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. `1/4m`
D. `2/7 m`

Answer

A

Explanation

`F_(1)=(GM.4m)/(x^(2)), F_(2)=(GM.3m)/((1-x)^(2))`
`F_(1)=1/3F_(2)`
`(4GMm)/(x^(2))=1/3(GMm)/((1-x)^(2))`
`4/(x^(2))=1/((1-x)^(2)) implies 2/x=1/(1-x)`
`2-2x=ximplies x2/3m`