A piece of wire is bent in the shape of a par | vedclass

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(B) Step $1$: Choosing the frame of reference and drawing the $FBD$. Solving the problem from the frame of the wire which is accelerating towards the

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$\frac{a}{2gk}$

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(B) Step $1$: Choosing the frame of reference and drawing the $FBD$. Solving the problem from the frame of the wire which is accelerating towards the right.So,a pseudo force on mass $m$ will act in the left direction.Pseudo force $= m 	imes a$.Step $2$: Equilibrium condition.At the equilibrium position,the acceleration of the bead will be zero in the chosen frame.Therefore,balancing the forces in the $x$ and $y$ directions,we get:$N \sin 	heta = ma \dots (1)$$N \cos 	heta = mg \dots (2)$Dividing these two equations,we get:$	an 	heta = \frac{a}{g} \dots (3)$Step $3$: Slope of the curve.The equation of the curve is $y = kx^2$.The tangent to the curve at the equilibrium point makes an angle $	heta$ with the $x$-axis.Therefore,the slope of the curve is $\frac{dy}{dx} = 	an 	heta$.$\frac{dy}{dx} = 2kx$.Equating the slopes: $2kx = \frac{a}{g}$.Therefore,$x = \frac{a}{2gk}$.

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