1. Equation of the wire and point $P(a,b)$

The wire is bent in the shape of the parabola $y=4Cx^2.$ Since the bead is at point $P(a,b)$ on the parabola, $b=4Ca^2.$

2. Condition for equilibrium in rotating frame

The wire rotates about the vertical $y$-axis with angular speed $\omega$. In the rotating frame, the bead is at rest on the wire, so the net effective force along the tangent to the wire must be zero.

The forces acting are:

• वजन downward: $mg$ along negative $y$-direction,
• centrifugal force horizontally outward: $m\omega^2 x$ along positive $x$-direction.

So effective force is $\vec F_{\text{eff}}=m\omega^2 x\,\hat i-mg\,\hat j.$

For equilibrium on the smooth wire, this force must be perpendicular to the tangent, i.e. its component along the tangent must vanish.

3. Slope of the tangent to the parabola

Given $y=4Cx^2,$ so $\frac{dy}{dx}=8Cx.$

At point $(a,b)$, slope of tangent is $\left(\frac{dy}{dx}\right)_{x=a}=8Ca.$

Thus a tangent vector is proportional to $\hat t = \hat i + (8Ca)\hat j.$

4. Set tangential component of effective force to zero

For equilibrium, $\vec F_{\text{eff}}\cdot \hat t =0.$

At $x=a$, $\vec F_{\text{eff}}=m\omega^2 a\,\hat i-mg\,\hat j.$

Now, $\left(m\omega^2 a\,\hat i-mg\,\hat j\right)\cdot \left(\hat i+8Ca\,\hat j\right)=0.$

So, $m\omega^2 a - mg(8Ca)=0.$

Cancelling $ma$, $\omega^2=8gC.$

Hence, $\omega=\sqrt{8gC}=2\sqrt{2gC}.$

5. Match with options

$\boxed{\omega=2\sqrt{2gC}}$ So the correct option is:

A

6. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They agree.