J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    If the radius of a spherical liquid (of surface tension S) drop increases from r to \[r+Ar,\]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, \[W=S\times 4\pi [{{(r+\Delta r)}^{2}}-{{(r)}^{2}}]\] \[=S\times 4\pi [{{r}^{2}}+{{(\Delta r)}^{2}}+2\Delta r\times r-{{r}^{2}}]\] \[=8\pi \,\Delta rSr\] [\[{{(\Delta r)}^{2}}\] is very small]


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