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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    The work done in increasing the size of a soap film from \[10\,cm\times 6\,m\] to \[10\,cm\times 11\,cm\] is\[3\times {{10}^{-4}}joule\]. The surface tension of the film is:

    A)  \[11.0\times {{10}^{-2}}\text{N/m}\] 

    B)  \[6.0\times {{10}^{-2}}\text{N/m}\]                      

    C)  \[3.0\times {{10}^{-2}}\text{N/m}\]  

    D)         \[1.5\times {{10}^{-2}}\text{N/m}\]

    Correct Answer: C

    Solution :

    Work done = Area increased\[\times \]surface tension so, surface tension, \[{{S}_{T}}=\frac{3\times {{10}^{-4}}}{2\times (10\times 11-10\times 6)\times {{10}^{-4}}}\] \[=3\times {{10}^{-2}}N\text{/}m\]


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