Haryana PMT Haryana PMT Solved Paper-2003

  • question_answer
    The radii of two soap bubbles are \[{{r}_{1}}\] and \[{{r}_{2}}\]\[({{r}_{2}}>{{r}_{1}}).\] When they come into contact, the radius of their common interface is :

    A)  \[{{r}_{2}}-{{r}_{1}}\]                                    

    B) \[\sqrt{r_{1}^{2}-r_{2}^{2}}\]

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

    D)  \[\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}}\]

    Correct Answer: D

    Solution :

                    If R be the radius of the interface, then \[\frac{4T}{r}=\frac{4T}{{{r}_{1}}}-\frac{4T}{{{r}_{2}}}\]                 or            \[\frac{1}{r}=\frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}}\]                                 \[=\frac{{{r}_{2}}-{{r}_{1}}}{{{r}_{1}}{{r}_{2}}}\]                 So,          \[r=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}}\]


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