A piece of wire is bent in the shape of a parabola $y = k x^{2}$ ( $y$ -axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowestpoint of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$ -axis with a constant acceleration $a$ . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$ -axis is

\frac{a}{g t}

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\frac{a}{2 g k}

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\frac{2 a}{g k}

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\frac{a}{4 g k}

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Solution

The correct option is B $\frac{a}{2 g k}$

Let $m$ be the mass of bead.
Balance the vertical component of the force on the bead,
$N sin θ = m g$
As the wire frame move +ve $x$ -axis, the pseudo force $m a$ will act on the bead towards -ve $x$ -axis

Now, balance the horizontal component of the force on the bead,
$\Rightarrow N cos θ = m a$
$tan θ = \frac{g}{a}$
$\Rightarrow cot θ = \frac{a}{g} = tan (90^{\circ} - θ) = \frac{d y}{d x} = 2 k x$
[Given, $y = k x^{2}$ ]
$∴ x = \frac{a}{2 k g}$

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