A) \[1\]
B) \[\operatorname{zero}\]
C) \[\frac{{{g}_{E}}}{{{g}_{M}}}\]
D) \[\frac{{{g}_{M}}}{{{g}_{E}}}\]
Correct Answer: A
Solution :
According to Millikan's oil drop experiment, electronic charge is given by \[q=\frac{6\pi nr({{v}_{1}}+{{v}_{2}})}{E}\]which is independent of g. So, \[\frac{electronic\text{ }charge\text{ }on\text{ }the\text{ }moon}{~electronic\text{ }charge\text{ }on\text{ }the\text{ }earth}=1\]You need to login to perform this action.
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