1. Set up the motion

The bead slides बिना friction on a fixed wire shaped as a quarter circle of radius $R$ in a vertical plane.

Let the bead move from the top point $A$ to the lower point $B$.

At any point, the forces on the bead are:

• Weight $mg$ vertically downward
• Normal reaction $N$ by the wire on the bead, along the radius of the circular path

Since the bead is constrained to move on a circular arc, the radial acceleration is $a_r = \frac{v^2}{R}$ towards the center.

2. Choose radial direction towards the center as positive

Let $\theta$ be the angle made by the radius to the bead with the horizontal, measured suitably so that:

• at $A$, $\theta = 90^\circ$
• at $B$, $\theta = 0^\circ$

The component of weight towards the center is $mg\sin\theta$

Applying Newton’s second law in the radial direction: $N + mg\sin\theta = \frac{mv^2}{R}$ So, $N = \frac{mv^2}{R} - mg\sin\theta$

3. Use energy conservation to find $v$

If the bead is released from near the top, starting essentially from rest, then after descending through height $R(1-\sin\theta)$, $\frac12 mv^2 = mgR(1-\sin\theta)$ Hence, $v^2 = 2gR(1-\sin\theta)$

Substitute into expression for $N$: $N = m\left(\frac{2gR(1-\sin\theta)}{R}\right) - mg\sin\theta$ $N = 2mg(1-\sin\theta) - mg\sin\theta$ $N = 2mg - 3mg\sin\theta$

4. Check sign of $N$ during motion

• Near the top $A$: $\sin\theta \approx 1$ $N \approx 2mg - 3mg = -mg < 0$ So the normal reaction on the bead is radially outward.

• Near the bottom $B$: $\sin\theta = 0$ $N = 2mg > 0$ So the normal reaction on the bead is radially inward.

Thus, as the bead moves from $A$ to $B$, the force of the wire on the bead is outward initially and inward later.

But the question asks for the force applied by the bead on the wire. By Newton’s third law, this force is opposite to the force of the wire on the bead.

Therefore, the force applied by the bead on the wire is:

• radially inward initially
• radially outward later

5. Correct option

Therefore, the correct answer is: $\boxed{\text{D: radially inwards initially and radially outwards later}}$