Two inclined frictionless tracks, one gradual and the other steep meet at `a` from where two stones are allowed to slide down from rest, one on each track as shown in Figure. Which of the following statement is correct ?
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Options

A. Both the stone reach the botton at the same time but not with the same speed.
B. Both the stone reach the bottom with the same speed and stone `I` reaches the bottom earlier than stone `II` Figure.
C. Both the stones reach the bottom with the same speed and stone `II` reaches the bottom earlier than stone `I`.
D. Both the stones reach the bottom at different times and with different speeds.

Answer

C

Explanation

As the given tracks are frictionless, hence mechanical energy will be conserved. As both the tracks having common height, `h`.
From conservation of mechanical energy.
`(1)/(2) mv^(2) = m gh`
(for both tracks `I` and `II`)
`v = sqrt(2gh)`
Hence, speed is same for both stones. For stone `1, a_(1)` = acceleration along inclined plane
=`g sin theta_(1)`
`II a_(2) = g sin theta_(2)` as `theta_(2) gt theta_(1)` hence, `a_(2) gt a_(1)`.
And both length for track `II` is alos less hence, stone `II` reaches earlier than stone `I`.