A) 2R/3
B) R/2
C) R/4
D) R/6
Correct Answer: A
Solution :
[a] If the contact is lost at P, the velocity v would be\[v=\sqrt{2g(R-H)}\](PE lost = KE gained) \[\frac{{{v}^{2}}}{R}=g\cos \theta \] (Centripetal force \[\frac{m{{v}^{2}}}{R}\]) \[\Rightarrow \]\[{{v}^{2}}=2g(R-h)=Rg\cos \theta \] Thus, \[\frac{2g(R-h)}{R}=g\times \frac{h}{R}\left( \because \cos \theta =\frac{h}{R} \right)\] \[\Rightarrow \]\[2g(R-h)=gh\]or\[2gR=3gh\Rightarrow h=\frac{2}{3}R\]You need to login to perform this action.
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