A) 2 : 1
B) 8:1
C) 4:1
D) 16 : 1
Correct Answer: A
Solution :
Let radius of big drop be R and of smaller be r. Then, volume of bigger drop\[=8\times \]volume of single drop Also, since shape of drop is assumed spherical, volume of a sphere of radius a is\[\frac{4}{3}\pi {{a}^{3}}\]. \[\therefore \] \[\frac{4}{3}\pi {{R}^{3}}=8\times \frac{4}{3}\pi {{r}^{3}}\] \[\Rightarrow \] \[R=2r\] Capacitance of a sphere of radius a is \[C=4\pi {{\varepsilon }_{0}}a\] \[\therefore \]Capacitance of big drop \[C=4\pi {{\varepsilon }_{0}}R\] Capacitance of small drop \[C=4\pi {{\varepsilon }_{0}}(2r)\] \[\therefore \] \[\frac{C}{C}=\frac{4\pi {{\varepsilon }_{0}}(2r)}{4\pi {{\varepsilon }_{0}}\,r}=\frac{2}{1}\]
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