A solid sphere of mass 0. 50 Kg is krpt on a horizontal surface .The coefficient of static friction between the surfaces in contact is 2 / 7 .The maximum force that can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface is

The correct option is B 3 . 3 N
The net torque experienced by the solid sphere due tothe two forces will be:

$\tau =FR-fR$

we also know that

$\tau =I\alpha$

So, equate the toque for the sphere:

$FR=I\alpha +fR$

We know that for a solid sphere $I=\frac{2}{5}M{R}^2$

$\Rightarrow F=\frac{2}{5}mR\alpha +\mu mg----\left(1\right)$

Since the solid sphere is also moving forward

$F=m{a}_c-\mu mg$

$a_c=\frac{F+\mu mg}{m}----\left(2\right)$

$\left(2\right)$ in $\left(1\right)$ use $a=r\alpha$

$F=\frac{2}{5}m\left(\frac{F+\mu mg}{m}\right)+\mu mg$

$F=\frac{2}{5}(F+\mu mg)+\mu mg$

$F=\frac{2}{5}F+\frac{2}{5}\times \frac{2}{7}\times 0.5\times 10+\frac{2}{7}\times 0.5\times 10$

$=\frac{3F}{5}=\frac{4}{7}+\frac{10}{7}=2$

$\therefore F=\frac{5\times 2}{3}=\frac{10}{3}=3.3N$