A force of magnitude mg acts tangentially at the highest point of a sphere of mass m kept on a rough horizontal plane. If the sphere rolls without slipping, 

The situation is shown in the figure. As force $F$ rotates the sphere point of the contact has a tendency to slip towards left so that static friction of sphere will act towards right

Let$a =$ linear acceleration, $r =$ radius of the sphere, then

angular acceleration $\frac{a}{r}$

Thus  $F+f=ma$

and  $\mathrm{Fr}-\mathrm{fr}=\mathrm{I}\alpha =\left[\frac{2}{5}{\mathrm{mr}}^2\right]\left(\frac{\mathrm{a}}{\mathrm{r}}\right)$

from above equations

$\mathrm{a}=\frac{10\mathrm{F}}{7\mathrm{m}}$