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. (a) ` t_p ltt Q`
B. (b) ` t_p gt t_q`
C. (c ) ` t_p =t_q`
D. (d) ` t_p/t_q= (length of at arc ACB)/()length of chord AB)`

Answer

A

Explanation

Horizontal velocity of particle (P) increases from (v) upto lowest point , after that it starts decreasing and again becomes (v) at (B). It means the average velocity of this particle is more than (v) whereas the particle (Q) moves with constant velocity (v) only . Therefore ot cover the same horizontal distance. the particle (P) will take samaller time than that of particle (Q) i.e. t_P lt t_Q`.