A) 4 times
B) 2 times
C) 32 times
D) 8 times
Correct Answer: B
Solution :
Volume of bigger drop\[=8\times \] volume of smaller drop \[\frac{4}{3}\pi {{R}^{3}}=8\times \frac{4}{3}\pi {{r}^{3}}\] \[\therefore \] \[R=2r\] Capacity of a smaller drop \[C=4\pi {{\varepsilon }_{0}}r\] Capacity of a bigger drop \[C'=4\pi {{\varepsilon }_{0}}R\] \[\therefore \] \[\frac{C'}{C}=\frac{4\pi {{\varepsilon }_{0}}R}{4\pi {{\varepsilon }_{0}}r}=\frac{2r}{r}\] Hence, \[C'=2C\]
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