IIT JEE 2009 P2 Q40

A piece of wire is bent in the shape of a parabola y=kx2(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 th eparabola 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

7 Answers

1
killerkiran ·

B

1
Manas ·

but i m getting A

11
Subash ·

A for sure

9
Celestine preetham ·

b

equate slope at point to a/g

1
greatvishal swami ·

B

1
nithin.yes ·

b is correct option

49
Subhomoy Bakshi ·

ma cos @ = mg sin @

tan @ = ag

slope = -tan @ = 2kx

so, -2kx= ag

so, x= - a2kg

hence option b!!!

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