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Question

A right circular cone of density ρ, floats just immersed with its vertex downwards, in a vessel containing two liquids of densities σ1 and σ2 respectively. The plane of separation of the two liquids cuts off from the axis of the cone at a fraction z (taken from vertex) of its length. Find z.

A
h(ρ+σ2σ1+σ2)1/3
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B
h(ρσ2σ1σ2)1/3
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C
h(ρσ2σ1+σ2)1/2
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D
h(ρσ2σ1σ2)
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Solution

The correct option is B h(ρσ2σ1σ2)1/3
VAB is the given cone. Let its height be h and semi-vertical angle α. Let the base AB of the cone be on the surface. CD is the plane of separation of two liquids, O and O are the centres of the base AB and plane of separation CD.


For equilibrium,
Weight of the cone = (Weight of liquid of density σ1 displaced) + (Weight of liquid of density σ2 displaced)
Now, radius of cone, r=OB=htanα
So, 13πh3tanαρg=13πz3tanασ1g+13π(h3z3)tanασ2g

or h3ρ=z3σ1+(h3z3)σ2
or h3(ρσ2)=z3(σ1σ2)
or z=h(ρσ2σ1σ2)1/3

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