A bead of mass $m$ is located on a parabolic wire with its axis vertical and vertex at the origin 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 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

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

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

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

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\frac{g}{2}

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Solution

The correct option is B $\frac{g}{\sqrt{2}}$
$m g (4 a - a) = \frac{1}{2} {\int ↺ m v}^{2}$

$V = \sqrt{6 a g}$
$x^{2} = 4 a y$
$2 x = 4 a \frac{d y}{d x}$
$\frac{d y}{d x} = \frac{x}{2 a} = \frac{\sqrt{4 a y}}{2 a} = 2 a / 2 a = 1$

at $(2 a, a)$
$t a n θ = 1$
$θ = 45^{\circ}$
$a_{t} = g sin 45^{\circ}$
$a_{1} = \frac{g}{\sqrt{2}}$

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The correct option is B g√2mg(4a−a)=12 ∫↺mv2 V=√6ag x2=4ay 2x=4adydx dydx=x2a=√4ay2a =2a/2a=1 at (2a,a) tan θ=1 θ=45∘ at=g sin45∘ a1=g√2