Question

A right circular cone of density $\rho ,$ floats just immersed with its vertex downwards, in a vessel containing two liquids of densities ${\sigma }_1$ and ${\sigma }_2$ 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. $h{\left(\frac{\rho +{\sigma }_2}{{\sigma }_1+{\sigma }_2}\right)}^{1/3}$
• B. $h{\left(\frac{\rho -{\sigma }_2}{{\sigma }_1-{\sigma }_2}\right)}^{1/3}$
• C. $h{\left(\frac{\rho -{\sigma }_2}{{\sigma }_1+{\sigma }_2}\right)}^{1/2}$
• D. $h\left(\frac{\rho -{\sigma }_2}{{\sigma }_1-{\sigma }_2}\right)$

Answer 1

The correct option is B $h{\left(\frac{\rho -{\sigma }_2}{{\sigma }_1-{\sigma }_2}\right)}^{1/3}$

$VAB$ is the given cone. Let its height be $h$ and semi-vertical angle $\alpha .$ Let the base $AB$ of the cone be on the surface. $CD$ is the plane of separation of two liquids, $O$ and $O^{\prime }$ are the centres of the base $AB$ and plane of separation $CD$.


$\therefore$ For equilibrium,
Weight of the cone = (Weight of liquid of density ${\sigma }_1$ displaced) + (Weight of liquid of density ${\sigma }_2$ displaced)
Now, radius of cone, $r=OB=h\tan \alpha$
So, $\frac{1}{3}\pi h^3\tan \alpha \rho g=\frac{1}{3}\pi z^3\tan \alpha {\sigma }_{1}g+\frac{1}{3}\pi (h^3-z^3)\tan \alpha {\sigma }_{2}g$

or $h^3\rho =z^3{\sigma }_1+(h^3-z^3){\sigma }_2$
or $h^3(\rho -{\sigma }_2)=z^3({\sigma }_1-{\sigma }_2)$
or $z=h{\left(\frac{\rho -{\sigma }_2}{{\sigma }_1-{\sigma }_2}\right)}^{1/3}$