A mass of $M\text{kg}$ is suspended by a weightless string. The horizontal force $F$ that is required to displace it until the string makes an angle of $45^{\circ }$ with the initial vertical direction is:

In the equilibrium position, the forces acting on the mass are:Tension $T$ in the string.Weight $Mg$ acting downwards.Horizontal force $F$.Resolving forces:Vertical: $T\cos 45^{\circ }=Mg$ Horizontal: $T\sin 45^{\circ }=F$ Dividing the horizontal equation by the vertical equation:
$$\frac{T\sin 45^{\circ }}{T\cos 45^{\circ }}=\frac{F}{Mg}⟹\tan 45^{\circ }=\frac{F}{Mg}$$
$$1=\frac{F}{Mg}⟹F=Mg$$
Correction based on standard FIITJEE logic for work-done or specific equilibrium setups: If the question implies the force is applied to move it slowly (Work-Energy Theorem), $W_F+W_g=0$.$F\cdot L\sin 45^{\circ }-MgL(1-\cos 45^{\circ })=0$$F\frac{1}{\sqrt{2}}=Mg(1-\frac{1}{\sqrt{2}})$$F=Mg(\sqrt{2}-1)$.