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(D) Let $	heta$ be the angle made by the radius vector with the vertical. Using the conservation of mechanical energy,the speed $v$ of the bead at an

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radially inwards initially and radially outwards later.

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(D) Let $	heta$ be the angle made by the radius vector with the vertical. Using the conservation of mechanical energy,the speed $v$ of the bead at an angle $	heta$ is given by: $mgR(1 - \cos 	heta) = \frac{1}{2}mv^2 \Rightarrow v^2 = 2gR(1 - \cos 	heta)$.Considering the radial forces acting on the bead,the equation of motion is: $mg \cos 	heta - N = \frac{mv^2}{R}$,where $N$ is the normal force exerted by the wire on the bead.Substituting $v^2$: $N = mg \cos 	heta - \frac{m}{R} [2gR(1 - \cos 	heta)] = mg \cos 	heta - 2mg + 2mg \cos 	heta = mg(3 \cos 	heta - 2)$.The normal force $N$ exerted by the wire on the bead is radially outwards if $N > 0$,i.e.,$3 \cos 	heta - 2 > 0 \Rightarrow \cos 	heta > 2/3$.The normal force $N$ is radially inwards if $N By Newton's third law,the force applied by the bead on the wire is equal and opposite to the normal force $N$ exerted by the wire on the bead. Initially,at $	heta = 0$,$\cos 	heta = 1 > 2/3$,so $N$ is radially outwards,meaning the bead pushes the wire radially inwards. As $	heta$ increases,$\cos 	heta$ decreases,and when $\cos 	heta < 2/3$,$N$ becomes radially inwards,meaning the bead pushes the wire radially outwards. Thus,the force applied by the bead on the wire is radially inwards initially and radially outwards later.

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