Answer:
t the apparent weight of the body decreases when it is submerged in water,
F'_{e} = mg - ρ_liquid g V
Explanation:
When a container is placed on a scale it falls due to the weight of the body and the needle attached to the spring shows the value of the weight that is proportional to the displacement of the spring.
When the body is immersed in a container with water, it exerts a push on the body, so its apparent weight decreases and the compression of the scale is less, therefore it indicates a smaller weight of the body.
Let's write the equilibrium equation for the body is placed on the scale without water
[tex]F_{e}[/tex] - W = 0
F_{e} = W
Let us propose the equilibrium equation for this case
F'_{e} + B - W = 0
where F'_{e} is the elastic force, B the push of the liquid and W the weight of the body
F'_{e} = W -B
The thrust of the liquid is given by the Archimedean principle which says that it is equal to the weight of the dislodged liquid
B = ρ g V
F'_{e} = mg - ρ_liquid g V (1)
we use the density concept
ρ_body = m / V
we substitute
F'_{e}= g V (ρ_body - ρ_agua) (2)
F'_{e} < F_{e}
We see from expression is 1 and 2 that the apparent weight of the body decreases when it is submerged in water,
To know the specific value of the decrease, the weight of the body in the air or its density and volume must be known