A bead of mass ( $m$ ) is located on a parabolic wire with its axis vertical and vertex at the origin is shown in figure and whose equation $x^{2} = 4 a y$ . 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 not equal to

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

$\frac{g}{2}$

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

$\frac{g}{\sqrt{5}}$

answer is A, B, D.

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Detailed Solution

$y = \frac{x^{2}}{4 a}; \frac{d y}{d x} = \frac{2 x}{4 a} = \frac{x}{2 a}$

If $y = a, x^{2} = 4 a y, x^{2} = 4 a a, x = 2 a$

$\frac{dy}{dx} = \frac{2 a}{2 a}; Tan θ = 1 \Rightarrow θ = 45^{\circ}$

$a_{t} = g \cos (90 - θ) = g \cos 45^{\circ} = \frac{g}{\sqrt{2}}$

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