WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    Two soap bubbles of radii \[x\] and \[y\] coalesce to constitute a bubble of radius z. Then z is equal to___

    A)  \[\sqrt{{{x}^{2}}+{{y}^{2}}}\]                    

    B)  \[\sqrt{x+y}\]

    C)  \[x+y\]               

    D)  \[\frac{x+y}{2}\]

    Correct Answer: A

    Solution :

                     Given that two soap bubbles coalesce to constitute a bubble of radius z. Now from the ideal gas law, we get \[pV={{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}}\]                 Hence, we have \[nRT={{n}_{1}}RT+{{n}_{2}}RT\]                 So,          \[n={{n}_{1}}+{{n}_{2}}\]                 Thus, we have                 \[{{p}_{1}}={{p}_{0}}+\frac{4T}{x},{{p}_{2}}={{p}_{0}}+\frac{4T}{y},p={{p}_{0}}+\frac{4T}{z}\] Assuming that the process is taking place in vacuum, we have \[{{p}_{1}}=\frac{4T}{x},{{p}_{2}}=\frac{4T}{y},p=\frac{4T}{z}\]                 Hence,                  \[pV={{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}}\]                 or            \[\frac{4T}{z}\left( \frac{4}{3}\pi {{z}^{3}} \right)=\frac{4T}{x}\left( \frac{4}{3}\pi {{x}^{3}} \right)+\frac{4T}{y}\left( \frac{4}{3}\pi {{y}^{3}} \right)\]                 Hence \[{{z}^{2}}={{x}^{2}}+{{y}^{2}}\Rightarrow z=\sqrt{{{x}^{2}}+{{y}^{2}}}\]


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