Archimedes' Problem: Iron Body Submerged in Water

[germangb]

i have the following problem:
-an iron (7800kg/m³) body is submerged into in water (density = 1000kg/m³)
his weight in the air is 0,13N.

what i did is:

E=dVg (Push force, density of the liquid, volume of the body sumerged and gravity, respectively)

then:
the volume of the iron body is 0,000633m³, then:

E = 1000*0,000633g = 1,60

so, the the apparent weight is 0,13-1,60= -1,47N
______________________________
did I do it right?

 

[supratim1]

how did you get volume of iron body as that? and how do you get 1000*0.000633*g = 1.60??

find volume correctly first. see that actual weight is 0.13N, therefore m = 0.13/g Kg. use density to find volume.

tell us what you get.

 

[Neandethal00]

i have the following problem:
-an iron (7800kg/m³) body is submerged into in water (density = 1000kg/m³)
his weight in the air is 0,13N.

what i did is:

E=dVg (Push force, density of the liquid, volume of the body sumerged and gravity, respectively)

then:
the volume of the iron body is 0,000633m³, then:

E = 1000*0,000633g = 1,60

so, the the apparent weight is 0,13-1,60= -1,47N
______________________________
did I do it right?

Use basic definition for density.
Volume(iron) = .13/(g*density)=1.699x10^-6 m3

Weight of iron in water = (weight in air) - dVg = .1133N

I'm open for correction.