CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • 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 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)  \[\frac{\sqrt{{{\rho }_{1}}\,{{\rho }_{2}}}}{\rho }\]

    Correct Answer: A

    Solution :

    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 \[=\frac{\rho -{{\rho }_{2}}}{{{\rho }_{1}}-\rho }\]


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