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Question
A cylinder made of copper and aluminium floats in mercury of density 13.6 gem-3, such that 0.26th part of it is below mercury. Find the density of solid.
Solution
Density of mercury = ρHg =13.6 g.cm-3
Density of solid cylinder = ρsolid = ?
0.26th part of the cylinder is below mercury
Let Vsolid = Volume of solid cylinder
Volume of mercury displaced by immersed part of the solid cylinder
`= "V"_"Hg" = 0.26 "V"_"solid"`
By law of floatation:
Weight of the solid cylinder = Weight of mercury displaced by immersed part of solid cylinder
`"V"_"solid" xx rho_"solid" xx "g" = "V"_"Hg" xx rho_"Hg" xx "g"`
`"V"_"solid" xx rho_"solid" = 0.26 "V"_"solid" xx 13.6`
`rho_"solid" = 0.26 × 13.6`
`rho_"solid" = 3.536` g cm-3
So, density of solid = 3.536 g cm-3
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