight at $$27$$°C $$is=$$ $$46$$ $$-$$ $$30$$ $$=$$ $$16$$ $$=$$ $$V1$$×$$1.24$$ρl×$$g-----(i)Loss$$ of weight at $$42$$°C $$is=$$ $$46$$ $$-$$ $$30.5$$ $$=$$ $$15.5$$ $$=$$ $$V2$$×$$1.2$$ρ$$1$$×$$g-----(ii)Now$$ dividing (i) by (ii), we get $$1615.5$$ $$=$$ $$V1V2$$×$$1.241.2$$ But $$V2V1$$ $$=$$ $$1+3$$α(t2-t1) $$=$$ $$15.5$$×$$1.2416$$×$$1.2$$ $$=$$ $$1.001042$$⇒$$3$$α(420-270) $$=$$ $$0.001042$$ ⇒α $$=$$ $$2.316$$×$$10-5/0C$$

	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 27°C it weighs 30 gm. When the temperature of liquid is raised to 42°C the metal piece weight 30.5 gm, specific gravity of the liquid at 42°C is 1.20, then the linear expansion of the metal will be

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