RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    The excess pressure inside a soap bubble A is twice that in another soap bubble B. The ratio of volumes of A and B is

    A)  1 : 2                                      

    B)  1 : 4

    C)  1 : 8                      

    D)         1 : 16

    Correct Answer: C

    Solution :

    Excess pressure inside the bubble is                 \[p=\frac{4T}{r}\] Here,     \[{{p}_{A}}=2{{p}_{B}}\]                 \[\frac{{{p}_{A}}}{{{p}_{B}}}=\frac{{{r}_{B}}}{{{r}_{A}}}\] So, radius of bubble \[B\] is\[{{r}_{B}}=2\] Ratio of volumes                 \[\frac{{{V}_{A}}}{{{V}_{B}}}={{\left( \frac{{{r}_{A}}}{{{r}_{B}}} \right)}^{3}}={{\left( \frac{1}{2} \right)}^{3}}\]                 \[\frac{{{V}_{A}}}{{{V}_{B}}}=\frac{1}{8}\]


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