BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    Pressure inside two soap bubbles are 1.01 and 1.02 atm. Ratio between their volumes is:

    A)  102 : 101                             

    B)         \[{{(102)}^{3}}:{{(103)}^{3}}\]

    C)         8 : 1                                      

    D)         2 : 1

    Correct Answer: C

    Solution :

    Let \[{{r}_{1}}\]and \[{{r}_{2}}\]be the radii of two soap bubble Excess pressure inside first soap bubble \[\frac{4T}{{{r}_{1}}}=1.01-1=0.01\,\text{atm}\] Excess pressure inside second soap bubble \[\frac{4T}{{{r}_{2}}}=1.02-1=0.02\,\text{atm}\]                 Therefor, \[\frac{4T/{{r}_{1}}}{4T/{{r}_{2}}}=\frac{0.01}{0.02}=\frac{1}{2}\]                 \[\Rightarrow \]               \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{1}{2}\] The ratio of their volumes is given by                                 \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{\frac{4}{3}\pi {{r}_{1}}^{3}}{\frac{4}{3}\pi {{r}_{2}}^{3}}\]                 \[\Rightarrow \]               \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\]                                 \[={{\left( \frac{2}{1} \right)}^{3}}\]                                 \[=8\]


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