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