J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    If the radius of a spherical liquid (of surface tension S) drop increases from r to r + \[\Delta \]r, the corresponding increase in the surface energy is

    A) \[8\pi r\Delta rS\]                           

    B)  \[4\pi r\Delta rS\]

    C)  \[16\pi r\Delta rS\]                        

    D)  \[2\pi r\Delta rS\]

    Correct Answer: A

    Solution :

                     Work done in blowing a liquid drop or soap bubble \[W=S\times 4\pi [r_{2}^{2}-r_{1}^{2}]\] Here, \[W=S\times 4\pi [{{(r+\Delta r)}^{2}}-{{(r)}^{2}}]\]                 \[=S\times 4\pi [{{r}^{2}}+{{(\Delta r)}^{2}}+2\Delta r\times r-{{r}^{2}}]\]                 \[=84\pi \Delta rSr\][\[{{(\Delta r)}^{2}}\]is very small]


You need to login to perform this action.
You will be redirected in 3 sec spinner