A) 3, 27
B) 27, 3
C) 27, 27
D) 3, 3
Correct Answer: B
Solution :
Key Idea: In coalescing into a single drop charge remains conserved. Also volume before and after coalescing remains same. Let R and r be the radii of bigger and each smaller drop respectively. In coalescing into a single drop, charge remains conserved. Hence, charge on bigger drop\[=27\times \]charge on smaller drop i.e., q'= 27q Now, before and after coalescing, volume remains same. That is,\[\frac{4}{3}\pi {{R}^{3}}=27\times \frac{4}{3}\pi {{r}^{3}}\] \[\therefore \] \[R=3r\] Hence, capacitance of bigger drop \[C'=4\pi {{\varepsilon }_{0}}R=4\pi {{\varepsilon }_{0}}(3r)\] \[=3(4\pi {{\varepsilon }_{0}}r)\] \[=3C\]
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