A block of mass m rests on a rough horizontal surface as shown in the figure. coefficient of friction between the block and the surface is meu. A force F=MG acting at an angle theta with the vertical side of the block pulls it in which of the following cases can the block be pulled along the surface?? open picture for options

For motion mgsin theta should be greater than equal to meu(mg -mgcos theta) now by solving we get cot theta by 2 greater than equal to meu

Dear Student,
Please find below the solution to your problem.

As shown in the figure the Normal reaction N on the block and Vertical component of the force F are opposite to weight.
We need only horizontal motion no vertical motion so vertical force should be balanced that is net force in vertical direction should be zero, i.e., N+FCosθ−Mg=0 or N=Mg−FCosθ , put F=Mg as given in the question. so by putting so we get N=Mg(1−cosθ) .
Now putting another condition to make F able to pull the block horizontally the Fsinθ should be greater than or eual to Staticfriction and we know that maximum value of static friction is μN
so we have Fsinθ≥μN
putting value of N we get Sinθ=μ(1−cosθ)
put cosθ=1−2Sin^2(θ/2)​ and sinθ=2sin2θ​Cos2θ​

Thanks and Regards