Three equal masses m are placed at the three corners of an equilateral triangle of side a. find the force exerted by this system on another particle of mass m placed at (a) the mid point of a side (b) at the center of the triangle.

Options

A. `(3GMm)/(4l^(2))`
B. `(4GMm)/(3l^(2))`
C. `(GMm)/(4l^(2))`
D. `(4GMm)/(l^(2))`

Answer

B

Explanation

Force on `m` due to masses at `Q` and `R` is zero. So, the net force is due to the mass at `P`.
Hence `F=(GMm)//(PL)^(2)`
Now, `PL=lsin60^@=l(sqrt(3))/2`
`F=(4GMm)/(3l^(2))`