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NEET Sample Paper NEET Sample Test Paper-43

  • question_answer
    Two spherical soap bubbles of radii \[{{r}_{1}}\] and \[{{r}_{2}}\] in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to -

    A) \[\frac{{{r}_{1}}+{{r}_{2}}}{2}\]                   

    B) \[\sqrt{{{r}_{1}}{{r}_{2}}}\]

    C) \[\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\]               

    D) \[r_{1}^{2}+r_{2}^{2}\]  

    Correct Answer: C

    Solution :

    Since the bubbles coalesce in vacuum and there is no change in temperature, hence its surface energy does not change. This means that the surface are remains unchanged. Hence, \[4\pi \,{{r}_{1}}^{2}+4\pi {{r}_{2}}^{2}=4\pi {{R}^{2}}\] \[\therefore \,\,\,\,\,\,\,\,R=\sqrt{{{r}_{1}}^{2}+{{r}_{2}}^{2}}\]


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