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 ( textelectronic charge on the moon/ textelectronic charge on the earth) to be

Question

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 ( textelectronic charge on the moon/ textelectronic charge on the earth) to be (A)  1  (B) 0 (C)  (gE/gM) (D)  (gM/gE)

Tardigrade Admin · Accepted Answer

According to Millikan's oil drop experiment, electronic charge is given by

q = \frac{6 πn r ( v _{1} + v _{2} )}{E}

which is independent of

g

.
So,

\frac{electronic charge on the moon}{electronic charge on the earth} = 1

Q. 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 $\frac{electronic charge on the moon}{electronic charge on the earth}$ to be

$1$

$0$

$\frac{g _{E}}{g _{M}}$

$\frac{g _{M}}{g _{E}}$

According to Millikan's oil drop experiment, electronic charge is given by $q = \frac{6 πn r ( v _{1} + v _{2} )}{E}$ which is independent of $g$ .
So, $\frac{electronic charge on the moon}{electronic charge on the earth} = 1$

Q. 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 earthelectronic charge on the moon​ to be

1

0

gM​gE​​

gE​gM​​

According to Millikan's oil drop experiment, electronic charge is given by q=E6πnr(v1​+v2​)​ which is independent of g. So, electronic charge on the earth electronic charge on the moon ​=1

Q. 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 $\frac{electronic charge on the moon}{electronic charge on the earth}$ to be

$1$

$0$

$\frac{g _{E}}{g _{M}}$

$\frac{g _{M}}{g _{E}}$

According to Millikan's oil drop experiment, electronic charge is given by $q = \frac{6 πn r ( v _{1} + v _{2} )}{E}$ which is independent of $g$ .
So, $\frac{electronic charge on the moon}{electronic charge on the earth} = 1$