Distance between the centres of two stars is $10a$. The masses of these two stars are $M$ and $16M$, and their radii $a$ and $2a$ respectively. A body of mass $m$ is to be projected from the surface of larger star towards the smaller star. If the minimum initial speed of the body is $x\sqrt{\frac{GM}{a}},$  then find $x$.(  $\sqrt{5}=2.24$)

Let’s locate point P where gravitation $al$ field intensity is zero. 
If it is at distance x from center of bigger star, then  $\frac{G\times 16M}{x^2}=\frac{G\times M}{{(10a-x)}^2}\Rightarrow x=8a$
If the projected object just crosses this point, then it is bound to reach the smaller star, potential at this point P
 $=\frac{G\times 16M}{8a}-\frac{GM}{2a}=-\frac{5}{2}\frac{GM}{a}$
Potential at surface of bigger star
$=\frac{G\times 16M}{8a}-\frac{GM}{8a}=-\frac{65}{8}\frac{GM}{a}$
Applying energy conservation,
$-\frac{65}{8}\frac{GMm}{a}+\frac{1}{2}m{v}^2=-\frac{5}{2}\frac{GMm}{a}$
Applying energy conservation,
$-\frac{65}{8}\frac{GMm}{a}+\frac{1}{2}m{v}^2=-\frac{5}{2}\frac{GMm}{a}$
$\Rightarrow v=\frac{3\sqrt{5}}{2}\times \sqrt{\frac{GM}{a}}=3.35\sqrt{\frac{GM}{a}}$