A small amount of mercury ‘Hg’ is taken in a container such that its expansion is equal to expansion of container made of glass then $γ_{H g} = \underline{x} γ_{g}$ . Where $x =$
$(γ_{H g} i s c o e f f i c i e n t o f v o l u m e e x p a n s i o n o f m e r c u r y; γ_{g} c o e f f i c i e n t o f v o l u m e e x p a n s i o n o f g l a s s)$

Question

A small amount of mercury ‘Hg’ is taken in a container such that its expansion is equal to expansion of container made of glass then $γ_{H g} = \underline{x} γ_{g}$ . Where $x =$

$(γ_{H g} i s c o e f f i c i e n t o f v o l u m e e x p a n s i o n o f m e r c u r y; γ_{g} c o e f f i c i e n t o f v o l u m e e x p a n s i o n o f g l a s s)$

Infinity Learn · Accepted Answer

if γHg is coefficient of volume expansion of mercury;γg coefficient of volume expansion of glass then practically found relation is  γHg $$=7$$γg

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A small amount of mercury ‘Hg’ is taken in a container such that its expansion is equal to expansion of container made of glass then γHg=x¯ γg . Where x= (γHg is coefficient of volume expansion of mercury;γg coefficient of volume expansion of glass)

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