A) \[{{n}^{1/3}}V\]
B) \[{{n}^{2/3}}V\]
C) V
D) nV
Correct Answer: B
Solution :
Key Idea: After coalescing, volume remains conserved. Volume of n identical drops = volume of one big drop i.e., \[n\times \frac{4}{2}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] or \[R={{n}^{1/3}}r\] ... (i) where R is radius of bigger drop and n is radius of each smaller drop. Potential of each smaller drop, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] ... (ii) Potential of bigger drop \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{nq}{R}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{nq}{{{n}^{1/3}}r}\] [from Eq. (ii)] \[V={{n}^{2/3}}V\] [from Eq. (ii)]
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