the weight of a solid is 80g in air and 50g in liquid of relative density 1.5. calculate the relative density of the solid.
2 answers
W(air) - W(liq.) = R.D.*VOL. 80-50=1.5 * V => 30/1.5=V =20cm3 . R.D.(solid) = 50/20 = 2.5
Loss of weight if solid in liquid = 50g- 30 g = 20g
Weight of equal volune of liquid by archimedes principle = 20 g
If the water displaced has a volume of V cm³ , then its weight= V×1.5 = 30 g
=》V= 20 cm³
Since , solid is completely immersed in the liquid , so volume of same solid = V
Therefore , The volume of water will be displaced by same solid = 20 cm³
Therefore , weight this displaced water =20 g
So we have specific gravity of solid = weight of solid in air / weight of water displaced = 80 g/ 20 g = 4
Hence , the relative density of the solid is 4.
Weight of equal volune of liquid by archimedes principle = 20 g
If the water displaced has a volume of V cm³ , then its weight= V×1.5 = 30 g
=》V= 20 cm³
Since , solid is completely immersed in the liquid , so volume of same solid = V
Therefore , The volume of water will be displaced by same solid = 20 cm³
Therefore , weight this displaced water =20 g
So we have specific gravity of solid = weight of solid in air / weight of water displaced = 80 g/ 20 g = 4
Hence , the relative density of the solid is 4.