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Question
What fraction of metal of density 3400 kgm-3 will be above the surface of mercury of density 13600 kgm-3, while floating in mercury?
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Solution
Density of metal =ρm = 3400 kgm-3
Density of mercury = pHg = 13600 kgm-3
Let volume of metal = x
and volume of metal inside mercury = y
By law of floatation:
Weight of mercury displaced by metal = wt. of metal
`"V"_"Hg" = rho_"Hg" xx "g" = "V"_"metal" xx rho_"metal" xx "g"`
`"y" xx 13600 = "x" xx 3400`
`"y" = 3400/13600 "x" = 1/4"x"`
Volume of metal inside mercury = Volume of mercury displaced
`"y" = 1/4 xx "Volume of metal"`
Volume of metal above the surface of mercury = x - y
`= "x" - 1/4 "x" = (1 - 1/4)"x" = (4 - 1)/"x" "x" = 3/4 "x"`
Fraction of metal above surface of mercury
`= 3/4 xx 1/"x" = 3/4`
`=> 3/4"th"` part of metal lies above the surface of mercury.
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