Three particles A, B and C of mass m, m and 2m lie in a straight line as shown. The only force acting on these particles is due to the gravitational attraction towards each other. Let the magnitude of net force on them be `F_(A), F_(B)` and `F_(C )` respectively. Then,`F_(A):F_(B):F_(C )` is equal to :

Options

A. `3:2:5`
B. `3:1:4`
C. `2:2:3`
D. `6:1:5`

Answer

A

Explanation

Taking rightward direction as positive,
`F_(A)=(G(m)(m))/(L^(2))hati+(G(m)(2m))/((2L)^(2))hati=(3)/(2)(Gm^(2))/(L^(2))hati`
`F_(B)=-(G(m)(m))/(L^(2))hati+(G(m)(2m))/(L^(2))hati=(Gm^(2))/(L^(2))`
`F_(C)=-(G(2m)(m))/((2L)^(2))hati-(G(2m)(m))/(L^(2))hati=-(5)/(2)(Gm^(2))/(L^(2))hati`
So, `|F_(A)| : |F_(B)| : |F_(C)|=(3)/(2):1:(5)/(2)=3:2:5`