A body of mass 1 kg rests on a horizontal floor with which it has a coefficient  of static friction  $1/\sqrt{3}$. It is desired to make the body move by applying the minimum possible force F newton . The value of F will be ……….
(Round off to the nearest integer)(Take, $g=10m{s}^{-2})$ 

Given ,mass of the body , m=1kg
Coefficient of static friction,  $\mu =1/\sqrt{3}$
Let's draw the free body diagram of the block 
               

           
Using the condition of the equilibrium in X-direction
                         $F\cos \theta =f=\varphi =\mu N\mathrm{....}\left(i\right)$
In  y- direction
               $F\sin \theta +N=mg\mathrm{....}\left(ii\right)$
$\Rightarrow N=mg-F\sin \theta$
Substituting the value of N in Eq.(i)
We  get 
         $F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }\Rightarrow F=\frac{\mu mg}{\sqrt{1+{\mu }^2}}$ 
Substituting the values in the above equation, we get 
$F=\frac{\frac{1}{\sqrt{3}}\times 10}{\sqrt{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}}\Rightarrow F=5N$
Hence, the body move by applying minimum possible force of 5N. so, the value of F will be 5.