Infinite number of masses, each of `1 kg`, are placed along the x-axis at `x = +- 1m, +- 2m, +-4m, +- 8m, +- 16m`.. The gravitational of the resultant gravitational potential in term of gravitaitonal constant `G` at the origin `(x = 0)` is

Options

A. `(G)/(2)`
B. G
C. 2G
D. 4G

Answer

C

Explanation

Aa V= GM `((1)/(r^(1)) + (1)/(r_(2))+ (1)/(r_(2))cdots)`
`=Gxx1 ((1)/(1) +(1)/(2) + (1)/(4) + (1)/(4) + (1)/(8) + (1)/(16))`
sum of GP = `(a)/(1-r)`
`V=G ((1)/(1- (1)/(2)))=2G`