Correct Answer - A::B::C

The force of attraction is zero at say x from the bigger star.

Then force on mass m due to bigger star = Force on mass m

due to small star

`(GM_Bm)/(x^2) = (GM_Sm)/((10a-x)^2) rArr (16M)/(x^2 = (M)/((10a -x))^2 rArr x = 8a`

If we throw a mass m from bigger star giving it such a velocity

that is sufficient to bring it to P, then later on due to greater

force by the star `M_s` it will pull it towards itself [without any

external energy thereafter].

The energy of the system (of these masses) initially

= Final energy when m is at P

`-(GM_BM_s)/(10a) - (GMB_m)/(2a) - (GM_Sm)/(8a) +(1)/(2) mv^2`

`=-(GM_BM_S)/(10a) - (GM_B m)/(8a) - (GM_Sm)/(2a)`

`[:. M_B = 16M, M_S = M)`

`:.v = (3)/(2)sqrt((5GM)/(a))`