Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    If value of surface tension of a liquid is \[70\text{ }dyne/cm\], then its value in \[N/m\] will be

    A) \[7\times {{10}^{3}}N/m\]          

    B)        \[7\times {{10}^{2}}N/m\]

    C) \[7\times {{10}^{-2}}N/m\]        

    D)        \[70\,\,N/m\]

    Correct Answer: C

    Solution :

    The unit dyne/cm is CGS system of unit. The CGS unit of force is dyne and SI unit is Newton. Also there are \[{{10}^{5}}\] dynes in one newton and \[100cm\] in\[1cm\]. \[\therefore \]  \[\frac{70\,\,dynes}{cm}=\frac{70\times {{10}^{-5}}}{{{10}^{-2}}}=7\times {{10}^{-2}}N/m\]


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