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

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

Tardigrade Admin · Accepted Answer

Since electronic charge (e=1.6 × 10-19 C ) is a universal constant. It does not depend on the value of acceleration due to gravity. ∴ Electronic charge on the moon = electronic charge on the earth or ( text electronic charge on themoon / text electronic charge on the earth )=1

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

1

0

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

Since electronic charge $(e = 1.6 \times 1 0^{- 19} C)$ is a universal constant. It does not depend on the value of acceleration due to gravity.
$∴$ Electronic charge on the moon = electronic charge on the earth
or $\frac{electronic charge on themoon}{electronic charge on the earth} = 1$

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 electronicchargeontheearthelectronicchargeonthemoon​ to be

gE​gM​​

1

0

gM​gE​​

Since electronic charge (e=1.6×10−19C) is a universal constant. It does not depend on the value of acceleration due to gravity. ∴ Electronic charge on the moon = electronic charge on the earth or electronic charge on the earth electronic charge on themoon ​=1

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

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

Since electronic charge $(e = 1.6 \times 1 0^{- 19} C)$ is a universal constant. It does not depend on the value of acceleration due to gravity.
$∴$ Electronic charge on the moon = electronic charge on the earth
or $\frac{electronic charge on themoon}{electronic charge on the earth} = 1$