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 \[V\rho g={{V}_{1}}{{\rho }_{1}}g+{{V}_{2}}{{\rho }_{2}}g\] \[({{V}_{1}}+{{V}_{2}})\rho g={{V}_{1}}{{\rho }_{1}}g+{{V}_{2}}{{\rho }_{2}}g\] \[\therefore \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{\rho }_{2}}-\rho }{\rho -{{\rho }_{1}}}\]You need to login to perform this action.
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