A) \[{{n}^{2/3}}V\]
B) \[{{n}^{1/3}}V\]
C) \[nV\]
D) \[V/n\]
Correct Answer: A
Solution :
Let the radius of each droplet be r units. So, the volume of each droplet is equal to \[\frac{4}{3}\pi {{r}^{3}}.\] Thus, n droplets have the total volume equal to \[n\left( \frac{4}{3}\pi {{r}^{3}} \right).\] Since, the number of the drop would be equal to the total volume of the droplets hence, \[\Rightarrow \] \[{{R}^{3}}=n{{r}^{3}}\] \[\Rightarrow \] \[R={{n}^{1/3}}r\] ?(i) The capacitance of each droplet is equal to, \[{{C}_{d}}=4\pi {{\varepsilon }_{0}}r\]and thus the charge of each droplet would equal \[{{q}_{d}}={{C}_{d}}{{V}_{d}}=4\pi {{\varepsilon }_{0}}r{{V}_{d}}\] ?(ii) The capacitance of the bigger drop would be equal to \[C=4\pi {{\varepsilon }_{0}}R.\] The potential of the bigger drop would be equal to V(say). Hence, \[V=\frac{n{{q}_{d}}}{C}=\frac{n(4\pi {{\varepsilon }_{0}}r{{V}_{d}})}{4\pi {{\varepsilon }_{0}}{{n}^{1/3}}r}={{n}^{2/3}}{{V}_{d}}\]
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