Mass `M` is distributed uniformly along a line of length `2L`. A particle of mass `m` is at a point that is at a distance `a` above the centre of the line on the its perpendicular bisector (Point `P` in figure). The gravitational force that the line exert on the particle is

Options

A. `(GMm)/(sqrt(L^(2)+a^(2)))`
B. `(GMm)/(a(L^(2)+a^(2)))`
C. `(GMm)/(asqrt(L^(2)+a^(2)))`
D. `(GMm)/(a(L^(2)+a^(2))^(2))`

Answer

C

Explanation

`dM=(M/(2L))dx F=(Gm(dM))/((a^(2)+x^(2))`
Components `F cos theta` (i.e. parallel to line) cancel each other. Net force will be perpendicular to the rod.
`F_(net)=int_(x=0)^(x=L)2F sin theta=int_(0)^(L)(2GmM)/(2L(a^(2)+x^(2)))(adx)/((a^(2)+x^(2))^(1//2))`
`int_(0)^(L)(GMma.dx)/(L(a^(2)+x^(2))^(3//2))=(GMm)/(asqrt(L^(2)+a^(2)))`