question_answer

A right circular cone of density p, floats just immersed with its vertex downwards in a vessel containing two liquids of densities \[{{\rho }_{1}}\]and \[{{\rho }_{2}}\] respectively, the planes of separation of the two liquids cuts off from the axis of the cone a fraction z of its length. Find z.   

A) \[{{\left( \frac{\rho +{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}} \right)}^{1/3}}\]

B) \[{{\left( \frac{\rho -{{\sigma }_{2}}}{{{\sigma }_{1}}-{{\sigma }_{2}}} \right)}^{1/3}}\]

C) \[{{\left( \frac{\rho -{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}} \right)}^{1/2}}\]

D) \[{{\left( \frac{\rho -{{\sigma }_{2}}}{{{\sigma }_{1}}-{{\sigma }_{2}}} \right)}^{1/2}}\]

Correct Answer: B

Solution :

[b] \[VAB\] is the given cone. Let its height be h  and semi-vertical angle a. Let the base AB of the cone be in the surface. CD is the surface of separation of two liquids, O and O? are the centres of the base AB and surface 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) or \[\frac{1}{3}\pi {{h}^{3}}\,{{\tan }^{3}}\,\,\alpha \rho g=\frac{1}{3}\pi {{z}^{3}}\,{{\tan }^{2}}\alpha {{\sigma }_{1}}g\] \[+\frac{1}{3}\pi ({{h}^{3}}-{{z}^{3}}){{\tan }^{2}}\alpha .{{\sigma }_{2}}g\] or \[{{h}^{3}}\rho ={{z}^{3}}{{\sigma }_{1}}+({{h}^{3}}-{{z}^{3}}){{\sigma }_{2}}\] or  \[{{h}^{3}}(\sigma -{{\sigma }_{1}})={{z}^{3}}({{\sigma }_{1}}-{{\sigma }_{2}})\] or \[z=h{{\left( \frac{\rho -{{\sigma }_{2}}}{{{\sigma }_{1}}-{{\sigma }_{2}}} \right)}^{1/3}}\]