A) \[{{(3)}^{1/2}}\,V\]
B) \[{{(3)}^{2/3}}\,V\]
C) \[{{(2)}^{2/3}}\,V\]
D) \[{{(2)}^{3/2}}\,V\]
Correct Answer: C
Solution :
Key Idea: Volume of small drops remain same as big drop on coalesing. The potential (V) on each drop with cahrge q and radius r is \[v=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Also, charge on big drop is double the charge on each small drop \[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q}{R}\] where, R is radius of big drop Also, Volume of 2 small drops = volume of one big drop \[2\times \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[R={{(2)}^{1/3}}r\] Putting this value of R in Eq. (i), we get \[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q}{{{(2)}^{1/3}}r}\] \[={{(2)}^{2/3}}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] \[\Rightarrow \] \[V'={{(2)}^{2/3}}V.\]
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