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VMMC VMMC Medical Solved Paper-2013

  • question_answer
    If the weight of a body in vacuum is \[\omega .{{\omega }_{1}}\]    and \[{{\omega }_{2}}\]are the weights when it is immersed     in a liquid of specific gravity\[{{\rho }_{1}}\] and \[{{\rho }_{2}}\] E     respectively, then find the relation among\[{{\omega }_{1}},{{\omega }_{2}}\]and \[{{\omega }_{2}}\].

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

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

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

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

    Correct Answer: B

    Solution :

     Loss in weight of liquid of density \[{{\rho }_{1}}=(w-{{w}_{1}})kg.\] if V is the volume of the body then \[V{{\rho }_{1}}=w-{{w}_{1}}\] \[\Rightarrow \] \[V=\frac{w-{{w}_{1}}}{{{\rho }_{1}}}\] Similarly, \[V{{\rho }_{2}}=w-{{w}_{2}}=V=\frac{w-{{w}_{2}}}{{{\rho }_{2}}}\] \[\frac{w-{{w}_{1}}}{{{\rho }_{2}}}=\frac{w-{{w}_{2}}}{{{\rho }_{2}}}\] \[w=\frac{{{w}_{1}}{{\rho }_{1}}-{{w}_{2}}{{\rho }_{2}}}{{{T}_{2}}-{{T}_{1}}}\]


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