A straight of length L extends from x=a to x=L+a. the gravitational force it exerts on a point mass 'm' at x=0, if the mass per unit length of the rod is `A+Bx^(2), ` is given by :

Options

A. `Gm[A(1/(a+L)-1/a)-BL]`
B. `Gm[A(1/a-1/a+L)-BL]`
C. `Gm[A(1/(a+L)-1/a)+BL]`
D. `Gm[A(1/a-1/(a+L))-BL]`

Answer

B

Explanation

`dm=rdx = (A+Bx^2)dx`
`F=int_(a)^(a+L) (GmDM)/x^2 =GM int_a^(a+L) ((A+Bx^2)dx)/x^2`
`=GM[int_a^(a+L) A/x^2 dx+a int_a^(a+L) Bdx]`
`F=4m[A(1/a-1/(a+L))]+BL " " =GM[A(L/(a(a+l)))+Bl]`