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 \[z=h{{\left( \frac{\rho -{{\sigma }_{2}}}{{{\sigma }_{1}}-{{\sigma }_{2}}} \right)}^{1/3}}\]You need to login to perform this action.
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