Two point masses A and B having masses in the ratio `4:3` are separated by a distance of 1m. When another point mass C of mass M is placed in between A and B, the force between A and C is `(1/3)^(rd)` of the force between B and C. Then the distance C from A is

Options

A. `(2)/(3)m`
B. `(1)/(3)m`
C. `(1)/(4)m`
D. `(2)/(7)m`

Answer

A

Explanation

(a) Let a point mass C is placed at a distance of x m from the point mass A as shown in the fiugure.

Here, `(M_(A))/(M_(B))=(4)/(3)`
Force between A and C is
`F_(AC)=(GMM_(A))/(x^(2))`....(i)
Force between B and C is
`F_(BC)=(GMM_(B))/((1-x)^(2))`....(ii)
According to given problem
`F_(AC)=(1)/(3)F_(BC)`
`:. (GM_(A)M)/(x^(2))=(1)/(3)((GM_(B)M)/((1-x)^(2)))" "` (Using (i) and (ii))
`(M_(A))/(x^(2))=(M_(B))/(3(1-x)^(2)) or (M_(A))/(M_(B))=(x^(2))/(3(1-x)^(2))`
`(4)/(3)=(x^(2))/(3(1-x)^(2)) or 4=(x^(2))/((1-x)^(2))`
`or 2=(x)/(1-x) or 2-2x=x`
` 3x=2 or x=(2)/(3)m`