A body weighs 40N in air,36.4N in a liquid and 36N in water.calculate the relative density of the liquid

Expert Answers
electreto05 eNotes educator| Certified Educator

The relative density (ρr) is the relationship between the density of a substance (ρs), with another density which is taken as reference. For liquids and solids, it is taken as reference the density of water (ρH2O).

ρr = ρ(s)/ρ(H2O)

Another thing is that the weight (W) of a body in a liquid, is the difference between its weight in air and the buoyancy force exerted by the liquid on the body.

W = W(air) – ρgV

ρgV   →   Buoyant force of the liquid. 


ρ   →   Density of the liquid.

g   →   Acceleration of gravity.

V   →  Volume of the body.

So, according to the data, we can write:

W(liq) = W(air) – ρ(liq) gV   →   ρ(liq) gV = W(air) – W(liq) = 40 – 36.4 = 3.6 N

ρ(liq) = 3.6/gV

W(H2O) = W(air) – ρ(H2O) gV   →   ρ(H2O) gV = W(air) – W(H2O) = 40 – 36 = 4 N

ρ(H2O) = 4/gV

Applying equation (1), we have:

ρr = ρ(liq)/ρ(H2O) = (3.6/gV)/(4/gV) = 3.6/4

ρr = 0.9

The relative density of the liquid is ρr = 0.9

gsenviro eNotes educator| Certified Educator

Relative density of the liquid is its density relative to the water and it can be calculated as a ratio of weight loss of body in liquid to weight loss of body in water. 

This can also be written as,

relative density of liquid = `(m_1-m_2)/(m_1-m_3)`

where, m1: weight of body in air

m2: weight of body in liquid and m3 is weight of body in water

hence, relative density of liquid = (40-36.4)/(40-36) = 3.6/4 = 0.9