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 \[ie,\] \[q'=27q\] 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 \[C'=4\pi {{\varepsilon }_{0}}R\] \[=4\pi {{\varepsilon }_{0}}R(3r)\] \[=3(4\pi {{\varepsilon }_{0}}r)\] \[=3C\]
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