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A piece of wire is bent in the shape of a parabola `y=kx^(2)` (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is:

A

`(a)/(gk)`

B

`(a)/(2gk)`

C

`(2a)/(gk)`

D

`(a)/(4gk)`

Text Solution

Verified by Experts

The correct Answer is:
B
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Knowledge Check

  • A piece of wire is bent in the shape of a parabola y = Kx^(2) (y - axis vorical) with a bead of mass m on it . The beat can side on the wire without friction , it stays the wire is now accleated parallel to the bead , where the bead can stay at rest with repect to the wire from the y - axis is

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    `(a)/(2gk)`
    C
    `(2a)/(gk)`
    D
    `(a)/(4gk)`

  • A straight wire carrying current is parallel to the y-axis as shown in Fig. The

    A
    magnetic field at the point P is parallel to the x-axis.
    B
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    C
    magnetic field are concentric circle with the wire passing through their common centre
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  • A wire frame ABCD has a soap film. The wire BC can slide on the frame without friction and it is in equilibrium in the position shown in the figure. Find m, if T is the surface tension of the liquid.

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    D
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