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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    The radius of a spherical drop of water is 1 mm. If surface tension of water be \[70\times {{10}^{-3}}N/m,\]the pressure difference inside and     outside the drop will be

    A)  \[70\text{ }N/{{m}^{2}}\]                           

    B)  \[140\text{ }N/{{m}^{2}}\]

    C)  \[280\text{ }N/{{m}^{2}}\]         

    D)  zero

    Correct Answer: C

    Solution :

                    The excess pressure p is given by \[P=\frac{4T}{R}\]where T is surface tension and R is radius of bubble. Given, \[t=70\times {{10}^{-3}}N/m,\,\,R=1mm={{10}^{-3}}m\] \[\therefore \]  \[p=\frac{4\times 70\times {{10}^{-3}}}{{{10}^{-3}}}=280N/{{m}^{2}}\]


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