Answer:
The density of the object [tex]\rho =[/tex] 8883 [tex]\frac{Kg}{m^{3} }[/tex]
Explanation:
Given data
[tex]F_{air}[/tex] = 302 N
[tex]F_{water}[/tex] = 268 N
[tex]F_{oil}[/tex] = 276 N
Buoyant force in water on the body is given by
[tex]F_B =[/tex] [tex]F_{air}[/tex] - [tex]F_{water}[/tex]
[tex]F_B =[/tex] 302 - 268
[tex]F_B =[/tex] 34 N
Volume of the object
[tex]V = \frac{F_B}{\rho g}[/tex]
[tex]V = \frac{34}{(1000)(9.81)}[/tex]
V = 3.465 × [tex]10^{-3}[/tex]
Mass of the object
[tex]m = \frac{F_{air} }{g}[/tex]
[tex]m = \frac{302}{9.81}[/tex]
m = 30.78 N
Now density of the body
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{30.78}{3.465 (10^{-3} )}[/tex]
[tex]\rho =[/tex] 8883 [tex]\frac{Kg}{m^{3} }[/tex]
Therefore the density of the object [tex]\rho =[/tex] 8883 [tex]\frac{Kg}{m^{3} }[/tex]
