Question

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Answer 1

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$\bold{(B)\:\frac{a}{2gk}}$

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step-by-step explanation:

The force acting on the bead as seen by the observer in the accelerated frame are:-

(a) N (normal due to surface)

(b) mg (due to earth's gravitation)

(c) ma ( pseudo force)

Let α is the angle which the tangent at P makes with the X- axis.

( refer to the attachments)

Now,

As the bead is in the equilibrium with respect to the wire,

therefore,

N sin α = ma

and

N cos α = mg

Therefore,

$\tan( \alpha ) = \frac{a}{g} ........... \:{ i}$

But,

$y = k {x}^{2}$

Therefore,

$\frac{dy}{dx} = \: 2kx \: = \tan( \alpha ) .....................{ii}$

Now,

From equation (i) and (ii),

we get,

$= > 2kx = \frac{a}{g}$

On simplifying,

we get,

$= > x = \frac{a}{2kg}$

Hence,

the bead can stay at rest with respect to the wire, from y- axis when

$x = \frac{a}{2kg}$

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