A length of path ACB is 1500 m and the length of the path ADB is 2100 m. Two particles start from point A simultaneously around the track ACBDA. One of them travels the track in clockwise sense and other in anti-clockwise sense and other in anti-clockwise with their respective constant speeds. After 12 s from the start, the first time they meet at the point B. After minimum time (in s) in which they meet first at point B, again will they meet at the point B is time t_min = (12)^(x) s.The value of x is

A length of path ACB is 1500 m and the length of the path ADB is 2100

1.	A length of path ACB is 1500 m and the length of the path ADB is 2100 m. Two particles start from point A simultaneously around the track ACBDA. One of them travels the track in clockwise sense and other in anti-clockwise sense and other in anti-clockwise with their respective constant speeds. After 12 s from the start, the first time they meet at the point B. After minimum time (in s) in which they meet first at point B, again will they meet at the point B is time t_min = (12)^(x) s.The value of x is
Answer» 2 3 4 5 Solution :The VELOCITY of first particle is `v_1 = 1500 /12 m s^(-1)` and the velocity of second particle is `v_2 = 2100 / 12 ms^(-1)`. Let after completing `n_1` and `n_2` trips, they will again meet at the point B. `therefore``(3600n_1)/v_1 = (3600n_2)/v_2` = t [total path length `= 1500 + 2100 = 3600`] or `n_1/n_2 = v_1/v_2 = (1500/12)/(2100/12) = 1500/2100 = 5/7` `therefore t_"min" = 3600 XX 5/v_1` `3600 xx 5/1500 xx 12 = 144 s = (12)^(x)` `therefore` x = 2