JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    If the work done in blowing a bubble of volume V is W, then the work done in Mowing a soap bubble of volume 2V will be

    A)  W                               

    B)  2W

    C)  \[\sqrt{2}W\]             

    D)  \[{{4}^{1/3}}W\]

    Correct Answer: B

    Solution :

     From the definition of surface tension (T), the surface tension of a liquid is equal to the work (W) required to increase the surface area  of the liquid film by unity at constant temperature, \[\therefore \]    \[W=T\times \Delta A\] Since, surface area of a sphere is \[4\pi {{R}^{2}}\] and there are two free surfaces, we have \[W=T\times \Delta A\]        ?(i) and volume of sphere \[=\frac{4}{3}\,\pi {{R}^{3}}\] i.e.,                  \[V=\frac{4}{3}\,\pi {{R}^{3}}\] \[\Rightarrow \]                        \[R={{\left( \frac{3V}{4\pi } \right)}^{1/3}}\]         ?(ii) From Eqs. (i) and (ii), we get \[W=T\times 8\pi \times {{\left( \frac{3V}{4\pi } \right)}^{2/3}}\] \[\Rightarrow \]            \[W\propto \,{{V}^{2/3}}\] \[\therefore \]    \[{{W}_{1}}\propto \,{{V}_{1}}^{2/3}\] and      \[{{W}_{2}}\propto \,{{V}_{2}}^{2/3}\] \[\therefore \]    \[\frac{{{W}_{2}}}{{{W}_{1}}}=\,{{\left( \frac{2V}{V} \right)}^{2/3}}\] \[\Rightarrow \]            \[{{W}_{2}}={{2}^{2/3}}{{W}_{1}}={{4}^{1/3}}W\]

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