Fluid Mechanics Viscosity Question 259
Question: 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.
Options:
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}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ VAB $ is the given cone. Let its height be hand 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}} $
BETA
