A) 2
B) 4
C) 6
D) 8
E) 16
Correct Answer: A
Solution :
Excess pressure is given by\[P=\frac{4T}{r}\] \[\Rightarrow \] \[r=\frac{4T}{P}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{1.02}{1.01}=\frac{102}{101}\] Ratio of volumes \[=\frac{\frac{4}{3}\pi r_{1}^{3}}{\frac{4}{3}\pi r_{2}^{3}}=\frac{{{(102)}^{3}}}{{{(101)}^{3}}}\approx 2\]
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