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KVPY Sample Paper KVPY Stream-SX Model Paper-2

  • question_answer
    A piece of granite floats at the interface of mercury and water contained in a beaker (Fig.). If the densities of granite, water and mercury are \[\rho ,\]\[{{\rho }_{1}}\]and \[{{\rho }_{2}}\] respectively, the ratio of the volume of granite in water to the volume in mercury is -

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

    B) \[({{\rho }_{2}}+\rho )/(\rho +{{\rho }_{1}})\]

    C) \[{{\rho }_{1}}{{\rho }_{2}}/\rho \]

    D) \[{{\rho }_{1}}/{{\rho }_{2}}\]

    Correct Answer: A

    Solution :

    [A]
    The weight of a floating body is equal to the weight of the displaced fluid. If V and v represent the total volume of the piece of granite and volume of granite in water respectively, we have \[V\rho g=v{{\rho }_{1}}g+(V+v){{\rho }_{2}}g\]
    Or,  \[v({{\rho }_{1}}-{{\rho }_{2}})=V(\rho -{{p}_{2}})\]
    Therefore, \[v/V=(\rho -{{p}_{2}})/({{\rho }_{1}}-{{\rho }_{2}})/({{\rho }_{1}}-{{\rho }_{2}})\]
    The ratio required in the question is \[v/(V-v)\]and is given by \[v/(V-v)=(\rho -{{p}_{2}})/[({{\rho }_{1}}-{{\rho }_{2}})/(\rho -{{\rho }_{2}})]\]
    or, \[v/(V-v)=(\rho -{{p}_{2}})/({{\rho }_{1}}-\rho )=(\rho -{{\rho }_{1}})\]


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