a piece of iron weighs 44.5gf in air, 39.5gf in water 40.3gf in parrafin oil. find density of iron density of paraffin oil.
a solid of density 5000kgm3 weighs 0.5kgf in air. if it floats in a liquid of density 8000kgm3, what is its apparent weight?
Let the volume of iron be V.
Its weight in air is = 44.5 gf = 0.445g N
Its weight in water is = 39.5 gf = 0.395g N
Its weight in paraffin is 40.3 gf = 0.403g N
If the volume of the iron piece is V and its density is d then,
Vdg = 0.445 g
=> Vd = 0.445 ……….(1)
We know density of water is 1000 kg/m3, so,
V(1000)g = 0.445g - 0.395g
=> 1000V = 0.05
=> V = 0.00005 …….(2)
(1) and (2) => d = 0.445/0.00005 = 8900 kg/m3
This is the density of iron.
If q is the density of paraffin then,
Vqg = 0.445g - 0.403g
=> 0.00005q = 0.042
=> q = 840 kg/m3
This is the density of paraffin..
Density of the solid is = 5000 kg/m3
Its weight in air is, W = 0.5 kgf = 0.5g N
The buoyant force (B) by the liquid is equal to its weight since it floats in the liquid.
B = 0.5g N
Thus, the apparent weight in the liquid is = W – B = 0
When a body floats, its apparent weight is zero.