Three masses, each equal to M, are placed at the three corners of a square of side a. the force of attraction on unit mass at the fourth corner will be

Options

A. `(Gm)/(3a^(2))`
B. `(Gm)/(sqrt(3a^(2)))`
C. `(3Gm)/(a^(2))`
D. `(Gm)/(a^(2))(sqrt2+1//2)`

Answer

d

Explanation

`F=(Gmxx1)/(a^(2))=(Gm)/(a^(2))`
`F_(R)=sqrt(F^(2)+F^(2))=sqrt2F`
`F_(R)=sqrt2(Gm)/(a^(2))`
`F_(R)` and `F_(2)` are in the same direction then,
`F_(R_(1))=F_(R)+F_(2)`
`=sqrt2(Gm)/(a^(2))+(Gm)/(2a^(2))=(Gm)/(a^(2))[sqrt2+(1)/(2)]`
`F_(R_(1))=(Gm)/(a^(2))[(2sqrt2+1)/(2)]`