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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    The surface tension and vapour pressure of water at\[20{}^\circ C\]is\[7.28\times {{10}^{-2}}N{{m}^{-1}}\]and\[2.33\times {{10}^{3}}Pa\]respectively. The radius of the smallest spherical droplet which can form without evaporating is

    A)  \[6.25\times {{10}^{-5}}m\]       

    B)  \[6.25\times {{10}^{-3}}m\]

    C)  \[7.45\times {{10}^{+3}}m\]                      

    D)  \[8.9\times {{10}^{-4}}m\]

    Correct Answer: A

    Solution :

                    Given \[T=7.28\times {{10}^{-2}}N/m\] \[p=2.33\times {{10}^{3}}Pa\] The drop will evaporate if the water pressure greater than the vapour pressure p. i.e.      \[p=\frac{2T}{R},\]where R = radius of drop \[\Rightarrow \] \[R=\frac{2T}{p}=\frac{2\times 7.28\times {{10}^{-2}}}{2.33\times {{10}^{3}}}=6.25\times {{10}^{-5}}m\]


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