A hollow vertical cylinder of radius R and height h has smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed vo tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will
Options:
A .  h2πR
B .  v0√2gh
C .  2πRh
D .  v02πR(√2hg)
Answer: Option D : D since the body has no initial velocity in the vertical direction. az=-g, vertical displacement z=-h. ∴z=at+12at2 ⇒−h=0+12(−g)t2 ⇒T=√2hg time taken to reach the bottom let,t be the time taken to complete one revolution. Then t=2πRv0 ∴ number of revolution=Tt=√2hg2πRv0=v02πR√2hg
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