CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The excess pressure inside one soap bubble is three times that inside a second soap bubble, then the ratio of their surface areas is

    A)  \[1:9\]                 

    B)         \[1:3\]                 

    C)  \[3:1\]                 

    D)         \[1:27\]

    E)  \[9:1\]

    Correct Answer: A

    Solution :

    Accordingly \[\frac{4T}{{{r}_{1}}}=3\times \frac{4T}{{{r}_{2}}}\] \[\Rightarrow \]               \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{1}{3}\] Ratio of surface areas                 \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{4\pi r_{1}^{2}}{4\pi r_{2}^{2}}=\frac{1}{9}\]


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