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A piece of wire is bent in the shape of a parabola $y = k x^{2} (y - a x i s v e r t i c a l)$ with a bead of mass ' $m$ ' on it. The bead can slide on the wire without friction. The wire is now accelerated parallel to the x-axis with an acceleration ‘ $a$ ’. The distance of the new equilibrium position of the bead is $\frac{a}{x g k}$ where the bead can stay at rest with respect to the wire from y-axis. The value of ' $x$ ' is

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answer is 2.

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Detailed Solution

There will be pseudo force in the -ve x direction on the bead

At equilibrium $\sum F = 0$

Therefore in the tangential line at the equilibrium position

$m a \cos θ = m g \sin θ$

At that point, slope of the curve is

$\frac{d y}{d x} = \tan θ$

$\Rightarrow \frac{d}{d x} (k x^{2}) = \tan θ$

Therefore from the above equations

$2 k x = \frac{a}{g}$

$\Rightarrow x = \frac{a}{2 g k}$

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