JCECE Medical JCECE Medical Solved Paper-2015

  • question_answer
    An elliptically shaped ring of dimensions shown in figure just touches, the horizontal surface of a liquid of surface tension S. The force required to pull the ring away from the liquid surface is

    A) \[2\pi (\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S\]

    B)  \[\pi ({{a}_{1}}+{{b}_{1}}+{{a}_{2}}+{{b}_{2}})S\]

    C)  \[\pi \left( \frac{{{a}_{1}}+{{a}_{2}}}{2}+\frac{{{b}_{1}}+{{b}_{2}}}{2} \right)S\]

    D)  \[\sqrt{2\pi }(\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S\]

    Correct Answer: A

    Solution :

     Internal mean radius \[{{r}_{1}}\sqrt{{{a}_{1}}{{b}_{1}}}\] Internal circumference of the ring \[=2\pi {{r}_{1}}=2\pi \sqrt{{{a}_{1}}{{b}_{1}}}\] External mean radius \[{{r}_{2}}=\sqrt{{{a}_{2}}{{b}_{2}}}\] External circumference of the ring \[=2\pi {{r}_{2}}=2\pi \sqrt{{{a}_{2}}{{b}_{2}}}\] Thus, force required \[F=2\pi \sqrt{{{a}_{1}}{{b}_{1}}S}+2\pi \sqrt{{{a}_{2}}{{b}_{2}}}\,S\] \[=2\pi (\sqrt{{{a}_{1}}{{b}_{1}}}+\sqrt{{{a}_{2}}{{b}_{2}}})S\]


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