Two particles each of mass 'm' are placed at `A` and `C` are such `AB=BC=L`. The gravitational force on the third particle placed at `D` at a distance `L` on the perpendicular bisector of the line `AC` is

Options

A. `(Gm^(2))/(sqrt(2)L^(2))` along BD
B. `(Gm^(2))/(sqrt(2)L^(2))` along DB
C. `(Gm^(2))/(L^(2))` along AC
D. None of these

Answer

B

Explanation

Here, `F_(0) = (Gm^(2))/((sqrt(2)L)^(2)) = (Gm^(2))/(2L^(2))`

`:.` Resultant force,
`F = 2F_(0) cos 45^(@)` (along DB)
`=2F_(0) xx (1)/(sqrt(2)) = 2 (Gm^(2))/(2L^(2)) xx (1)/(sqrt(2)) = (Gm^(2))/(sqrt(2)L^(2))` (along DB)