an object is solid throughout, when in ethyl alcohol, its weight is 15.2N, when in water weight is 13.7N?

Question

what is the volume of the object

Answer 1

let wight actual be mg and volume v

upthrust = v *density (medium) g (9.8 m/s^2)

equation is 2 cases

mg - v d(alco) g = 15.2
mg - v d(water) g = 13.7
subtract
v g [d(w) - d(alco)] =1.5

v = 1.5 / 9.8 [1000 - d(alcohol)]
v = 0.1530/ [1000 - d(alcohol)] meter^3

put the density of alcohol in MKS
-----------------------------------
internet: density of methyl alcohol = 0.789 gr/cm^3
d (alco) = 0.789* 10^-3 /(10^-2)^3 kg/m^3
d (alco) = 0.789 *1000 = 789 kg/m^3
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v = 0.1530/ [1000 - 789] meter^3
v = 0.1530/ 211 meter^3
v = 7.25* 10^-4 m^3

Answer 2

object actual weight = w
the OSTENSIBILITY weight of object in ethyl alcohol = w' = 15.2N
the OSTENSIBILITY weight of object in water = w" = 13.7N

when we put an object in a (fluid) liquid if it float in the liquid, there is a force thet liquid will exert on the object called BUOYANCY force and its magnitude is:
Fb = γV where γ=ρg and V = volume of the object

i) for object in ethyl alcohol applying newton's second law
w - γ1V - w' = 0

ii) for object in water applying newton's second law
w - γ2V -w" = 0

subtractiong two above equations;

w' - w" = V (γ2 - γ1) =>
V = (w' - w") / (γ2 - γ1)
V = (w' - w") / (ρ2g - ρ1g)

ρ2 = 1000 Kg/m^3
ρ1 = 789 Kg/m^3
V = 1.5 / [ (1000 - 789)*9.8 ]

► V = 0.000725 m^3 = 725 cm^3

Answer 3

Water Specific Gravity is 1.000 (1,000kg/m³)
The Alcohol has a Specific Gravity of 0.785 (785kg/m³)

The Density of water is much greater than that of ethyl alcohol.
The buoyancy force of the water is greater than that of the alcohol

Volume of object = 0.785m³