Question

is a body of mass 2 kg and density 8000 kg M -3 is completely immersed in a liquid of density 800 kg and -3 calculate the weight of the liquid displaced by it​

Answer 1

Answer: The force exerted by a liquid on an object dipped in it will be

F = ρgV

where ρ is the density of the liquid, V is th volume of the object immersed in the liquid and g is the acceleration due to gravity.

So

here

ρ = 800 kg/m3

g =10 m/s2

and V = mass/density = 2/8000 = 2.5 X 10-4 m3

thus,

F = 800 X 10 X 2.5 X 10-4 = 8 X 2.5 X 10-1

or

F = 2 N

Answer 2

Explanation:

Here, mass of the body in air, M=2.0kg

density of the body, d=9000kgm3

Density of the liquid, dl=800kgm3

volume of the body, V=Md=2.0kg8000kgm3

Volume of liquid displaced when the body is completely dipped in it = volume of the body =2.5×10−4m3

mass of the liquid displaced, m=Vdl=(2.5×10−4m3)(800kgm3)=0.2kg

weight of the liquid displaced, Wl=mg=(0.2kg)(10ms2)=2N

Force of buoyancy, FB=Wl=2N

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