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A particle P is sliding down a frictionl...

A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at `t=0`. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at `t=0` along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let `t_P` and `t_Q` be the respective times taken by P and Q to reach the point B. Then:

A

`t_(p) lt t_(Q)`

B

`t_(p) = t_(Q)`

C

`t_(p) gt t_(Q)`

D

`(t_(p))/(t_(Q)) = ("Length of are" ACB)/("Length of" AB)`

Text Solution

Verified by Experts

The correct Answer is:
a
As `A` the horizontal speeds of both the masses is the same .The velocity of `Q` remain the same horizontal as so force it acting on the horizontal dipection but it case of P as shown at any internadiate position the horizontal velocity first increase (the as `N sin theta)` reaches a maximum value at O and then decreases this it always remain greater than v Therefore `t_(p) lt t_(Q)`

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