A block of mass $1kg$ is at rest on a horizontal table. The co-efficient of static friction between the block and the table is $0.5$. The magnitude of the force acting upwards at an angle of ${60}^{\circ }$ from the horizontal that will just start the block moving is

The correct option is D $\frac{20}{2+\sqrt{3}}N$
draw free body diagram of the given problem


Apply the equilibrium of force in vertical direction.
$R+F\sin {60}^{\circ }=mg$
$R=mg-\frac{\sqrt{3}F}{2}$
Find the value of $F$ acting on the block.
If block just starts moving, then the horizontal component of force will be equal to limiting friction
i.e., $F\cos {60}^{\circ }=f=\mu R$
$F+\frac{\sqrt{3}F}{2}=10$
$F=\frac{20}{2+\sqrt{3}}$