Question

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 volume of the parts of the sphere in upper and lower liquids is:

• A. $ \frac{\rho -{{\rho }_{2}}}{{{\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}}} $
• E. $ \frac{\sqrt{{{\rho }_{1}}\,{{\rho }_{2}}}}{\rho } $

Answer 1

Answer: (A)

Let $ {{V}_{1}} $ and $ {{V}_{2}} $
be the volumes, then $ {{V}_{1}}+{{V}_{2}}=V $
As ball is floating. Weight of ball = upthrust on ball due to two liquids
$ V\rho g={{V}_{1}}{{\rho }_{1}}g+{{V}_{2}}{{\rho }_{2}}g $
$ \Rightarrow $ $ V\rho ={{V}_{1}}{{\rho }_{1}}+(V-{{V}_{1}}){{\rho }_{2}} $
$ \Rightarrow $ $ {{V}_{1}}=\left( \frac{\rho -{{\rho }_{2}}}{{{\rho }_{1}}-{{\rho }_{2}}} \right)V $
Fraction in upper part $ =\frac{{{V}_{1}}}{V}=\frac{\rho -{{\rho }_{2}}}{{{\rho }_{1}}-{{\rho }_{2}}} $
Fraction in lower part $ =1-\frac{{{V}_{1}}}{V}=1-\frac{\rho -{{\rho }_{2}}}{{{\rho }_{1}}-{{\rho }_{2}}} $
$ =\frac{{{\rho }_{1}}-\rho }{{{\rho }_{1}}-{{\rho }_{2}}} $
$ \therefore $ Ratio of lower and upper parts $ =\frac{\rho -{{\rho }_{2}}}{{{\rho }_{1}}-\rho } $