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A frictionless wire is fixed between A a...


A frictionless wire is fixed between `A` and `B` inside a circle of radius `R`. A small bead slips along the wire. Find the time taken by the bead to slip from `A` to `B`.

A

`2sqrt(R//g)`

B

`gR//sqrt(g cos theta)`

C

`(2sqrt(gR))/(g cos theta)`

D

`(2sqrt(gR cos theta))/g`

Text Solution

Verified by Experts

The correct Answer is:
A

`AB = 2 R cos theta`
acceleration along AB
`a = g cos theta`
`u =0` from A to B

`2 R cos theta =0+(1)/(2) (g cos theta)t_(2)`
`t=2 sqrt((R)/(g))`
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