Answer to Question #85087 in Mechanics | Relativity for jason p

Question

An object weighing 303 N in air is immersed
in water after being tied to a string connected
to a balance. The scale now reads 272 N .
Immersed in oil, the object appears to weigh
279 N .
Find the density of the object.
Answer in units of kg/m
3
.

Answer 1

Weight of the object in the water P_1 equals to the difference of weight of the object in the air and buoyant force:

P_1=P_0-ρ_1 gV (1)

From (1) we can find the object’s volume:

V=(P_0-P_1)/(ρ_1 g) (2)

Weight of the object in the air:

P_0=ρgV (3)

Density of the object from (3):

ρ=P_0/gV (4)

Let’s substitute (2) into (4):

ρ=(P_0 ρ_1)/(P_0-P_1 )

ρ=(303∙1000)/(303-272)=9774 (kg/m^3)