JEE MAIN - Physics (2007 - No. 17)
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
$${{electro\,\,ch\arg e\,\,on\,\,the\,\,moon} \over {electronic\,\,ch\arg e\,\,on\,\,the\,\,earth}}\,\,to\,be$$
$${{electro\,\,ch\arg e\,\,on\,\,the\,\,moon} \over {electronic\,\,ch\arg e\,\,on\,\,the\,\,earth}}\,\,to\,be$$
$${g_M}/{g_E}$$
$$1$$
$$0$$
$${g_E}/{g_M}$$
Vysvetlenie
electronic charge does does not depend on acceleration due to gravity as it is a universal constant.
So, electronic charge on earth
$$=$$ electronic charge on moon
$$\therefore$$ Required ratio $$=1.$$
So, electronic charge on earth
$$=$$ electronic charge on moon
$$\therefore$$ Required ratio $$=1.$$