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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    Two mercury drops (each of radius' r') merge to from bigger drop. The surface energy of the bigger drop, if T is the surface tension, is :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[4\eta {{r}^{2}}T\]

    B) \[2\eta {{r}^{2}}T\]

    C) \[{{2}^{8/3}}\eta {{r}^{2}}T\]

    D) \[{{2}^{5/3}}\eta {{r}^{2}}T\]

    Correct Answer: C

    Solution :

                                    \[2.\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\]                 \[R={{2}^{1/3}}r\]                 \[S.E.=T.4\pi {{R}^{2}}\]                 \[T4\pi {{2}^{2/3}}{{r}^{2}}\]                 \[T{{.2}^{8/3}}\pi {{r}^{2}}.\]      


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