A) \[2:1\]
B) \[8:1\]
C) \[108:101\]
D) \[{{(102)}^{2}}:{{(103)}^{2}}\]
Correct Answer: B
Solution :
From the question we have Excess pressure inside 1st bubble is \[=(1.01-1)\]atmosphere \[=0.01\] atmosphere Excess pressure inside IInd bubble is \[=(1.02-1)=0.02\]atmosphere While we know excess pressure are given by \[\frac{4T}{{{r}_{1}}}\]and\[\frac{4T}{{{r}_{2}}}\]respectively So,\[\frac{\frac{4T}{{{r}_{1}}}}{\frac{4T}{{{r}_{2}}}}=\frac{0.01}{0.02}\] or \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{1}{2}\] Let volumes of first and second bubbles are \[{{V}_{1}}\] and \[{{V}_{2}}\] respectively So, \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{\frac{4}{3}\pi r_{1}^{3}}{\frac{4}{3}\pi r_{2}^{3}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}={{\left( \frac{2}{1} \right)}^{3}}=\frac{8}{1}\] Hence, \[{{V}_{1}}:{{V}_{2}}=8:1\]You need to login to perform this action.
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