If g $_{E}$ and g $_{M}$ are the accelerations due to gravity on the surfaces of the Earth and the Moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the Moon/ electronic charge on the Earth) to be:

1

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g_{E} / g_{M}

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g_{M} / g_{E}

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Solution

The correct option is A $1$
Formula for Millikan's oil drop experiment,
$q = \frac{6 π η r (v_{1} + v_{2})}{E}$
As from above formula the electronic charge is independent of acceleration. So it will be same on earth and moon. Thus $q_{m} / q_{E} = 1$

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If gE and gM are the accelerations due to gravity on the surfaces of the Earth and the Moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the Moon/ electronic charge on the Earth) to be:

The correct option is A 1Formula for Millikan's oil drop experiment,q=6πηr(v1+v2)EAs from above formula the electronic charge is independent of acceleration. So it will be same on earth and moon. Thus qm/qE=1

If g $_{E}$ and g $_{M}$ are the accelerations due to gravity on the surfaces of the Earth and the Moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the Moon/ electronic charge on the Earth) to be:

The correct option is A $1$
Formula for Millikan's oil drop experiment,
$q = \frac{6 π η r (v_{1} + v_{2})}{E}$
As from above formula the electronic charge is independent of acceleration. So it will be same on earth and moon. Thus $q_{m} / q_{E} = 1$