question_answer

A solid sphere of volume V and density\[\rho \]floats at the interface of two immiscible liquids of densities\[{{\rho }_{1}}\]and\[{{\rho }_{2}}\]respectively. If\[{{\rho }_{1}}<\rho <{{\rho }_{2}},\]then the ratio of volumes of the parts of the sphere in upper and lower liquids is

A)  \[\frac{\rho -{{\rho }_{1}}}{{{\rho }_{2}}-\rho }\]                             

B)  \[\frac{{{\rho }_{2}}-\rho }{\rho -{{\rho }_{1}}}\]

C)  \[\frac{\rho +{{\rho }_{1}}}{\rho +{{\rho }_{2}}}\]                           

D)  \[\frac{\rho +{{\rho }_{2}}}{\rho +{{\rho }_{1}}}\]

Correct Answer: B

Solution :

                 As \[{{\rho }_{1}}<\rho <{{\rho }_{2}}\] According to question, the volume of solid sphere is V and density is\[\rho \]. Suppose\[{{V}_{1}}\]is the volume of the part of the sphere immersed in a liquid of density\[{{\rho }_{1}}\]and\[{{V}_{1}}\]is the volume of the part of the sphere immersed in a liquid of density\[{{\rho }_{2}}\]. Then                      \[V={{V}_{1}}+{{V}_{2}}\]            As the sphere is floating therefore its weight will be equal to the up thrust force on it. So, The weight of sphere = up thrust due to both liquids \[V\rho g={{V}_{1}}{{\rho }_{1}}g+{{V}_{2}}{{\rho }_{2}}g\] \[({{V}_{1}}+{{V}_{2}})\rho g={{V}_{1}}{{\rho }_{1}}g+{{V}_{2}}{{\rho }_{2}}g\] \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{\rho }_{2}}-\rho }{\rho -{{\rho }_{1}}}\]