At what height from the center of the Earth the acceleration due to gravity will be 1/4th of its value as at the earth.

Explanation

Given: Height from the centre of the Earth, R. =R+h
The acceleration due to gravity at that height, g. = `g//4`
  Formula: g=`(GM)/(R^(2))`
`(g)/(g)=((R)/(R))^(2)`
=`((R+h)/(R))^(2)=[(R)/(R)+(h)/(R)]^(2)`
=`[1+(h)/(R)]^(2)`
4=`[1+(h)/(R)]^(2)` Apply root on both sides
`2=[1+(h)/(R)]`
or  h=R
R’=2R
From the centre of the Earth, the object is placed at twice the radius of the earth 
(ii)" How many electrons are passing per second in a circuit in which there is a current of 5A."
"Current I = 5A
Timet - 1 second"
"Charge of electron (e) = `1.6xx10^(-19)C`
I=`(q)/(t)=(ne)/(t)`"
"Number of electron", n =`(It)/(e)`
=`(5xx1)/(1.6xx10^(-19))`
n=`3.125xx10^(19)`