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?

Answer 1

Density of metal =ρ_m = 3400 kgm^-3
Density of mercury = p_Hg = 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.