A right circular cone of density - floats just immersed with its vertex downwards- in a vessel containing two liquids of densities -1 and -2 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- hp-2-1-21 - 3B- h-2-1-21 - 2C- h-2-1-21 - 3D- h-2-1-2

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Floating

A right circu...

Question

A right circular cone of density $ρ,$ floats just immersed with its vertex downwards, in a vessel containing two liquids of densities $σ_{1}$ and $σ_{2}$ 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 $z$ .

h {(\frac{ρ + σ_{2}}{σ_{1} + σ_{2}})}^{1 / 3}

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h {(\frac{ρ - σ_{2}}{σ_{1} - σ_{2}})}^{1 / 3}

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h {(\frac{ρ - σ_{2}}{σ_{1} + σ_{2}})}^{1 / 2}

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h (\frac{ρ - σ_{2}}{σ_{1} - σ_{2}})

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Solution

The correct option is B $h {(\frac{ρ - σ_{2}}{σ_{1} - σ_{2}})}^{1 / 3}$
$V A B$ is the given cone. Let its height be $h$ and semi-vertical angle $α .$ Let the base $A B$ of the cone be on the surface. $C D$ is the plane of separation of two liquids, $O$ and $O^{'}$ are the centres of the base $A B$ and plane of separation $C D$ .

$∴$ For equilibrium,
Weight of the cone = (Weight of liquid of density $σ_{1}$ displaced) + (Weight of liquid of density $σ_{2}$ displaced)
Now, radius of cone, $r = O B = h tan α$
So, $\frac{1}{3} π h^{3} tan α ρ g = \frac{1}{3} π z^{3} tan α σ_{1} g + \frac{1}{3} π (h^{3} - z^{3}) tan α σ_{2} g$

or $h^{3} ρ = z^{3} σ_{1} + (h^{3} - z^{3}) σ_{2}$
or $h^{3} (ρ - σ_{2}) = z^{3} (σ_{1} - σ_{2})$
or $z = h {(\frac{ρ - σ_{2}}{σ_{1} - σ_{2}})}^{1 / 3}$

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Floating

Standard XII Physics

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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.