BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    A soap bubble in vacuum has a radius 3 cm and another soap bubble in vacuum has radius 4 cm If two bubbles coalesce under isothermal condition, then the radius of the new bubble will be :

    A)  7 cm                                     

    B)         5 cm

    C)         4.5 cm                                 

    D)         2.3 cm

    Correct Answer: B

    Solution :

    Key Idea: An isothermal process obeys Boyles law. Since,    process    is isothermal the total pressure of air inside bubble is same as excess of pressure given by    \[P=\frac{4T}{R}\]  where T is surface tension and R is radius. Also, an isothermal process obeys Boyles law. Hence, PV= constant. Let R be the radius of coalesce system, then \[{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}=PV\] \[\left( \frac{4T}{{{R}_{1}}} \right)\left( \frac{4}{3}\pi R_{1}^{3} \right)+\left( \frac{4T}{{{R}_{2}}} \right)\left( \frac{4}{3}\pi R_{2}^{3} \right)=\left( \frac{4T}{R} \right)\left( \frac{4}{3}\pi {{R}^{3}} \right)\]    \[\Rightarrow \]               \[r_{1}^{2}+r_{2}^{2}={{R}^{2}}\] Given,  \[{{R}_{1}}=3cm,\,{{R}_{4}}=4\,cm\] \[\therefore \]  \[R=\sqrt{{{3}^{2}}+{{4}^{2}}}\] \[\Rightarrow \]               \[R=5\,cm\]


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