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A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is $x^{2} = 4 a y$ . The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point $y = 4 a$ on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by $y = a$ is

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$\frac{g}{\sqrt{2}}$

$\frac{3 g}{\sqrt{2}}$

$\frac{g}{3}$

$\frac{g}{4}$

answer is A.

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Detailed Solution

$x^{2} = 4 a y; \frac{d y}{d x} = \frac{1}{2 a} x$ at point $(2 a, 2 a), \frac{d y}{d x} = 1; θ = 45^{\circ}$

Component of weight along tangential direction is $m g \sin θ$
Tangential acceleration is $g \sin θ$

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