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:

Options

A. `t_pltt_Q`
B. `t_p=t_Q`
C. `t_pgtt_Q`
D. `t_p/t_Q=` length of arc ACB/length of arc AB

Answer

A

Explanation

At A the horizontal speeds of both the masses is the same. The velocity of Q remains the same in horizontal as no force is acting in the horizontal direction. But in case of P as shown at any intermediate position, the horizontal velocity first increases (due to `N sin theta`), reaches a max value at O and then decreases. Thus it always reamains greater than v. Therefore `t_pltt_Q`.