A) \[1:16\]
B) \[8:1\]
C) \[1:4\]
D) \[1:64\]
E) \[32:1\]
Correct Answer: D
Solution :
Excess pressure inside a spherical drop of water \[P=\frac{2T}{R}\] Given: \[{{p}_{1}}=4{{p}_{2}}\] \[\frac{2T}{{{R}_{1}}}=4\times \frac{2T}{{{R}_{2}}}\] Or \[{{R}_{2}}=4{{R}_{1}}\] Now, \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{4\pi R_{1}^{3}{{d}_{1}}}{4\pi R_{2}^{3}{{d}_{2}}}\] Or \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{R_{1}^{3}}{R_{2}^{3}}\] Or \[\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{1}{64}\]
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