a block of wood floats in brine solution of density 1.20g cm^-3, such that 3/8 of its value is above brine. calculate the density of wood
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According to Archimedes' principle.
Mass of volume displaced is by object is buoyancy force.
Let the volume of wood is V.
then the volume submerged in liquid is V/4
let the density of wood be ρ.
let the density of liquid be ρl=0.8cm−3.
weight of wood ρVg.
buoyancy force =4ρlVg
now weight of block is equal to buoyancy force.
ρVg=4ρlVg
So ρ=40.8=.2gcm−3
then the volume submerged in oil is 60% of is 3V/5
weight of wood ρVg
buoyancy force =5ρoil3Vg
now weight of block is equal to buoyancy force.
ρVg=5ρoil3Vg
So ρoil=3ρ×5=30.2×5=0.33gcm−3
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