Question

physics

An object floats with half its volume beneath the surface of the water. The weight of the displaced water is 2000 N. What is the weight of the object?
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Answer #1

Let V be the volume of the object inside the water surface. Now the buoyant force is equal to the weight of the water displaced.


Hence, Vρg = 2000N


Or, V = 2000/ρg = 0.2041m3.


Thus, the total volume of the object V1is 2* 0.2041 = 0.4082 m3.


We have the relation, F = (ρf - ρb)Vg


Where, ρfis the density of the fluid and ρbis the density of the object or the body and V is the volume of the displaced liquid. Here, F = 2000 N.


Therefore, 2000 = (1000-ρb)9.8*0.2041


Or, ρb   = 0.0899 m3.


Now the true weight of the object is given as V1ρbg = 0.4082* 0.0899*9.8 = 0.359 N.


The weight of the displaced water = Vρg. Here, ρ is the density of water displaced = 1000 kg.m3. and g is the acceleration due to gravity = 9.8 m.s–2. Given that the weight of the displaced water is 2000 N.


answered by: physics.dr.tang
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