A) 0.75 cm
B) 0.75 m
C) 7.5 cm
D) 7.5 m
Correct Answer: A
Solution :
[a] : Pressure outside the bigger drop \[={{P}_{1}}\] Pressure inside the bigger drop \[={{P}_{2}}\] Radius of bigger drop, \[{{r}_{1}}=3cm\] Excess pressure\[={{P}_{2}}-{{P}_{1}}=\frac{4S}{{{r}_{1}}}=\frac{4S}{3}\] Pressure inside small drop \[={{P}_{3}}\] Excess pressure\[={{P}_{3}}-{{P}_{2}}=\frac{4S}{{{r}_{2}}}=\frac{4S}{1}\] Pressure difference between inner side of small drop and outer side of bigger drop \[={{P}_{3}}-{{P}_{1}}=\frac{4S}{3}+\frac{4S}{1}=\frac{16S}{3}\] This pressure difference should exist in a single drop of radius r. \[\therefore \]\[\frac{4S}{r}=\frac{16S}{3}\]or\[r=\frac{3}{4}cm=0.75cm\]
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