A) 16
B) 8
C) 4
D) 2
Correct Answer: B
Solution :
The pressure inside a soap bubble is the sum of excess pressure and atmospheric pressure i.e., \[(P+p).\] Here, For first soap bubble \[P+{{p}_{1}}=1.01\,atm\]For second soap bubble \[P+{{p}_{2}}=1.02\,atm\] \[\therefore \] \[P+{{p}_{1}}=1.01P\Rightarrow {{p}_{1}}=1.01P-P\] \[{{p}_{1}}=0.01P\] and \[P+{{p}_{2}}=1.02P\,\Rightarrow {{p}_{2}}=1.02P-P\] \[\therefore \] \[{{p}_{2}}=0.02P\] For I soap bubble excess of pressure \[{{p}_{1}}=\frac{4T}{{{R}_{1}}}\] For II soap bubble excess of pressure \[{{p}_{2}}=\frac{4T}{{{R}_{2}}}\] Where \[{{R}_{1}}\]and \[{{R}_{2}}\]are the radius of soap bubbles, So, \[{{R}_{1}}=4T/{{p}_{1}}\]and \[{{R}_{2}}=4T/{{p}_{2}}\] \[\therefore \] \[\frac{Volume\,I\,soap\,bubble}{Volume\,of\,II\,soap\,bubble}\] \[=\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{\frac{4}{3}\pi R_{1}^{3}}{\frac{4}{3}\pi R_{2}^{3}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}}\] \[\Rightarrow \] \[\frac{{{V}_{1}}}{{{V}_{2}}}={{\left( \frac{4T/{{p}_{1}}}{4T/{{p}_{2}}} \right)}^{3}}={{\left( \frac{{{p}_{2}}}{{{p}_{1}}} \right)}^{3}}={{\left( \frac{0.02}{0.01} \right)}^{3}}\] \[={{\left( \frac{2}{1} \right)}^{3}}=\frac{8}{1}\] \[{{V}_{1}}:{{V}_{2}}=8:1\]
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