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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 anticlockwise sense 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, will they again meet at the point B is time tmin=(12)xs. The value of x is __________

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answer is 2.

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Detailed Solution

The velocity of first particle is v1=150012ms−1 and the velocity  of second particle is v2=210012ms−1. Let after completing n1and n2 trips, they will again meet at the point B.∴ (1500+2100)n1v1=(1500+2100)n2v2=t Or  n1n2=v1v2=150012210012=15002100=57∴ tmin=3600×5v1=3600×51500×12=144s=(12)x∴ x=2
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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 anticlockwise sense 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, will they again meet at the point B is time tmin=(12)xs. The value of x is __________