# A body weighs 0.5 kg in air, 0.3 kg in water and 0.2 kg in liquid. What is the relative density of the liquid?

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Hello!

A weight, by the definition, is the pressure force on a base.

When a body floats in a liquid (and doesn't touch a bottom), two forces acting on it: the gravity force downwards and the buoyant force upwards. Equilibrium means that these forces are equal in magnitude.

If we consider a liquid to be a base for a body, then by Newton's Third Law body's weight is equal in magnitude to a buoyant force, and therefore to the gravity force and to the body's weight in the air.

In another words, for a body not touching a bottom the weight is the same as in a liquid. It isn't our situation.

But if a body touches a bottom, the force of pressure on a bottom is considered its weight. And there are three forces acting on a body:

1) the gravity force downwards, which is the same as the body's weight in the air (`W_a`);
2) the buoyant force (`F_w` for water and `F_l` for another liquid) upwards;
3) the bottom reaction force upwards, it is the same in magnitude as the body's weight in water (`W_w` ) or in another liquid (`W_l` ).

Therefore `W_a=F_w+W_w` and `W_a=F_l+W_l.`

All three weights are given: `W_a=0.5kg,` `W_w=0.3kg` and `W_l=0.2kg.`

Also we need Archimedes' principle: `F_w=rho_w*V*g` and `F_l=rho_l*V*g` where `V` is the volume of a body (suppose it is immersed completely). `rho_w` is the density of water and `rho_l` is the density of liquid.

Then the relative (to the water's density) density of the liquid is

`rho_l/rho_w=F_l/F_w=(W_a-W_l)/(W_a-W_w)=(0.5-0.2)/(0.5-0.3)=(0.3)/(0.2)` =1.5  (this is the answer).

The liquid is 1.5 times more dense than water and pushes a body up stronger. Less weight is remained for the bottom.