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Three particles A, B and C of mass m, m ...

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 :

A

`3:2:5`

B

`3:1:4`

C

`2:2:3`

D

`6:1:5`

Text Solution

Verified by Experts

The correct Answer is:
A
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`
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